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$$ y_t = θy_{t−1} + u_t \\ t = 1,...,T; $$ I need to derive a formula for $y_t$ and show that $$ E\left[\frac{\Sigma y_{t-1}u_t}{ \Sigma(y_{t-1})^2}\right] \neq 0 $$ | I am having trouble understanding the estimation of an AR process. In some textbooks, the AR(1) process is defined as follows: $y_{t}=\theta y_{t-1}+ϵ_t$ (which does not contain a constant). So the OLS estimator is biased. I am confused about the cause of the bias. It is explained that $y_{t-1}$ is dependent on $ϵ_{t-1}$ although it is independent of $ϵ_t$. However in linear regression, if the equation does not contain a constant, we cannot make sure the expectation of disturbance is zero. So the OLS estimator is bias without a constant. Does it mean that the OLS estimator is unbiased if I add a constant in AR process? | Consider this equation : $$\sqrt{\left( \frac{dy\cdot u\,dt}{L}\right)^2+(dy)^2}=v\,dt,$$ where $t$ varies from $0$ to $T$ , and $y$ varies from $0$ to $L$. Now how to proceed ? This equation arises out of following problem : A cat sitting in a field suddenly sees a standing dog. To save its life, the cat runs away in a straight line with speed $u$. Without any delay, the dog starts with running with constant speed $v>u$ to catch the cat. Initially, $v$ is perpendicular to $u$ and $L$ is the initial separation between the two. If the dog always changes its direction so that it is always heading directly at the cat, find the time the dog takes to catch the cat in terms of $v, u$ and $L$. See my solution below : Let initially dog be at $D$ and cat at $C$ and after time $dt$ they are at $D'$ and $C'$ respectively. Dog velocity is always pointing towards cat. Let $DA = dy, \;AD' = dx$ Let $CC'=udt,\;DD' = vdt$ as interval is very small so $DD'$ can be taken straight line. Also we have $\frac{DA}{DC}= \frac{AD'}{ CC'}$ using triangle property. $\frac{dy}{L}= \frac{dx}{udt}\\ dx = \frac{dy.udt}{L}$ $\sqrt{(dx)^2 + (dy)^2} = DD' = vdt \\ \sqrt{(\frac{dy.udt}{L})^2 + (dy)^2} = vdt $ Here $t$ varies from $0-T$, and $y$ varies from $0-L$. Now how to proceed? | eng_Latn | 18,700 |
When I was learning implicit differentiation in my class I was told to not think of $dy/dx$ as a fraction. Now I am doing integration by $u$-substitution and we treat $du/dx$ as a fraction and solve for $dx$. Why is this? | In the book Thomas's Calculus (11th edition) it is mentioned (Section 3.8 pg 225) that the derivative $dy/dx$ is not a ratio. Couldn't it be interpreted as a ratio, because according to the formula $dy = f'(x)dx$ we are able to plug in values for $dx$ and calculate a $dy$ (differential). Then if we rearrange we get $dy/dx$ which could be seen as a ratio. I wonder if the author says this because $dx$ is an independent variable, and $dy$ is a dependent variable, for $dy/dx$ to be a ratio both variables need to be independent.. maybe? | Consider this equation : $$\sqrt{\left( \frac{dy\cdot u\,dt}{L}\right)^2+(dy)^2}=v\,dt,$$ where $t$ varies from $0$ to $T$ , and $y$ varies from $0$ to $L$. Now how to proceed ? This equation arises out of following problem : A cat sitting in a field suddenly sees a standing dog. To save its life, the cat runs away in a straight line with speed $u$. Without any delay, the dog starts with running with constant speed $v>u$ to catch the cat. Initially, $v$ is perpendicular to $u$ and $L$ is the initial separation between the two. If the dog always changes its direction so that it is always heading directly at the cat, find the time the dog takes to catch the cat in terms of $v, u$ and $L$. See my solution below : Let initially dog be at $D$ and cat at $C$ and after time $dt$ they are at $D'$ and $C'$ respectively. Dog velocity is always pointing towards cat. Let $DA = dy, \;AD' = dx$ Let $CC'=udt,\;DD' = vdt$ as interval is very small so $DD'$ can be taken straight line. Also we have $\frac{DA}{DC}= \frac{AD'}{ CC'}$ using triangle property. $\frac{dy}{L}= \frac{dx}{udt}\\ dx = \frac{dy.udt}{L}$ $\sqrt{(dx)^2 + (dy)^2} = DD' = vdt \\ \sqrt{(\frac{dy.udt}{L})^2 + (dy)^2} = vdt $ Here $t$ varies from $0-T$, and $y$ varies from $0-L$. Now how to proceed? | eng_Latn | 18,701 |
I'm working on solving this inhomogeneous problem with the method of undetermined coefficients using tips from this website: . There is an equation halfway down that looks almost exactly the same as mine, so I referenced it for help, but I'm still not getting the right answer. The equation is this: $$y''-2y'+y=te^t$$ I started out with my "guess" equation as: $Y_P(t)=e^t(At+B)$ Then I took the first and second derivatives: $$Y_P'(t)=e^tAt+Be^t+Ae^t$$ $$Y_P''(t)=e^tAt+2Ae^t+Be^t$$ After finding that I plugged in the three equations to the LHS of the original equation and then separated like terms: $$e^tt(A-2A+A)+e^t(2A+B-2A-2B+B)=te^t$$ In each of the parentheses I end up with a value of $0$, which can't be right. What am I doing wrong? | In order to solve $$y''-2y'+y=te^t$$ I did: $$y = (at+b)e^t\implies\\ y' = ate^t+ae^t+be^t\implies \\y''= ate^t+2ae^t+be^t$$ but when I put these in the differential equation I'm getting zeros eveywhere. Does somebody know what I'm doing wrong? | Find the general solution of $xy''-(1+x)y'+y=x^2$ knowing that the homogeneous equation has the following solution: $e^{ax}$, where $a$ is a parameter you have to find. I have found that $a=1$ or $a=1/x$. | eng_Latn | 18,702 |
I have a very long SQRT() equation, that doesn't fit in a line. I couldn't simplify it, since it involves terms in addition. Is there an alternative way to represent a root as 'SQRT{some function}' ? \documentclass{article} \begin{document} \begin{equation} \Delta\sigma_{V,ij} = \sqrt{\Delta\sigma_{x,ij}^2 + \Delta\sigma_{y,ij}^2 + \Delta\sigma_{z,ij}^2 - (\Delta\sigma_{x,ij}\cdot\Delta\sigma_{y,ij} + \Delta\sigma_{y,ij}\cdot\Delta\sigma_{z,ij} + \Delta\sigma_{z,ij}\cdot\Delta\sigma_{x,ij}) + 3 \cdot (\Delta\tau_{xy,ij}^2 + \Delta\tau_{yz,ij}^2 + \Delta\tau_{zx,ij}^2)} \end{equation} \end{document} | How can continue the equation of the rectangle red in next line? \documentclass[11pt,a4paper,twocolumn]{article} ... \begin{equation*} \left| \dot{r} \right| = \sqrt{{- \, {e}^{-t} \left( \cos t + \sin t \right)}^{2} \, + \, {{e}^{-t} \left( \cos t + \sin t \right)}^{2} \, + \, {(- \, {e}^{-t})}^{2}} \end{equation*} | Consider this equation : $$\sqrt{\left( \frac{dy\cdot u\,dt}{L}\right)^2+(dy)^2}=v\,dt,$$ where $t$ varies from $0$ to $T$ , and $y$ varies from $0$ to $L$. Now how to proceed ? This equation arises out of following problem : A cat sitting in a field suddenly sees a standing dog. To save its life, the cat runs away in a straight line with speed $u$. Without any delay, the dog starts with running with constant speed $v>u$ to catch the cat. Initially, $v$ is perpendicular to $u$ and $L$ is the initial separation between the two. If the dog always changes its direction so that it is always heading directly at the cat, find the time the dog takes to catch the cat in terms of $v, u$ and $L$. See my solution below : Let initially dog be at $D$ and cat at $C$ and after time $dt$ they are at $D'$ and $C'$ respectively. Dog velocity is always pointing towards cat. Let $DA = dy, \;AD' = dx$ Let $CC'=udt,\;DD' = vdt$ as interval is very small so $DD'$ can be taken straight line. Also we have $\frac{DA}{DC}= \frac{AD'}{ CC'}$ using triangle property. $\frac{dy}{L}= \frac{dx}{udt}\\ dx = \frac{dy.udt}{L}$ $\sqrt{(dx)^2 + (dy)^2} = DD' = vdt \\ \sqrt{(\frac{dy.udt}{L})^2 + (dy)^2} = vdt $ Here $t$ varies from $0-T$, and $y$ varies from $0-L$. Now how to proceed? | eng_Latn | 18,703 |
$x'=x^2- \frac{2}{t^2}$ Hmm I have not idea how to do it... Maybe any smart substitution could work? | Solve differential equation: $x'=x^2- \frac{2}{t^2}$ Maybe is it sth connected with homogeneous equation? I have no idea how to solve it. | I am able to solve simple differential equations like : $$\dfrac{dy}{dx} = 3x^2 + 2x$$ We simply bring $dx$ to other site and integrate. But how do we find solutions of differential equations like : $$\frac{d^2x}{dt^2} + \dfrac{kx}{m} = 0$$ ? We have been told the solution is $x(t) = A\cos ( \omega t + \phi_o)$ where $\omega\ =\sqrt{\dfrac{k}{m}}$ but how do we actually find it? | eng_Latn | 18,704 |
Solve the hanging rope shape using the variational principle? I know how to write down the local ordinary differential equation (ODE) via Newton's force law, balancing between the left and right rope tension and the gravity exerted on a infinitesimal piece of rope. The solution is which has the form $$ y (x)= a \cosh \left(\frac{x}{a} \right) = \frac{a}{2}\left(e^\frac{x}{a} + e^{-\frac{x}{a}}\right) $$ with appropriate boundary conditions. But I only aware the way to solve from ODE. Could we solve the hanging rope shape using the variational principle? This means that we input the shape of the rope minimize the potential energy $V$ (because there is no kinetic $T$) for lagrangian $L= T-V$ (question 1: but this seems maximize the $L$ strangely?). Suppose the length of the role is fixed say $\ell$. The two hanging points out horizontal apart at $x=-l$ and $x=l$. Then we have a constraint on the length of the role is fixed say $\ell$. Then we have a constraint $$ \ell = \int_{-l}^l \sqrt{1 + y'(x)^2} dx $$ Then we minimize potential energy $V$ (but this maximizes the lagrangian $L$ strangely?) $$ V= \rho g \int_{-l}^l y(x) \sqrt{1 + y'(x)^2} dx $$ with $ \rho$ the constant density and gravity constant $g$. I suppose we are solving the eq of motion from Variational Principle with a Constraint like $$\boxed{ V - \lambda(\ell - \int_{-l}^l \sqrt{1 + y'(x)^2} dx )= \rho g \int_{-l}^l y(x) \sqrt{1 + y'(x)^2} dx - \lambda(\ell - \int_{-l}^l \sqrt{1 + y'(x)^2} dx )} $$ How do we obtain $y(x)$ then by minimization or maximization ? See these two related questions show no answers (!!!): | Shape of a string/chain/cable/rope/wire? The height of a string in a gravitational field in 2-dimensions is bounded by $h(x_0)=h(x_l)=0$ (nails in the wall) and also $\int_0^l ds= l$. ($h(0)=h(l)=0$, if you take $h$ as a function of arc length) . What shape does it take? My try so far: minimise potential energy of the whole string, $$J(x,h, \dot{h})=\int_0^l gh(x) \rho \frac{ds}{l}=\frac{g \rho }{l}\int_0^l h(x) \sqrt{1+\dot{h}^2} dx$$ With the constraint $$\int_0^l \sqrt{1+\dot{h}^2} dx- l=0$$ If it helps, it's evident that $\dot{h}(\frac{l}{2})=0$. Generally, this kind of equation is a case of a constrained variational problem, meaning that the integrand in $$\int_0^l \frac{g \rho }{l}h(x) \sqrt{1+\dot{h}^2} +\lambda(\int_0^l \sqrt{1+\dot{h}^2} dx- l)dx$$ Must satisfy the Euler Lagrange equation. The constraint must also be satisfied. But, in truth, by this point I am clueless. $\lambda$ is worked through $\nabla J = \lambda \nabla(\int_0^l \sqrt{1+\dot{h}^2} dx- l)$. I have tried this , but get nonsensical answers. Is this method the best? If so, in what ways am I going about it wrongly thusfar? | Elastic collision in two dimensions Suppose a particle with mass $m_1$ and speed $v_{1i}$ undergoes an elastic collision with stationary particle of mass $m_2$. After the collision, particle of mass $m_1$ moves with speed $v_{1f}$ in a direction of angle $\theta$ above the line it was moving previously. Particle with mass $m_2$ moves with speed $v_{2f}$ in a direction of angle $\phi$ below the line which particle with mass $m_1$ was moving previously. Using equations for conservation of momentum and kinetic energy, how can we prove these two equations $\frac{v_{1f}}{v_{1i}}=\frac{m_1}{m_1+m_2}[\cos \theta \pm \sqrt{\cos^2 \theta - \frac{m_1^2-m_2^2}{m_1^2}}]$ and $\frac{\tan(\theta +\phi)}{\tan(\phi)}=\frac{m_1+m_2}{m_1-m_2}$ ? EDIT. Here is what I've done: For the first one, set the $xy$ coordinate system so that the positive direction of the $x$ axis points toward the original path of the particle with mass $m_1$. So we have three equations: $m_1v_{1i}=m_1v_{1f}\cos \theta + m_2v_{2f} \cos \phi$ $0=m_1v_{1f}\sin \theta - m_2v_{2f}\sin \phi$ $m_1v_{1i}^2=m_1v_{1f}^2+m_2v_{2f}^2$. From the second one, we get: $v_{2f}=\frac{m_1v_{1f}\sin \theta}{m_2 \sin \phi}$ Plotting this into third equation, we get $v_{1i}^2=v_{1f}^2(1+\frac{m_1 \sin^2 \theta}{m_2 \sin^2 \phi})$ (1) From the first equation, we have $\cos \phi =\frac{m_1(v_{1i}-v_{1f}\cos \theta)}{m_2v_{2f}}$ which after applying the equation we have for $v_2f$ becomes $\sin^2 \phi = \frac{1}{1+\frac{(v_{1i}-v_{1f}\cos \theta)^2}{\sin^2 \theta \times v_1f^2}}$ Plotting this into equation (1), gives us an equation in terms of $m_1$, $m_2$, $v_{1f}$, $v_{1i}$ and $\theta$, but it is too far from what I expected. For the second one, assigning the $xy$ coordinate in a way that the positive direction of the $x$ axis points toward the final path of the particle $m_2$, will give us three equations (two for conservation of linear momentum and one for conservation of kinetic energy), but I don't know what to do next. | eng_Latn | 18,705 |
I have been trying to solve the following problem, but I am getting stuck... we have: $-y^2dx + x^2dy = 0$ given: $u(x,y) = \frac{1}{(x-y)^2}$ both have partial derivative: $\frac{-2xy}{(x-y)^3}$ so we have the exact equation: $\frac{-y^2}{(x-y)^2}dx + \frac{x^2}{(x-y)^2}dy = 0$ now, I try integrate the dx expression: $\int\frac{-y^2}{(x-y)^2}dx$ This gives me: $\frac{y^2}{(x-y)} + g(y)$ Differentiating with respect to y gives: $\frac{-y(y-2x)}{(y-x)^2} +g'(y)$ I have checked my work for the above integral and derivative and also used an online calculator to verify that they are correct. Every other problem of this type that I have tried has had something meaningful come of: $\frac{-y(y-2x)}{(y-x)^2} +g'(y) = \frac{x^2}{(x-y)^2}$ However, I have not been able to get anywhere using this to try to find a value for $g(y)$ | I was doing some practice problems that my professor had sent us and I have not been able to figure out one of them. The given equation is: $-y^2dx +x^2dy = 0$ He then asks us to verify that: $ u(x, y) = \frac{1}{(x-y)^2}$ is an integrating factor. I multiplied through to get: $\frac{-y^2}{(x-y)^2}dx + \frac{x^2}{(x-y)^2}dy = 0$ However, the partial derivatives of these do not equal each other so I am a bit confused... | I was doing some practice problems that my professor had sent us and I have not been able to figure out one of them. The given equation is: $-y^2dx +x^2dy = 0$ He then asks us to verify that: $ u(x, y) = \frac{1}{(x-y)^2}$ is an integrating factor. I multiplied through to get: $\frac{-y^2}{(x-y)^2}dx + \frac{x^2}{(x-y)^2}dy = 0$ However, the partial derivatives of these do not equal each other so I am a bit confused... | eng_Latn | 18,706 |
I have to show the following equation: $\int_{0}^{\pi} t \cdot f(sin \; t) \; dt = \frac{\pi}{2} \int_{0}^{\pi} f(sin \; t) \; dt$ with $f : [0, 1] \rightarrow \mathbb{R}$ is continuous. I transformed both sites to the following: $\int_{0}^{\pi} t \cdot f(sin \; t) \; dt = \pi \cdot F(sin \; \pi)$ and $\frac{\pi}{2} \int_{0}^{\pi} f(sin \; t) \; dt = \frac{\pi}{2} F(sin \; \pi) - \frac{\pi}{2} F(0)$ Now I don't know how to go on, because I do know nothing about $f(x)$ or $F(x)$. Is my approach correct or is there a other possibility to show the equation? | Let $f(\sin x)$ be a given function of $\sin x$. How would I show that $\int_0^\pi xf(\sin x)dx=\frac{1}{2}\pi\int_0^\pi f(\sin x)dx$? | Consider this equation : $$\sqrt{\left( \frac{dy\cdot u\,dt}{L}\right)^2+(dy)^2}=v\,dt,$$ where $t$ varies from $0$ to $T$ , and $y$ varies from $0$ to $L$. Now how to proceed ? This equation arises out of following problem : A cat sitting in a field suddenly sees a standing dog. To save its life, the cat runs away in a straight line with speed $u$. Without any delay, the dog starts with running with constant speed $v>u$ to catch the cat. Initially, $v$ is perpendicular to $u$ and $L$ is the initial separation between the two. If the dog always changes its direction so that it is always heading directly at the cat, find the time the dog takes to catch the cat in terms of $v, u$ and $L$. See my solution below : Let initially dog be at $D$ and cat at $C$ and after time $dt$ they are at $D'$ and $C'$ respectively. Dog velocity is always pointing towards cat. Let $DA = dy, \;AD' = dx$ Let $CC'=udt,\;DD' = vdt$ as interval is very small so $DD'$ can be taken straight line. Also we have $\frac{DA}{DC}= \frac{AD'}{ CC'}$ using triangle property. $\frac{dy}{L}= \frac{dx}{udt}\\ dx = \frac{dy.udt}{L}$ $\sqrt{(dx)^2 + (dy)^2} = DD' = vdt \\ \sqrt{(\frac{dy.udt}{L})^2 + (dy)^2} = vdt $ Here $t$ varies from $0-T$, and $y$ varies from $0-L$. Now how to proceed? | eng_Latn | 18,707 |
derivative of inseparable equation Is there a way to find the derivative of an equation where I cannot separate the independent variable from the dependent variable. For example, what is $dy/dx$ given \begin{equation*} \cos(y) = \dfrac{x}{x+y} \end{equation*} | Implicit Differentiation Help I have to use implicit differentiation to find $\frac{dy}{dx}$ given: $$x^2 \cos(y) + \sin(2y) = xy$$ I don't even know where to begin, I missed the class where we went over implicit differentiation, and because of that, I am completely stuck. Thank you everyone. Edit: I don't know how to make the equation look all nice and whatnot, so sorry about that | Using equations to draw out complex objects How do people come up with equations of curves to draw out complex objects? Some popular examples would include: & . This explains the rationale for the batman curve nicely. But other than trial and error, I can't see a reasonable way of drawing the much more complicated PSY curve. | eng_Latn | 18,708 |
ODE Cauchy problem. Solution not unique? Consider the following Cauchy problem: $$ \frac{\partial y}{\partial t} = \frac{-t + \sqrt{t^2+4y}}{2} \\ y(2) = -1 $$ I propose two valid solutions: $$ y_1(t) = 1 - t \\ y_2(t) = \frac{-t^2}{4} $$ Why does this not contradict the Existence and Uniqueness Theorem related to this kind of questions? | $y'=$ ${-x+\sqrt{(x^2+4y)}}\over 2$ , $y(2)=-1$ Prove that the solutions $y_1$ and $y_2$ for the initial value problem $y'=$${-x+\sqrt{(x^2+4y)}}\over 2$ , $y(2)=-1$ are: $y_1=1-x$ $y_2={{-x^2}\over 4}$ And explain why the two solutions not contradiction with the theory of existence and uniqueness. ( sorry I don't speak English well ) My answer: For the initial condition : $y_1=1-x$ Satisfy the initial condition $y(2)=-1$ $y_1=1-2=-1$ And Satisfy the differential equation ${y_1}^{\prime}=-1$ ${{-x+\sqrt{x^2+4y}}\over 2}={{-x+\sqrt{x^2+4(1-x)}}\over 2}$ $={{-x+\sqrt{(x-2)^2}}\over 2}$ $={{-x+x-2}\over 2}=-1$ L.h.s.=R.h.s. $y^{\prime}={{-x+\sqrt{x^2+4y}}\over 2}$ Also $y_2$ Satisfy the initial condition $y_2={{-x^2}\over4}={-4\over4}=-1$ And satisfy the differential equation. $y_2^{\prime}={-x\over 2}$ ${{-x+\sqrt{x^2+4y}}\over 2}={{-x+\sqrt{x^2+4({-x^2\over 4})}}\over 2}={-x\over 2}$ And for second part of exercise: ${\partial f\over \partial y}={1\over \sqrt{x^2+4y}}$ and this not defined when $x=2, \ y=-1$ So $f(x,y)$ no satisfy lipshtize condition in any rectangular has (2,-1), Lipshtize condition including uniqueness of solution, so no contradict with the existence and uniqueness. True ? If this also wrong I well be delet it , I answered by the same way of my teacher :( | Why is $\frac{\operatorname dy'}{\operatorname dy}$ zero, since $y'$ depends on $y$? I know that $\frac{dy'(x)}{dy}=0$ (where $y'=\frac{dy(x)}{dx}$). The reason explained is that $y'$ does not depend explicitly on $y$. But intuitively, $y'$ depends on $y$, since if you vary $y$ you will modify $y'$. Why is my reasoning wrong (my reasoning sounds like it's related to functional calculus, instead of standard calculus)? I tried to write $\frac{dy'(x)}{dy}=\frac{d}{dy}\frac{dy}{dx}=\frac{d}{dx}\frac{dy}{dy}=\frac{d}{dx}1=0$, but this proof doesn't convince me. I think that other way to see this is saying that if a function is of the form $f(y)$, it will not dependend on the variable $y'$. But the same way you write $f(y)=y^2$, you could write $f(y)=\frac{d}{dx}y$, which clearly depends on $y'$. So I don't know if there are some types of operations which are restricted (for example, taking limits): Note: The problem raises in the context of Classical Mechanics, where: $\frac{\partial L(\dot{x},t)}{\partial x}=0$. | eng_Latn | 18,709 |
Create extra bullets, offset according to angle of spaceship? I am trying to create a weapons upgrade for my Spaceship, using libGosu and Chingu (in Ruby). This question is about how to deal with this issue specifically in Ruby, and more specifically in libGosu. In the player class I have tried several variations of the following: def fire Bullet.create(:x => @x, :y => @y, :angle => @angle) Bullet.create(:x => @x + Gosu::offset_x(90, 25), :y => @y + Gosu::offset_y(@angle, 0), :angle => @angle) end It sort of works, but not exactly how it ideally should. For reference, this is what the Bullet class looks like: class Bullet < Chingu::GameObject def initialize(options) super(options.merge(:image => Gosu::Image["assets/laser.png"])) @speed = 7 self.velocity_x = Gosu::offset_x(@angle, @speed) self.velocity_y = Gosu::offset_y(@angle, @speed) end def update @y += self.velocity_y @x += self.velocity_x end end How should I construct "def fire" in order to get the extra bullets to align properly when the spaceship rotates? In the first image (above) the bullets are separated as intended. In the second image the bullets are clumping together when the spaceship is rotated. | Shooting many bullets toward the mouse position I try to shoot a bullet toward the mouse position by this code. float deltaX = mousePosition.x - (aircraft.getPosition().x + aircraft.getLocalBounds().width / 2); float deltaY = mousePosition.y - (aircraft.getPosition().y + aircraft.getLocalBounds().height / 2); float angleRadian = (float)(std::atan2(deltaY, deltaX)); float angleDegree = (float)(angleRadian * 180 / PI); It works fine with one bullet but when I add another one bullet to shoot together I use this code to specify start shooting point of each bullets. sf::Vector2f offset1(aircraft.getPosition()); offset1.x += -10;//start shooting point first bullet. sf::Vector2f offset2(aircraft.getPosition()); offset2.x += 10;//start shooting point second bullet. this is a picture of this problem. When I rotate aircraft to picture 2 the bullets are not the same as the first picture. How can I solve this problem? | Ball motion with air resistance coupled differential equation for fourth-order Runge-Kutta I've created a MATLAB function for solving coupled differential equation with the fourth-order Runge-Kutta method based on what is provided (Simultaneous Equations of First Order). Here the function: function [y] = rk4_c(f, g, h, x, y, z, n) % Runge-Kutta % Implementation of the fourth-order method for coupled equations % h = dt % x is the time here for ii=1:(n-1) k1 = h * f(x(ii), y(ii), z(ii)); l1 = h * g(x(ii), y(ii), z(ii)); k2 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1); l2 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1); k3 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2); l3 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2); k4 = h * f(x(ii) + h, y(ii) + k3, z(ii) + l3); l4 = h * g(x(ii) + h, y(ii) + k3, z(ii) + l3); y(ii+1) = y(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4); z(ii+1) = z(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end I would like to use this function for solving motion equations for a ball with air resistance. So starting with the basic equations (I hope they are correct): $$md^2x/dt^2 = F_{D_x}$$ and $$md^2y/dt^2 = F_{D_y}-mg$$ with $$F_{D_x} = -Dvvcos(\theta)$$ and $$F_{D_y} = -Dvvsin(\theta)$$ where $D$ is a drag coefficient. From my understanding, and from what I read, both second-order equation can be converted to two coupled equations that could be used with my coupled Runge-Kutta method function. Can someone help me understand how to convert my equation to a set of coupled equations that could be used with my MATLAB function? EDIT I was able to use my rk4_c function, however it only works for solving the velocities. I'm now trying to see how I could solve for the positions x and y. I'm kind of learning about Runge-Kutta methods at the same time, so sorry for the questions that may sound trivials. % Runge Kutta code to solve projectile motion with quadratic drag % dVx/dt = -(D/m)*vx*sqrt(vx^2+vy^2) % dVy/dt = -(D/m)*vy*sqrt(vx^2+vy^2) - g clc clear all % Constant D = 0.24; % m = 2; % kg g = 9.80665; % m/s^2 % Define function handles fVx = @(t,vx,vy) -(D/m)*vx*sqrt(vx^2+vy^2); fVy = @(t,vx,vy) -(D/m)*vy*sqrt(vx^2+vy^2) - g; % Initial conditions v0 = 200; % m/s theta = 30*pi/180; % rad t(1) = 0; vx(1) = v0*cos(theta); vy(1) = v0*sin(theta); % Step size h = 0.01; % s tFinal = 2; N = ceil(tFinal/h); % RK4 simultaneous coupled loop for ii = 1:N % Update time t(ii+1) = t(ii) + h; % Update vx and vy k1 = h * fVx(t(ii), vx(ii), vy(ii)); l1 = h * fVy(t(ii), vx(ii), vy(ii)); k2 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1); l2 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1); k3 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2); l3 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2); k4 = h * fVx(t(ii) + h, vx(ii) + k3, vy(ii) + l3); l4 = h * fVy(t(ii) + h, vx(ii) + k3, vy(ii) + l3); vx(ii+1) = vx(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4); vy(ii+1) = vy(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end % Plot the solution figure(1) plot(t, vx, t,vy) xlabel('Time (s)') ylabel('Velocities (m/s)') legend('Vx', 'Vy') | eng_Latn | 18,710 |
Could someone help solve this PDE? Given, $U_t-U_{xx}-2U_x=0$ Using method of separation of variables, find ALL possible solution. My answer : $U(x,t)=X(x)T(t)$ $T'(t)/T(t)$=$[X''(x)+2X'(x)]/X(x)$=$\lambda$ From here I'm stuck. Could someone show me some working steps for me to proceed further. | PDE separation of variables Hi could someone guide me this problem It says , $ u_t - u_{xx}-2 u_x=0 $ Use the method of separation of variables to find all possible solutions. Could someone help me out for this problem. I'm beginner at PDE. I would be much appreciated if you able to show some partial work so that I can understand. Thanks in advance for taking my consideration | Equation of motion of a free particle We know that the equation of motion of particle can be derived from the respective action. But in the book I am reading, the author is saying: ... timelike worldline of a massive particle is parametrised by proper time, $x^a=x^a(\tau)$. The velocity (tangent) vector is $u^a \equiv \dot{x}^a(\tau)$. This is normalised as $u^au_a =-c^2$. How can one derive e.o.m? | eng_Latn | 18,711 |
Another solution to the problem of finding the interior points and limit points of $A=\{(x,y):x^2+2y^2<1\}$ Find out all interior points and limit points of $A=\{(x,y):x^2+2y^2<1\}$ ($(x,y) \in \mathbb R^2$) I've got an answer from original post but as i'm not familiar with this subject, I cannot understand the solution. Is there any other solution? | interior points and limit points of $\{(x,y): x^2+2y^2 < 1\}$ Find out all interior points and limit points of $ A =\{(x,y): x^2+2y^2 < 1\}$ I understand this problem graphically, but i'm not quite sure how to prove the answer rigorously using mathematical word. My try : Let k be an element of A, and suppose t is an element of $\{(x,y):x^2+2y^2 =1\}$, which makes minimum $||t-k||$. Then $N(k,\delta)\subset A$ when $\delta = \||t-k||$. I suppose this will work, but I want more clear and detailed proof. Thank you in advance. | Compute $P(Y<3X)$ using joint PDF I'm given a joint pdf $f_{X,Y}(x,y)=2e^{-x-y}, 0<x<y, 0<y $ and asked to compute $P(Y<3X)$. To do this, I let $Y=3X$ (the boundary) and found that the region of integration is under this line. To find $P(Y<3X)$, it seems to me that the integral for this region be written as $P(Y<3X)=\int_0^\infty \int_0^{3x} 2e^{-x-y} dy dx $ However this isn't right, as the boundaries are actually $P(Y<3X)=\int_0^\infty \int_x^{3x} 2e^{-x-y} dy dx $ Why is this? I'm having a hard time seeing how to construct these boundaries (especially the fact that the $y$ boundary goes from $x$ to $3x$ instead of $0$ to $3x$) Thank you. | eng_Latn | 18,712 |
Newton method to find $\frac{1}{\sqrt{a}}$ What function should I use to find $\frac{1}{\sqrt{a}}$ without using division? | Using Newton's method calculate $ \frac{1}{\sqrt{a}} $ without division Suggest algorithm for the numerical calculation $ \frac{1}{\sqrt{a}} \ a > 0 $ without division, use Newton's method. My idea is: $$ \frac{1}{\sqrt{a}} = (\sqrt{a})^{-1} = a^{-\frac{1}{2}}$$ $$ x = a^{-\frac{1}{2}} \Rightarrow f(x)=x^2 - a^{-1}$$ And next step is calulate $ f'(x)=2x $, but after substitution I got: $$ x_{n+1} = \frac{x_n^2+a^{-1}}{2x_n}$$ And know I still have divion, I guess. Maybe you've better idea on this task? | Calculate Euler equations of fluid dynamics without division? I'm working on the calculation of the with the . Unfortunately I'm not allowed to do a division. So I'm wondering if there's a form which does not need a division. At the moment the Euler equations look like this: $$ \frac{\partial}{\partial t} \begin{pmatrix} \rho \\ \rho v_1 \\ \rho v_2 \\ \rho v_3 \\ \rho E \end{pmatrix} = -\mathrm{div} \begin{pmatrix} \rho v_1 & \rho v_2 & \rho v_3 \\ \rho v_1^2 + p & \rho v_1 v_2 & \rho v_1 v_3 \\ \rho v_2 v_1 & \rho v_2^2 + p & \rho v_2 v_3 \\ \rho v_3 v_1 & \rho v_3 v_2 & \rho v_3^2 + p \\ (\rho E + p) v_1 & (\rho E + p) v_2 & (\rho E + p) v_3 \end{pmatrix} $$ As you can see, I first need to calculate $\frac{\rho v_1}{\rho}$ to get $v_1$ so I can calculate e.g. $\rho v_1^2$ | eng_Latn | 18,713 |
Differential Equations Recipes I have many questions I'd like to ask today. I'm currently studying for my A-levels which is a qualification mainly based in England. I'm taking A-level Further Maths, which involves the study of First and Second Order Differential Equations. It seems we are just being taught to apply recipes to these types of equations and are given no intuition as to why these things even work? This isn't just for teaching in England, I've watched many American lectures and lecture notes and the same thing is done. Let's take a look at Non-Linear Separable First Order Differential Equations. We have defined these equations to take the form $$ N(y)\frac{dy}{dx} = M(x)$$ We are taught to solve these equations by multiplying $\ dx $ to both sides and then integrate by their respective variables. Like so: $$ \int N(y)\ dy = \int M(x) \ dx $$ I have just accepted this for some time, but I'd like to get to the bottom of wtf is going on. We are taught from Year 12/Calculus 1 days that $\frac{dy}{dx} $ is not a ratio but can be treated like one in many cases. Ok... then could you explain really what we just did when solving the above differential equation? I can accept the fact that we can treat $\frac{dy}{dx}$ as a ratio but it's not really a ratio. So, can we be told what we are really doing? Is it some kind of hidden secret? I just want to know why we multiplied both sides by $\ dx$, what are we actually doing? I can assume that the reason we are not told is because it involves higher level math, that someone taking Calculus would not understand. But will actual reasoning behind why these recipes work be exposed to us during University math courses? | What am I doing when I separate the variables of a differential equation? I see an equation like this: $$y\frac{\textrm{d}y}{\textrm{d}x} = e^x$$ and solve it by "separating variables" like this: $$y\textrm{d}y = e^x\textrm{d}x$$ $$\int y\textrm{d}y = \int e^x\textrm{d}x$$ $$y^2/2 = e^x + c$$ What am I doing when I solve an equation this way? Because $\textrm{d}y/\textrm{d}x$ actually means $$\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$$ they are not really separate entities I can multiply around algebraically. I can check the solution when I'm done this procedure, and I've never run into problems with it. Nonetheless, what is the justification behind it? What I thought of to do in this particular case is write $$\int y \frac{\textrm{d}y}{\textrm{d}x}\textrm{d}x = \int e^x\textrm{d}x$$ $$\int \frac{\textrm{d}}{\textrm{d}x}(y^2/2)\textrm{d}x = e^x + c$$ then by the fundamental theorem of calculus $$y^2/2 = e^x + c$$ Is this correct? Will such a procedure work every time I can find a way to separate variables? | Find the general solution of $y'= a^{x + y}$ where y is the function How should I approach this problem? Should I put ln logarithm on both sides and move factor of variable of lnx on right side in front of the ln, or do somehing else? I understand general solving and types of solving these differential equations of first order, but I have no idea what my next step should be. | eng_Latn | 18,714 |
Why do materials obey Hooke's law? Why do materials extend proportionally to the force exerted on them (Hooke's law)? I thought that when materials are compressed or extended under force, their atoms become closer or further apart; the inter-atomic forces are essentially electrostatic; electrostatic forces (and indeed most other forces) follow an inverse-square law; So I would naively conclude that springs should follow an inverse square law. But clearly in most situations the law is linear. Where is my logic flawed? | Can Hooke's law be derived? Can we derive from the theory of elasticity? I know it is not a fundamental law and therefore can be derived from more basic considerations. | Kepler problem in time: how do two gravitationally attracted particles move? Two particles with initial positions and velocities $r_1,v_1$ and $r_2,v_2$ are interacting by the inverse square law (with G=1), so that $$ {d^2r_1\over dt^2} = - { m_2(r_1-r_2)\over |r_1-r_2|^3} $$ $$ {d^2r_2\over dt^2} = - { m_1(r_2-r_1)\over |r_1-r_2|^3} $$ (the inverse square law along the line of separation). What is the complete solution of these differential equations? What is the position of the two objects as a function of time? After reading a lot on Wikipedia, I've come to the definition of center of mass and relative coordinates: $$R(t) ~=~ \frac{m_1 r_1 + m_2 r_2}{m_1+m_2}$$ $$\ddot{r}(t) r(t)^2 ~=~ (m_1+m_2)G$$ Where $R$ is the center of mass, and $r$ is the displacement between the particles... Is this correct? How do I proceed to solve the differential equation? | eng_Latn | 18,715 |
Special character with latex I'am trying to write an equation where thee is this special caracter: \^{e} \begin{equation} \^{e} = 1 \end{equation} But the probleme is \^{e} is invisible, I see only “= 1” | How to typeset the symbol “^” (caret/circumflex/hat) I need to display the symbol '^' d <- dist(fascores, method = "euclidean")^2 How do I do that? | Find the value of $y(1)$ of the ODE $y'+y=|x|$. Let $y$ be the solution of $$y'+y=|x|$$ for $x\in\mathbb{R}$ and $y(-1)=0$. Then the value of $y(1)$ is $\frac{2}{e}-\frac{2}{e^2}$ $\frac{2}{e}-2e^2$ $2-\frac{2}{e}$ $2-2e$ I don't know what to do with the absolute value function in this problem. So I started like the regular first order equation by calculating integrating factor, and got $$ye^x=\int |x|e^x dx+C.$$ Now I got stuck as how to tackle the absolute value function? Help me to solve this. Thanks! | eng_Latn | 18,716 |
In Relativistic Quantum Mechanics, Bjorken and Drell state that expanding the square root in the equation $$-\hbar^2\frac{\partial^2\psi}{\partial t^2}=\sqrt{-\hbar^2c^2\boldsymbol{\nabla}^2+m^2c^4}\psi$$ would result in a non-local wave equation. What exactly do they mean by this? One could argue that in order to compute $\psi$'s spatial derivatives at a certain point in space-time up to arbitrary order, one has to know the behavior of $\psi$ in points arbitrarily far from the point of interest, even if there is no world line linking those points. But if we were to solve a local equation such as the Dirac equation, we could do that nonetheless. Accordingly, I think there must be a deeper reason for the non-locality that Bjorken and Drell warn for. | The Wikipedia has a derivation of the Klein-Gordon equation. It gets to this step: $$\sqrt{\textbf{p}^2 c^2 + m^2 c^4} = E$$ and inserts the QM operators to get $$\left( \sqrt{ (-i \hbar \nabla)^2 c^2 + m^2 c^4 } \right) \psi = i\hbar\frac{\partial}{\partial t} \psi$$ The article then says This, however, is a cumbersome expression to work with because the differential operator cannot be evaluated while under the square root sign. In addition, this equation, as it stands, is nonlocal. To fix this, the first equation is squared instead to get $$\textbf{p}^2 c^2 + m^2 c^4 = E^2$$ after which the QM operators are inserted and the expression is simplified to get $$-\hbar^2 c^2 \nabla^2 \psi + m^2 c^4 \psi = -\hbar^2 \frac{\partial^2}{\partial t^2} \psi$$ A couple things I don't understand. First, are the solutions to this differential equation not exactly the same as the solutions to the first differential equation? Both sides of the starting equation were squared, so it seems to me that regardless of the particular form of the resulting differential equation, both of them should have the exact same set of solutions. Secondly, why is the first differential equation cumbersome to work with? It seems like it would in fact be easier to work with, since the operator under the square root could be expanded in terms of a Taylor series and then you have an equation that is first order in time. And finally, can someone elaborate on what nonlocal means? The linked article on the Wikipedia page didn't entirely help me understand it. | Take a sponge ball and compress it. The net force acting on the body is zero and the body isn't displaced. So can we conclude that there is no work done on the ball? | eng_Latn | 18,717 |
I have been given this question from my university to complete and I am having some trouble. The lecturer did not do a very good job of explaining the concept and I could not find any materials online to help so I came here. The question is- What happens if you apply the method of steepest descent to $f(x)=(x_1^2)+(x_2^2)+(x_3^2)$ Thank you. | I am totally stumped for what to do here, any help would be appreciated. 1) What happens if you apply the method of steepest descent to $f(x)=x_1^2+x_2^2+x_3^2$? 2)Lagrange multipliers method- Find the dimensions of the rectangular box open at the top of greatest internal volume, given the surface area of the five faces is $108 \, \mathbb{cm}^2$. Thanks | I always use align in my documents, and avoid equation. Is there anything wrong with that? My reasoning behind this: align > equation, so why not use it? | eng_Latn | 18,718 |
Let $X(t) = \left[\begin{matrix}a(t) & b(t) \\c(t) & d(t)\end{matrix}\right]$ and let $U$ be a nonsingular matrix. How do you solve $$\frac{d}{dt} X(t)=UX(t)$$ I presume there is some general method to solve these kinds of ODEs but I cannot find anything about it, online. BTW: ----------------- I know that you can get 4 ODE equations for the four unknown functions of $t$. The problem is that each ODE equation includes other functions as so $$\frac{d a}{d t}=u_{11}a+u_{12}c$$ | Solve the following system of equations $ \begin{cases} x_1^{'}(t)=x_1(t)+3x_2(t) \\ x_2^{'}(t)=3x_1(t)-2x_2(t)-x_3(t) \\ x_3^{'}=-x_2(t)+x_3(t)\end{cases} $. First, I create the column vectors $X$ and $X^{'}$. Then the matrix $$A= \begin{bmatrix} 1 & 3 & 0 \\ 3 & -2 & -1 \\ 0 & -1 & 1 \\ \end{bmatrix} $$ Now, I find the eigenvalues, $-4,3,1$ and their corresponding eigenvectors $(-3,5,1)^T (-3,-2,1)^T (1,0,3)^T$. I'm just not sure how to take it from here and solve the system of differential equations. I want a diagonal matrix $D$ so that I can read the solutions easy, but I'm not sure how to do it. EDIT Building on @Francisco 's answer, I'd have that: $$X=c_1 (-3,5,1)^T e^{-4t} + c_2 (-3,-2,1)^T e^{3t} + c_3 (1,0,3)^T e^{t} $$. But I believe this could be written in a simpler form. | I am able to solve simple differential equations like : $$\dfrac{dy}{dx} = 3x^2 + 2x$$ We simply bring $dx$ to other site and integrate. But how do we find solutions of differential equations like : $$\frac{d^2x}{dt^2} + \dfrac{kx}{m} = 0$$ ? We have been told the solution is $x(t) = A\cos ( \omega t + \phi_o)$ where $\omega\ =\sqrt{\dfrac{k}{m}}$ but how do we actually find it? | eng_Latn | 18,719 |
The first equation of motion is $v = u + at$. The second equation of motion is $s = ut + \frac{at^2}{2}$. If we divide the second equation of motion by time $t$, why don't we get the first equation of motion where has $1/2$ come from? | I'm preparing for my exam, but I have difficulties in perceiving why there is a $\frac{1}{2}$ in the distance formula $d=\frac{1}{2}at^2$? | Consider this equation : $$\sqrt{\left( \frac{dy\cdot u\,dt}{L}\right)^2+(dy)^2}=v\,dt,$$ where $t$ varies from $0$ to $T$ , and $y$ varies from $0$ to $L$. Now how to proceed ? This equation arises out of following problem : A cat sitting in a field suddenly sees a standing dog. To save its life, the cat runs away in a straight line with speed $u$. Without any delay, the dog starts with running with constant speed $v>u$ to catch the cat. Initially, $v$ is perpendicular to $u$ and $L$ is the initial separation between the two. If the dog always changes its direction so that it is always heading directly at the cat, find the time the dog takes to catch the cat in terms of $v, u$ and $L$. See my solution below : Let initially dog be at $D$ and cat at $C$ and after time $dt$ they are at $D'$ and $C'$ respectively. Dog velocity is always pointing towards cat. Let $DA = dy, \;AD' = dx$ Let $CC'=udt,\;DD' = vdt$ as interval is very small so $DD'$ can be taken straight line. Also we have $\frac{DA}{DC}= \frac{AD'}{ CC'}$ using triangle property. $\frac{dy}{L}= \frac{dx}{udt}\\ dx = \frac{dy.udt}{L}$ $\sqrt{(dx)^2 + (dy)^2} = DD' = vdt \\ \sqrt{(\frac{dy.udt}{L})^2 + (dy)^2} = vdt $ Here $t$ varies from $0-T$, and $y$ varies from $0-L$. Now how to proceed? | eng_Latn | 18,720 |
Why do we need to use chain rule of differentiation For example, here we have $y = \sin(x^2 + 5x -6)$. So, it’s answer will be $\cos(t)\cdot(2x+5)$, where $t= x^2 + 5x -6$. But I want to know what if we solve it directly I.e $\cos(2x+5)$ wrong? What’s the reason? Why choose the 1st approach? | how to prove the chain rule? I have just learnt about the chain rule but my book doesn't mention a proof on it. I tried to write a proof myself but can't write it. So can someone please tell me about the proof for the chain rule in elementary terms because I have just started learning calculus. | Solve differential equation, $x'=x^2-2t^{-2}$ Solve differential equation: $x'=x^2- \frac{2}{t^2}$ Maybe is it sth connected with homogeneous equation? I have no idea how to solve it. | eng_Latn | 18,721 |
McLaurin expansion for $f(x)=e^{\sin{x}}$ of the 4:th order. By the 4:th order, they mean using the 4:th derivative. But the differentiation gets a bit ugly quite fast, so instead of computing all the derivatives , I should be able to use the standard expansions: $$e^t=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+B(t) \\ \sin{x}=x-\frac{x^3}{3!}+B(x),$$ Where $B(t),B(x)$ are functions that are bounded close to $x=0.$ Correct me if I'm wrong, but both of the expansions above are of 4:th order since for the sine function we have two terms vanishing because they become zero? I'm not sure how I should fuse the separate expansions of $e^t$ and $\sin{x}$ to one expansion. Setting $t=\sin{x}$ seems to give me an expression tedious tedious to work with. | Mclaurins with $e^{\sin(x)}$ To evaluate $e^{\sin(x)}$ I use the standard series $e^t$ and $\sin(t)$, combining them gives me: $e^t = 1+t+\dfrac{t^2}{2!}+\dfrac{t^3}{3!}+\dfrac{t^4}{4!}+O(t^5)$ $\sin(t) = t-\dfrac{t^3}{3!}+O(t^5)$ $\therefore e^{\sin(x)} = 1+\sin(x)+\dfrac{\sin^2(x)}{2!}+\cdots+O(x^5)$ $\iff e^{\sin(x)}=1+\left(x-\dfrac{x^3}{6}\right)+\dfrac{\left(x-\frac{x^3}{6}\right)^2}{2}+\dfrac{x^3}{6}+O(x^5) = 1+x +\dfrac{x^2}{2}-\dfrac{x^4}{6} + O(x^5)$ In the last step, I only evaluate up to $\sin^3x$ term, everything above has a grade equal to or greater than $x^5$ However, I'm wrong. According to Wolfram, the series expansion for $e^{\sin x} = 1+x+\dfrac{x^2}{2}-\dfrac{x^4}{8}+O(x^5)$. What did I do wrong? I can't seem to find what or where. | find an approximate solution, up to the order of epsilon The question is to find an approximate solution, up to the order of epsilon of following problem. $$y'' + y+\epsilon y^3 = 0$$ $$y(0) = a$$ $$y'(0) = 0$$ I tried to solve the given problem using perturbation theory. $$y(t) = y_0(t) + \epsilon y_1(t) + \epsilon^2 y_2(t) + \cdots$$ $$t = w(\epsilon)t = (1 + \epsilon w_1 + \epsilon^2 w_2 + \cdots )t$$ However, i failed to find an appropriate approximate solution... help me!! | eng_Latn | 18,722 |
Solve y′−∫x0y(t)dt=2 How i solve this? I have not idea how to approach this differential equation. y′−∫x0y(t)dt=2 I get the probably familiar DE: y′′−y=0. please someone can solve this exercise | Solve $y'-\int_0^xy(t)dt=2$ I have not idea how to approach this differential equation. $$y'-\int_0^xy(t)dt=2$$. Basically, I did, $$F''(t)-F(x)+F(0)=2 \;\;\;\;\;\;\; F'=y$$ I am stuck. Thank You. | Double integral with transformation; possible error in limits I'm asked to show that the following integral is true with the transformation $u = x+y$ and $y=uv$: $$\int_0^1 \int_0^{1-x} e^{y/(x+y)} dx dy = \frac{e-1}2 $$ I found the determinant of the jacobian of the transformation is $-u$, and the exponent is just $v$ so the integral becomes $$\iint_R |-u|e^v du dv = \frac {u^2}2 e^v|_R$$ $$=\frac{(x+y)^2}2e^{y/(x+y)}|_{x=0}^{1-x}\;|_{y=0}^1$$ And plugging it all in doesn't work (AFAICT). For one, the x's don't all cancel. I've looked at it 100 times. It's definitely written dxdy not dydx even though there is definitely an x in the integral. Maybe this is a typo and why my calculation isn't working out? I tried reversing the order to dydx and it didn't work. I tried changing the limit of integration to $1-y$ and it didn't work out. I'm not even sure how to figure out the domain of integration in the uv-plane. Flipping things around yields y=uv and x=u-uv. But if my domain is bounded by x=0 and y=0 and y=1 (ignoring the last limit for now), then I have uv=o=u and uv=1. !?! Am I just making an algebra mistake plugging everything in? Is there a way to handle a variable in it's own integral that I'm unaware of? | eng_Latn | 18,723 |
Consider the following system of differential equations Consider the following model. Find the general solution $\frac{dx}{dt}=-2y$, $\frac{dy}{dt}=8x$ So far this is what I have: $$\frac{dx}{dt}=-2y$$ $$x'=-2y$$ $$y=-\frac{x'}{2}$$ $$y=-\frac{x''}{2}$$ and then $$\frac{dy}{dt}=8x$$ $$y'=8x$$ $$-\frac{x''}{2}=8x$$ $$-x''=16x$$ $$x''+16x=0$$ Is this correct so far? What step should I take next? | Differential Equations: solve the system Solve the following system: $$dx/dt=-.2(y-2)$$ $$dy/dt=.8(x-2)$$ This is what I have so far, but I got stuck.. $$\begin{eqnarray} dx/dt&=&-.2y-.4\\ x'&=&-.2y-.4\\ x'+.4&=&-.2y\\ y&=&-x/2'-2\\ y&=&-x''/2-2\\ \\ dy/dt&=&.8(x-2)\\ dy/dt&=&.8x-1.6\\ y'&=&.8x-1.6\\ -x''-2&=&.8x-1.6\\ -x''&=&.8x+.4\\ x''+16x+.8&=&0 \end{eqnarray}$$ when I used the quadratic formula I got stuck. Is this correct so far? Can someone finish it off for me? | How to integrate $\frac{1}{\sqrt{1+x^2}}$ using substitution? How you integrate $\frac{1}{\sqrt{1+x^2}}$ using following substitution? $1+x^2=t$ $\Rightarrow$ $x=\sqrt{t-1} \Rightarrow dx = \frac{dt}{2\sqrt{t-1}}dt$... Now I'm stuck. I don't know how to proceed using substitution rule. | eng_Latn | 18,724 |
Find an equation in spherical coordinates for the surface represented by the rectangular equation The rectangular equation is $$x^2+y^2-8z^2=0$$ $$x^2+y^2=8z^2$$ Know in the relationship between rectangular and spherical coords. we can manipulate our given to fit the form: $$x^2+y^2+z^2=9z^2$$ $$\rho^2=x^2+y^2+z^2, \space z=\rho\cos(\phi)$$ $$\rho^2=9\rho^2\cos^2(\phi)$$ $$1=9\cos^2(\phi)$$ $$\frac{1}{3}=\cos(\phi)$$ $$\arccos\left(\frac{1}{3}\right)=1.23 \space rads$$ And so the equation in spherical coords. is $\phi=1.23$ I know my math is correct but I have the wrong answer so ilm not sure where I went wrong. | Find an equation in spherical coordinates for the surface represented by the rectangular equation The rectangular equation is $$x^2+y^2-8z^2=0$$ $$x^2+y^2=8z^2$$ Know in the relationship between rectangular and spherical coords. we can manipulate our given to fit the form: $$x^2+y^2+z^2=9z^2$$ $$\rho=x^2+y^2+z^2, \space z=\rho\cos(\phi)$$ $$\rho^2=9\rho^2\cos^2(\phi)$$ $$1=9\cos(\phi)$$ $$\frac{1}{3}=\cos(\phi)$$ $$\arccos(\frac{1}{3})=1.23 \space rads$$ And so the equation in spherical coords. is $\phi=1.23$ I know my math is correct but I have the wrong answer so I'm not sure where I went wrong. | Ball motion with air resistance coupled differential equation for fourth-order Runge-Kutta I've created a MATLAB function for solving coupled differential equation with the fourth-order Runge-Kutta method based on what is provided (Simultaneous Equations of First Order). Here the function: function [y] = rk4_c(f, g, h, x, y, z, n) % Runge-Kutta % Implementation of the fourth-order method for coupled equations % h = dt % x is the time here for ii=1:(n-1) k1 = h * f(x(ii), y(ii), z(ii)); l1 = h * g(x(ii), y(ii), z(ii)); k2 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1); l2 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1); k3 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2); l3 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2); k4 = h * f(x(ii) + h, y(ii) + k3, z(ii) + l3); l4 = h * g(x(ii) + h, y(ii) + k3, z(ii) + l3); y(ii+1) = y(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4); z(ii+1) = z(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end I would like to use this function for solving motion equations for a ball with air resistance. So starting with the basic equations (I hope they are correct): $$md^2x/dt^2 = F_{D_x}$$ and $$md^2y/dt^2 = F_{D_y}-mg$$ with $$F_{D_x} = -Dvvcos(\theta)$$ and $$F_{D_y} = -Dvvsin(\theta)$$ where $D$ is a drag coefficient. From my understanding, and from what I read, both second-order equation can be converted to two coupled equations that could be used with my coupled Runge-Kutta method function. Can someone help me understand how to convert my equation to a set of coupled equations that could be used with my MATLAB function? EDIT I was able to use my rk4_c function, however it only works for solving the velocities. I'm now trying to see how I could solve for the positions x and y. I'm kind of learning about Runge-Kutta methods at the same time, so sorry for the questions that may sound trivials. % Runge Kutta code to solve projectile motion with quadratic drag % dVx/dt = -(D/m)*vx*sqrt(vx^2+vy^2) % dVy/dt = -(D/m)*vy*sqrt(vx^2+vy^2) - g clc clear all % Constant D = 0.24; % m = 2; % kg g = 9.80665; % m/s^2 % Define function handles fVx = @(t,vx,vy) -(D/m)*vx*sqrt(vx^2+vy^2); fVy = @(t,vx,vy) -(D/m)*vy*sqrt(vx^2+vy^2) - g; % Initial conditions v0 = 200; % m/s theta = 30*pi/180; % rad t(1) = 0; vx(1) = v0*cos(theta); vy(1) = v0*sin(theta); % Step size h = 0.01; % s tFinal = 2; N = ceil(tFinal/h); % RK4 simultaneous coupled loop for ii = 1:N % Update time t(ii+1) = t(ii) + h; % Update vx and vy k1 = h * fVx(t(ii), vx(ii), vy(ii)); l1 = h * fVy(t(ii), vx(ii), vy(ii)); k2 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1); l2 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1); k3 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2); l3 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2); k4 = h * fVx(t(ii) + h, vx(ii) + k3, vy(ii) + l3); l4 = h * fVy(t(ii) + h, vx(ii) + k3, vy(ii) + l3); vx(ii+1) = vx(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4); vy(ii+1) = vy(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end % Plot the solution figure(1) plot(t, vx, t,vy) xlabel('Time (s)') ylabel('Velocities (m/s)') legend('Vx', 'Vy') | eng_Latn | 18,725 |
python add float numbers and keep largest decimal part In Python, I am adding float numbers. I have numbers client_balance = 40360.7416622703 fund_manager_balance = 676.600516394647 uits= 6.72791159564433 here is my python code def update_client_balance(self, record): if record.client_id.account_name != ACCOUNT_MANAGER_NAME: self.client_balance += record.units def update_manager_balance(self, record): if record.client_id.account_name == ACCOUNT_MANAGER_NAME: self.fund_manager_balance += record.units After execution of these functions I have client_balance = 40360.7416622703 fund_manager_balance = 683.328427990291 Now I want to validate the result of these functions manually so (client_balance + fund_manager_balance)(before function execution) -uits - (client_balance + fund_manager_balance)(after function execution) By Value 40360.7416622703 + 676.600516394647 + 6.72791159564433 - (40360.7416622703 + 683.328427990291) = -0.00000000000727595761418343 It should give me Zero value but it's giving me -0.00000000000727595761418343 Please help me how I can solve this? | Is floating point math broken? Consider the following code: 0.1 + 0.2 == 0.3 -> false 0.1 + 0.2 -> 0.30000000000000004 Why do these inaccuracies happen? | Ball motion with air resistance coupled differential equation for fourth-order Runge-Kutta I've created a MATLAB function for solving coupled differential equation with the fourth-order Runge-Kutta method based on what is provided (Simultaneous Equations of First Order). Here the function: function [y] = rk4_c(f, g, h, x, y, z, n) % Runge-Kutta % Implementation of the fourth-order method for coupled equations % h = dt % x is the time here for ii=1:(n-1) k1 = h * f(x(ii), y(ii), z(ii)); l1 = h * g(x(ii), y(ii), z(ii)); k2 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1); l2 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1); k3 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2); l3 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2); k4 = h * f(x(ii) + h, y(ii) + k3, z(ii) + l3); l4 = h * g(x(ii) + h, y(ii) + k3, z(ii) + l3); y(ii+1) = y(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4); z(ii+1) = z(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end I would like to use this function for solving motion equations for a ball with air resistance. So starting with the basic equations (I hope they are correct): $$md^2x/dt^2 = F_{D_x}$$ and $$md^2y/dt^2 = F_{D_y}-mg$$ with $$F_{D_x} = -Dvvcos(\theta)$$ and $$F_{D_y} = -Dvvsin(\theta)$$ where $D$ is a drag coefficient. From my understanding, and from what I read, both second-order equation can be converted to two coupled equations that could be used with my coupled Runge-Kutta method function. Can someone help me understand how to convert my equation to a set of coupled equations that could be used with my MATLAB function? EDIT I was able to use my rk4_c function, however it only works for solving the velocities. I'm now trying to see how I could solve for the positions x and y. I'm kind of learning about Runge-Kutta methods at the same time, so sorry for the questions that may sound trivials. % Runge Kutta code to solve projectile motion with quadratic drag % dVx/dt = -(D/m)*vx*sqrt(vx^2+vy^2) % dVy/dt = -(D/m)*vy*sqrt(vx^2+vy^2) - g clc clear all % Constant D = 0.24; % m = 2; % kg g = 9.80665; % m/s^2 % Define function handles fVx = @(t,vx,vy) -(D/m)*vx*sqrt(vx^2+vy^2); fVy = @(t,vx,vy) -(D/m)*vy*sqrt(vx^2+vy^2) - g; % Initial conditions v0 = 200; % m/s theta = 30*pi/180; % rad t(1) = 0; vx(1) = v0*cos(theta); vy(1) = v0*sin(theta); % Step size h = 0.01; % s tFinal = 2; N = ceil(tFinal/h); % RK4 simultaneous coupled loop for ii = 1:N % Update time t(ii+1) = t(ii) + h; % Update vx and vy k1 = h * fVx(t(ii), vx(ii), vy(ii)); l1 = h * fVy(t(ii), vx(ii), vy(ii)); k2 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1); l2 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1); k3 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2); l3 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2); k4 = h * fVx(t(ii) + h, vx(ii) + k3, vy(ii) + l3); l4 = h * fVy(t(ii) + h, vx(ii) + k3, vy(ii) + l3); vx(ii+1) = vx(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4); vy(ii+1) = vy(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end % Plot the solution figure(1) plot(t, vx, t,vy) xlabel('Time (s)') ylabel('Velocities (m/s)') legend('Vx', 'Vy') | eng_Latn | 18,726 |
Show that the solution of the system remains in a region for all future time $t \geq 0$ Consider the system of equations \begin{cases} \dot{x} = x(1- x^2 - y^2)\\ \dot{y} = y(4- x^2 - y^2) \end{cases} Define $(x(t;(x_0,y_0)), y(t;(x_0,y_0)))$ as the solution of this system with initial condtition $(x_0,y_0) \in \mathbb{R^2}$, so $(x(0;(x_0,y_0)), y(0;(x_0,y_0))) = (x_0 , y_0)$. Suppose that the initial condition $(x_0,y_0)$ lies in the region $S \subset \mathbb{R^2}$, with $S = \{(x,y) : x>0 , y>0$ and $1 < x^2 + y^2 < 4 \}$. So $(x(0;(x_0,y_0)), y(0;(x_0,y_0))) = (x_0 , y_0) \in S$. Show that $(x(t;(x_0,y_0)), y(t;(x_0,y_0))) \in S$, for all $t \geq 0$ and $\lim_{t \rightarrow \infty} (x(t;(x_0,y_0)), y(t;(x_0,y_0))) = (0,2)$. Well, for $(x,y) \in S$ we have $\dot{x} <0$ and $\dot{y} > 0$. Then, $x(t)$ is a monotonic decreasing function of time, and $y(t)$ is a monotonic increasing function of time. A Lemma in my textbook says that in this case any solution $x(t)$, $y(t)$ must leave a region at some later time. This confuses me, since I have to show that a solution remains in the region for all $t \geq 0$. Maybe I am misinterpreting something. Furthermore $(0,2)$ is clearly a critical point of the system. How can be showed that the limit approaches $(0,2)$? | Specific system of differential equations I have the following system of equations: \begin{eqnarray}\frac{dx}{dt} = x(1 - x^2 - y^2) \\ \frac{dy}{dt} = y(4 - x^2 - y^2) \end{eqnarray} I want to prove that if a solutions starts (at time $t = 0$) in the region $$ G = \{(x, y) \in \mathbb{R}^2 : x > 0, y > 0 \text{ and } 1 < x^2 + y^2 < 4\} $$ that it remains there for all $t \geq 0$. I have already proven that, from $G$, a solution cannot cross the $x$-axis where $0 < x < 2$ or the $y$-axis where $0 < y < 2$ (since those line segments are orbits), but the pieces of the circles with radius 1 and 2 (where $x > 0, y > 0$) are the lines I haven't done yet. In this question we can focus on just the circle with radius 1 since the proof for the other circle is completely analogous to it. In my book, there is a similar problem where they show that if a solution crosses a certain line, that solution attains its maximum or minimum value for $x$ and $y$, and then they derive a contradiction. However, in this case, for $n \geq 1$ the $n$-th derivative of $x$ equals 0 on the circle with radius 1. (If the second derivative were negative, $x$ would attain its maximum on the circle, a contradiction since $\frac{dx}{dt} < 0$ in $G$, so $x$ is decreasing.) I don't see why a solution can't cross the circle at all. On the circle, $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} > 0$, so after crossing the circle, the solution would go back up to $G$. Is the circle itself an orbit? EDIT I think I have found the answer. Using polar coordinates, I found $$ \frac{dr}{dt} = r - r^3 + 3r \sin^2(\theta) $$ Now, $$ \frac{dr}{dt} > 0 \Leftrightarrow 0 < r - r^3 + 3r \sin^2(\theta) \leq r - r^3 + 3r = 4r - r^3 $$ since $0 \leq \sin^2(\theta) \leq 1$. Note that $0 < r \leq 1$ satisfies $4r - r^3 > 0$, and thus $\frac{dr}{dt} > 0$. Therefore, if a solution goes from $G$ to the edge of the unit circle, $r$ must, on the edge, either be decreasing or constant (i.e., its derivative is less than or equal to 0), contradicting the fact that the derivative is positive. Can someone tell me if this is correct? | If $x_1,x_2,\ldots,x_n$ are the roots for $1+x+x^2+\ldots+x^n=0$, find the value of $\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}$ Let $x_1,x_2,\ldots,x_n$ be the roots for $1+x+x^2+\ldots+x^n=0$. Find the value of $$P(1)=\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}$$ Source: IME entrance exam (Military Institute of Engineering, Brazil), date not provided (possibly from the 1950s) My attempt: Developing expression $P(1)$, replacing the 1 by $x$, follows $$P(x)=\frac{(x_2-x)\cdots (x_n-x)+\ldots+(x_1-x)\cdots (x_{n-1}-x)}{(x_1-x)(x_2-x)\cdots (x_n-x)}$$ As $x_1,x_2,\ldots,x_n$ are the roots, it must be true that $$Q(x)=(x-x_1)\cdots(x-x_n)=1+x+x^2+\ldots+x^n$$ and $$Q(1)=(1-x_1)\cdots(1-x_n)=n+1$$ Therefore the denominator of $P(1)$ is $$(-1)^{n} (n+1).$$ But I could not find a way to simplify the numerator. Another fact that is probably useful is that $$1+x^{n+1}=(1-x)(x^n+x^{n-1}+\ldots+x+1)$$ with roots that are 1 in addition of the given roots $x_1,x_2,\ldots,x_n$ for the original equation, that is $$x_k=\text{cis}(\frac{2k\pi}{n+1}),\ \ k=1,\ldots,n.$$ This is as far as I could go... Hints and full answers are welcomed. | eng_Latn | 18,727 |
solve the following first order but not of first degree ordinary differential equation | Solve $x=y\frac{dy}{dx}-\left(\frac{dy}{dx}\right)^{2}$ | Order and degree of a differential equation | eng_Latn | 18,728 |
Solving two non-linear equations with two unknowns | Error/warning when using NSolve for simple equation | Prove that equation has exactly 2 solutions | eng_Latn | 18,729 |
Why Wolfram|Alpha can solve an equation but not Mathematica? | Can Reduce *really* not solve for x here? | Why my differential equations become True? | eng_Latn | 18,730 |
Acceleration closed formula in strong gravitational field We know that acceleration of a falling body relative to earth is G*M_earth/(R_earth)^2 What would the acceleration formula if the earth's mass is same as sun's mass but with same earth's radius? Is there algabreac simple closed formula according to GR for the acceleration of falling body relative to an observer in rest relative to the earth?, or we just must solve GR differential equations numerically? | What is the weight equation through general relativity? The gravitational force on your body, called your weight, pushes you down onto the floor. $$W=mg$$ So, what is the weight equation through general relativity? | The formal solution of the time-dependent Schrödinger equation Consider the time-dependent (or some equation in Schrödinger form) written down as $$ \tag 1 i \partial_{0} \Psi ~=~ \hat{ H}~ \Psi . $$ Usually, one likes to write that it has a formal solution of the form $$ \tag 2 \Psi (t) ~=~ \exp\left[-i \int \limits_{0}^{t} \hat{ H}(t^{\prime}) ~\mathrm dt^{\prime}\right]\Psi (0). $$ However, this form for the solution of $(1)$ is actually built by the method of successive approximations which actually returns a solution of the form $$ \tag 3 \Psi (t) ~=~ \hat{\mathrm T} \exp\left[-i \int \limits_{0}^{t} \hat{H}(t^{\prime})~\mathrm dt^{\prime}\right]\Psi (0), \qquad t>0, $$ where $\hat{\mathrm T}$ is the time-ordering operator. It seems that $(3)$ doesn't coincide with $(2)$, but formally $(2)$ seems to be perfectly fine: it satisfies $(1)$ and the initial conditions. So where is the mistake? | eng_Latn | 18,731 |
$f(x, y(x))=0$ Show that $y'=-\frac{\partial f / \partial x}{\partial f/ \partial y}.$ | Proving $f' = - \frac{\partial{F}/\partial{x}}{\partial{F}/\partial{y}}$ | Find the general solution of $y'= a^{x + y}$ where y is the function | eng_Latn | 18,732 |
Two different solutions to integral | Getting different answers when integrating using different techniques | Two different solutions to integral | eng_Latn | 18,733 |
Explain $\Delta x = v_0t + \tfrac{1}{2}gt^2$ please? | Why there is a $\frac{1}{2}$ in the distance formula $d=\frac{1}{2}at^2$? | Why my differential equations become True? | yue_Hant | 18,734 |
Taking the second derivative of a parametric curve I understand that for the parametric equations $$\begin{align*}x&=f(t)\\ y&=g(t)\end{align*}$$ If $F(x)$ is the function with parameter removed then $\displaystyle F'(x) = \frac{\text{d}y}{\text{d}t}\big/\frac{\text{d}x}{\text{d}t}$ But the procedure for taking the second derivative is just described as " replace $y$ with dy/dx " to get $$\frac{\text{d}^2y}{\text{d}x^2}=\frac{\text{d}}{\text{d}x}\left(\frac{\text{d}y}{\text{d}x}\right)=\frac{\left[\frac{\text{d}}{\text{d}t}\left(\frac{\text{d}y}{\text{d}t}\right)\right]}{\left(\frac{\text{d}x}{\text{d}t}\right)}$$ I don't understand the justification for this step. Not at all. But that's all my book says on the matter then it launches in to plugging things in to this formula, and it seems to work well enough, but I don't know why. I often find answers about question on differentials are beyond my level, I'd really like to get this, it'd mean a lot to me if someone could break it down. | Explanation behind Second Derivative of a Parametric Equation Formula I am looking to find out how the second derivative formula works. I can blindly apply it but I don't have a grasp of what is going on or why. The first derivitive has the formula $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. I told this was because if you divide them you cancel out the dts and get dy/dx. The second derivative has the formula $\frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{ \frac{d}{dt}\left(\frac{dy}{dx}\right) }{\frac{dx}{dt}}$. Is this something that can be explained with only a knowledge of calculus or is this something I would have to take on faith until I learn more. Any help would be greatly appreciated. | Solve differential equation, $x'=x^2-2t^{-2}$ Solve differential equation: $x'=x^2- \frac{2}{t^2}$ Maybe is it sth connected with homogeneous equation? I have no idea how to solve it. | eng_Latn | 18,735 |
What does the (2) after Unix system calls names mean? I was looking through some documentations for my university task and noticed that all of the syscalls there are followed by (2), like execv(2), fork(2) or fcntl(2). What do they represent? | What does the number in parentheses shown after Unix command names in manpages mean? For example: man(1), find(3), updatedb(2)? What do the numbers in parentheses (Brit. "brackets") mean? | What type of differential equation is $f'(x) = f(x/2)$ and how do you solve it? I have the following different equation $$f'(x) = f(x/2)$$ with $f(0)=10$. What type of DE is this, and how would you solve it? It seems $f(x)$ is likely to be some relative of $e^x$, since $f'(x) = f(x)$, which is close, but I don't even know what that type of DE is called with that $f(x/2)$ feature, so I'm not having any luck searching for a tutorial. The best candidate I've found was a "delay differential equation", but that seems more suited to $f(x-3)$ than $f(x/2)$. | eng_Latn | 18,736 |
Having trouble in solving second-order linear ordinary differential equation $(2x^2+1)y''-4xy'+4y=0$ $y_1(x) = x$ is given and I have to find $y_2(x)$ and solve equation. thanks. | Solve the second order ODE $(2x^2 + 1)y'' − 4xy' + 4y = 0$ I need help solving this past exam question, the professor posted exam questions with out solutions to last year's exam, I tried reduction of order, however that seemed to complicate things. And then I tried Euler-Cauchy method however I realized that won't work due to the $(2x^2 + 1)y''$. Given $y_1=x$ is a solution $$(2x^2 + 1)y'' − 4xy' + 4y = 0$$ Find the second solution $y_2$ for equation. | Solving a implicit, recursive function Assuming $F(0,0)=0$, can I solve $F(F(x,y),y)=0$ for $y=f(x)$ near $(0,0).$ | eng_Latn | 18,737 |
Calculation of double integral for the dedicatied region D Find the double integral which represents the intersection of the cylinder and the sphere happened over $$D:\{(x,y):\, -1\le x\le 1 \,\,\text{ and }\,\, 1-\sqrt{1-x^2} \le y\le 1+\sqrt{1-x^2}\, \}$$ The integral is $$2\iint_D \sqrt{f_x^2+f_y^2+1}\, d\sigma$$ where $f(x,y) = \sqrt{4-x^2-y^2}$. First of all I calculated $f_x^2=\frac{x^2}{-x^2-y^2+4}$, $f_y^2=\frac{y^2}{-x^2-y^2+4}$. However, I am not sure how to take into account the region $D$. Any help to finilize calculation are welcomed. | Surface area inside cylinder Find the surface area of the part $\sigma$: $x^2+y^2+z^2=4$ that lies inside the cylinder $x^2+y^2=2y$ So, the surface is a sphere of $R=2$. It looks there should be double integral to calculate the surface, but how, which way? | Why is $\frac{\operatorname dy'}{\operatorname dy}$ zero, since $y'$ depends on $y$? I know that $\frac{dy'(x)}{dy}=0$ (where $y'=\frac{dy(x)}{dx}$). The reason explained is that $y'$ does not depend explicitly on $y$. But intuitively, $y'$ depends on $y$, since if you vary $y$ you will modify $y'$. Why is my reasoning wrong (my reasoning sounds like it's related to functional calculus, instead of standard calculus)? I tried to write $\frac{dy'(x)}{dy}=\frac{d}{dy}\frac{dy}{dx}=\frac{d}{dx}\frac{dy}{dy}=\frac{d}{dx}1=0$, but this proof doesn't convince me. I think that other way to see this is saying that if a function is of the form $f(y)$, it will not dependend on the variable $y'$. But the same way you write $f(y)=y^2$, you could write $f(y)=\frac{d}{dx}y$, which clearly depends on $y'$. So I don't know if there are some types of operations which are restricted (for example, taking limits): Note: The problem raises in the context of Classical Mechanics, where: $\frac{\partial L(\dot{x},t)}{\partial x}=0$. | eng_Latn | 18,738 |
Can we differentiate a differential equation to determine its order and degree? It is known that we must need to convert the differential equation in polynomial equation of differential coefficients. But Can we differentiate a differential equation (whose degree is not defined) to determine its order and degree ? Example to show my doubt clearly: $y''=e^{y'}$ Above differential equation has its degree undefined. Differentiating it with respect to $x$ $y'''=(y'')^2$ So we may conclude that this is third order differential equation with degree $=1$ I know that on differentiating a differential equation number of arbitrary constants of solution equation increases hence we should not differentiate a differential equation in general. But I did not found a reference which states that we cannot differentiate a differential equation to determine its order and degree so I want to confirm my thoughts. | Order and degree of a differential equation Here is a question in my book Find the order and degree of the differential equation $$y=1+\frac{dy}{dx}+\frac{1}{2!}{\left(\frac{dy}{dx}\right)}^2+\frac{1}{3!}{\left(\frac{dy}{dx}\right)}^3+\cdots$$ At first sight we can conclude that the order is $1$ and the degree is undefined as as the power of $\frac{dy}{dx}$ continues to increase and has no limit.However my book gives the following solution Rewrite the DE as $$y=\exp\left({\frac{dy}{dx}}\right)$$ $$\implies \frac{dy}{dx}=\ln y$$ whose order and degree is 1 . Now ,I completely agree with this solution however I find it rather counterintuitive to my first line of thought .If the book is correct how can it be justified to prove my intuition was wrong? | How to determine the direction of a wave propagation? In the textbook, it said a wave in the form $y(x, t) = A\cos(\omega t + \beta x + \varphi)$ propagates along negative $x$ direction and $y(x, t) = A\cos(\omega t - \beta x + \varphi)$ propagates along positive $x$ direction. This statement looks really confusing because when it says the wave is propagating along $\pm$ x direction, to my understanding, we can drop the time term and ignore the initial phase $\varphi$ while analyzing the direction, i.e. $y(x, 0) = A\cos(\pm\beta x)$, however, because of the symmetry of the cosine function, $\cos(\beta x)\equiv \cos(-\beta x)$, so how can we determine the direction of propagation from that? I know my reasoning must be incorrect but I don't know how to determine the direction. So if we don't go over the math, how to figure out the direction of propagation from the physical point of view? Why $-\beta x$ corresponding to the propagation on positive x direction but not the opposite? | eng_Latn | 18,739 |
Integration of $\sin(2x)$ via two different methods When I integrate $\sin(2x)$ first using the trig substitution of $\sin(2x) = 2 \sin x\cos x$ and then using u-substitution I get $\sin^2(x) +C$. Integrating the same expression using u-substitution directly, I get $-\cos(2x)/2+C$. These answers do not seem to be equivilent. When I let $x=0$ in the first example's answer I get $=0$ but when I let $x=0$ in the seccond answer I get $-1/2$. Clearly I must be doing something wrong. If I had made these integrations definite integrals over $0$ to $\pi/2$ seems like I would have gotten different answsers. Can you assist? | Integral of $\sin x \cos x$ using two methods differs by a constant? $$ \int \sin \theta\cos \theta~d \theta= \int \frac {1} 2 \sin 2\theta~ d \theta=-\frac {1} 4 \cos 2\theta$$ But, if I let $$ u=\sin \theta , \text{ then }du=\cos \theta~d\theta $$ Then $$ \int \sin \theta\cos \theta~d \theta= \int u ~ du =\frac { u^2 } 2 =\frac {1} 2 \sin^2 \theta $$ Since $$ \sin^2 \theta =\frac {1} 2 - \frac {1} 2 \cos 2\theta$$ The above can be written as $$ \int \sin \theta\cos \theta~d \theta= \frac {1} 2 \sin^2 \theta =\frac {1} 2 \left( \frac {1} 2 - \frac {1} 2 \cos 2\theta \right)=\frac {1} 4-\frac {1} 4 \cos 2\theta $$ Why are the two results differ by the constant $1/4$? Thank you. | Notation of the second derivative - Where does the d go? In school I was taught that we use $\frac{du}{dx}$ as a notation for the first derivative of a function $u(x)$. I was also told that we could use the $d$ just like any variable. After some time we were given the notation for the second derivative and it was explained as follows: $$ \frac{d(\frac{du}{dx})}{dx} = \frac{d^2 u}{dx^2} $$ What I do not get here is, if we can use the $d$ as any variable, I would get the following result: $$ \frac{d(\frac{du}{dx})}{dx} =\frac{ddu}{dx\,dx} = \frac{d^2 u}{d^2 x^2} $$ Apparently it is not the same as the notation we were given. A $d$ is missing. I have done some research on this and found some vague comments about "There are reasons for that, but you do not need to know..." or "That is mainly a notation issue, but you do not need to know further." So what I am asking for is: Is this really just a notation thing? If so, does this mean we can actually NOT use d like a variable? If not, where does the $d$ go? I found related question, but it does not really answer my specific question. So I would not see it as a duplicate, but correct me if my search has not been sufficient and there indeed is a similar question out there already. | eng_Latn | 18,740 |
decoupling and integrating differential equations I am having trouble with the process of decoupling. If I have $$\frac{dx}{dt}=-x+y$$ $$\frac{dy}{dt}=-x-y$$ I am trying to figure out how to solve for $x(t)$ and $y(t)$ by decoupling the system so that I only have one variable but I can't seem to get anywhere | Getting equation from differential equations I have: $\dfrac {dx} {dt}$=$-x+y$ $\dfrac {dy}{dt}$=$-x-y$ and I am trying to find $x(t)$ and $y(t)$ given that $x(0)=0$ and $y(0)=1$ I know to do this I need to decouple the equations so that I only have to deal with one variable but the decoupling is what I am having trouble with Do I set them equal to each other and then just move like terms to separate sides getting two different equations and then integrate? | How do "Dummy Variables" work? I do not understand how dummy variables work in math. Suppose we have: $$I_1 = \int_{0}^{\infty} e^{-x^2} dx$$ How is this equivalent to: $$I_2 = \int_{0}^{\infty} e^{-y^2} dy$$ How does this dummy variable system work? Since $y$ is the dependent variable for $I_1$ How can $y$ itself be and independent variable for $I_2$ ?? Thanks! | eng_Latn | 18,741 |
Find CRS (coordinate reference system) based on known location and coordinates? Is there any tool or any tutorial which would help me to figure out CRS (coordinate reference system) from related information? In my particular case I know only the system is in meters, has origin (probably of zone) at E 91°30' and N 50°20' (not 100% sure about exactly 50°20') and it is presumably commonly used in Russia. I would be glad for general solution if any. | Identifying Coordinate System of Shapefile when Unknown? I have a Shapefile but its coordinate system is Unknown, and there is no *.prj file. How can I identify it now? | Kepler problem in time: how do two gravitationally attracted particles move? Two particles with initial positions and velocities $r_1,v_1$ and $r_2,v_2$ are interacting by the inverse square law (with G=1), so that $$ {d^2r_1\over dt^2} = - { m_2(r_1-r_2)\over |r_1-r_2|^3} $$ $$ {d^2r_2\over dt^2} = - { m_1(r_2-r_1)\over |r_1-r_2|^3} $$ (the inverse square law along the line of separation). What is the complete solution of these differential equations? What is the position of the two objects as a function of time? After reading a lot on Wikipedia, I've come to the definition of center of mass and relative coordinates: $$R(t) ~=~ \frac{m_1 r_1 + m_2 r_2}{m_1+m_2}$$ $$\ddot{r}(t) r(t)^2 ~=~ (m_1+m_2)G$$ Where $R$ is the center of mass, and $r$ is the displacement between the particles... Is this correct? How do I proceed to solve the differential equation? | eng_Latn | 18,742 |
How to construct a bijective mapping from a 4D cuboid to a 4D sphere? I thought about the parameterizing the 4D sphere using $$x_1= r\cos(\theta_1), x_2= r\sin(\theta_1)\cos(\theta_2), x_3= r\sin(\theta_1)\sin(\theta_2)\cos(\theta_3),\\ x_4= r\sin(\theta_1)\sin(\theta_2)\sin(\theta_3),$$ where $x_1,x_2,x_3,x_4$ are my 4D space coordinates. But what will be the range of values of $\theta_1, \theta_2, \theta_3$ to make the mapping from $(r,\theta_1, \theta_2, \theta_3)$ to $(x_1,x_2,x_3,x_4)$ bijective ? | Analogue of spherical coordinates in $n$-dimensions What's the analogue to spherical coordinates in $n$-dimensions? For example, for $n=2$ the analogue are polar coordinates $r,\theta$, which are related to the Cartesian coordinates $x_1,x_2$ by $$x_1=r \cos \theta$$ $$x_2=r \sin \theta$$ For $n=3$, the analogue would be the ordinary spherical coordinates $r,\theta ,\varphi$, related to the Cartesian coordinates $x_1,x_2,x_3$ by $$x_1=r \sin \theta \cos \varphi$$ $$x_2=r \sin \theta \sin \varphi$$ $$x_3=r \cos \theta$$ So these are my questions: Is there an analogue, or several, to spherical coordinates in $n$-dimensions for $n>3$? If there are such analogues, what are they and how are they related to the Cartesian coordinates? Thanks. | Ball motion with air resistance coupled differential equation for fourth-order Runge-Kutta I've created a MATLAB function for solving coupled differential equation with the fourth-order Runge-Kutta method based on what is provided (Simultaneous Equations of First Order). Here the function: function [y] = rk4_c(f, g, h, x, y, z, n) % Runge-Kutta % Implementation of the fourth-order method for coupled equations % h = dt % x is the time here for ii=1:(n-1) k1 = h * f(x(ii), y(ii), z(ii)); l1 = h * g(x(ii), y(ii), z(ii)); k2 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1); l2 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1); k3 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2); l3 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2); k4 = h * f(x(ii) + h, y(ii) + k3, z(ii) + l3); l4 = h * g(x(ii) + h, y(ii) + k3, z(ii) + l3); y(ii+1) = y(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4); z(ii+1) = z(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end I would like to use this function for solving motion equations for a ball with air resistance. So starting with the basic equations (I hope they are correct): $$md^2x/dt^2 = F_{D_x}$$ and $$md^2y/dt^2 = F_{D_y}-mg$$ with $$F_{D_x} = -Dvvcos(\theta)$$ and $$F_{D_y} = -Dvvsin(\theta)$$ where $D$ is a drag coefficient. From my understanding, and from what I read, both second-order equation can be converted to two coupled equations that could be used with my coupled Runge-Kutta method function. Can someone help me understand how to convert my equation to a set of coupled equations that could be used with my MATLAB function? EDIT I was able to use my rk4_c function, however it only works for solving the velocities. I'm now trying to see how I could solve for the positions x and y. I'm kind of learning about Runge-Kutta methods at the same time, so sorry for the questions that may sound trivials. % Runge Kutta code to solve projectile motion with quadratic drag % dVx/dt = -(D/m)*vx*sqrt(vx^2+vy^2) % dVy/dt = -(D/m)*vy*sqrt(vx^2+vy^2) - g clc clear all % Constant D = 0.24; % m = 2; % kg g = 9.80665; % m/s^2 % Define function handles fVx = @(t,vx,vy) -(D/m)*vx*sqrt(vx^2+vy^2); fVy = @(t,vx,vy) -(D/m)*vy*sqrt(vx^2+vy^2) - g; % Initial conditions v0 = 200; % m/s theta = 30*pi/180; % rad t(1) = 0; vx(1) = v0*cos(theta); vy(1) = v0*sin(theta); % Step size h = 0.01; % s tFinal = 2; N = ceil(tFinal/h); % RK4 simultaneous coupled loop for ii = 1:N % Update time t(ii+1) = t(ii) + h; % Update vx and vy k1 = h * fVx(t(ii), vx(ii), vy(ii)); l1 = h * fVy(t(ii), vx(ii), vy(ii)); k2 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1); l2 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1); k3 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2); l3 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2); k4 = h * fVx(t(ii) + h, vx(ii) + k3, vy(ii) + l3); l4 = h * fVy(t(ii) + h, vx(ii) + k3, vy(ii) + l3); vx(ii+1) = vx(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4); vy(ii+1) = vy(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end % Plot the solution figure(1) plot(t, vx, t,vy) xlabel('Time (s)') ylabel('Velocities (m/s)') legend('Vx', 'Vy') | eng_Latn | 18,743 |
An ODE with Hyperbolic function as solution Consider the following ODE, $$ x \ddot{x} - \dot{x}^2 - 1 = 0. $$ I can guess that the solution must be $$ x = a \cosh \left(\frac{t - b}{a} \right) $$ How to solve this ODE? | Finding solution of non linear DE $x\ddot{x}-\dot{x}^2=1$ I am looking for help on how to find the solution of the following differential equation, $$x\ddot{x}-\dot{x}^2=1,$$ which comes from solving the Euler-Lagrange equations for the lagrangian $L=x\sqrt{\dot{x}^2+1}$. It is from an exercise in my field theory course. My professor provided me with the solution $x(t)=c_1\cosh{\left(\frac{t-c_2}{c_1}\right)}$, which is a solution when plugged into the equation, but does not give any derivation. Is there a way to solve this DE other than an educated guess? If so how? | $e^{mx}$ in solving second order differential equations In a book I am reading on differential equations, the author writes about the solution to a homogenous, linear, second order differential equation with constant coefficients. The author says something like, "Let us suppose that the solution is of the form $y=e^{mx}$ " . After this the author introduces the characteristic equation of a differential equation of the form mentioned above, and proceeds to describe how to solve it w/ undetermined coefficients, variation of parameters, etc. How did mathematicians first come up with this "assumption" that the solution was of the form $y=e^{mx}$? And, are there any other forms of solutions for these types of equations? | eng_Latn | 18,744 |
A set couple ODE's $$\frac{d^2x}{{\rm dt}^2}=-GM\frac{x}{{(x^2+y^2)}^\frac{3}{2}}$$ $$\frac{d^2y}{{\rm dt}^2}=-GM\frac{y}{{(x^2+y^2)}^\frac{3}{2}}$$ How can I solve those equations, I tried using wolfram alpha but it didn't work. | Solving $\textbf{r}''(t)=\frac{GM}{(r(t))^3}\textbf{r}(t)$ I have the following problem that I don't know how to solve: Let $\textbf{r}(t)$ be a curve and let $r(t):= |\textbf{r}(t)|$. Let the curve $\textbf{r}(t)$ be defined by: $$\textbf{a}(t)=\frac{GM}{(r(t))^3}\textbf{r}(t)$$ Where $\textbf{a}(t) = \textbf{r}''(t)$ is the acceleration of the particle and $G,M \in \mathbb{R}$. Suppose that $\textbf{r}(t)$ and $\textbf{r}'(t)$ are not parallel. The objective is to determine the trajectory $\textbf{r}(t)$ in polar coordinates. So basically we have the following equation: $$\textbf{r}''(t)=\frac{GM}{(r(t))^3}\textbf{r}(t)$$ The thing is that I don't know how I should proceed from now on. If we'd have just normal scalar function this would be a differential equation and I think I would be able to solve it, but like this I have no idea how to approach the problem. How can I solve this? Edit: If $$\textbf{r}(t):=(x(t),y(t))$$ then the equation turns into the following system of equations: $$x''=\frac{GM}{(\sqrt{x^2 + y^2})^3}x$$ $$y''=\frac{GM}{(\sqrt{x^2 + y^2})^3}y$$ I don't know if this helps but I think that this turns thing easier perhaps. | The formal solution of the time-dependent Schrödinger equation Consider the time-dependent (or some equation in Schrödinger form) written down as $$ \tag 1 i \partial_{0} \Psi ~=~ \hat{ H}~ \Psi . $$ Usually, one likes to write that it has a formal solution of the form $$ \tag 2 \Psi (t) ~=~ \exp\left[-i \int \limits_{0}^{t} \hat{ H}(t^{\prime}) ~\mathrm dt^{\prime}\right]\Psi (0). $$ However, this form for the solution of $(1)$ is actually built by the method of successive approximations which actually returns a solution of the form $$ \tag 3 \Psi (t) ~=~ \hat{\mathrm T} \exp\left[-i \int \limits_{0}^{t} \hat{H}(t^{\prime})~\mathrm dt^{\prime}\right]\Psi (0), \qquad t>0, $$ where $\hat{\mathrm T}$ is the time-ordering operator. It seems that $(3)$ doesn't coincide with $(2)$, but formally $(2)$ seems to be perfectly fine: it satisfies $(1)$ and the initial conditions. So where is the mistake? | yue_Hant | 18,745 |
Partial Differentiation without chain rule in Euler Lagrange Equations The Euler-Lagrange equations for a bob attached to a spring are $$ \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial v} = \frac{\partial L}{\partial x} $$ But $v$ is a function of $x$. Is it or is it not. Because normal thinking says that $x$ is a function of $t$ and $v$ is a function of $t$ but it is not necessary that $v$ be a function of $x$. But mathematically \begin{align} x&=f(t) \\ t&=f^{-1}(x) \\ v &= g(t) \\ &=g(f^{-1} (x)).\end{align} So chain rule should be applied in the Euler Lagrange equations. Then why is it not? Is it because on applying the chain rule on $f^{-1}$ and on applying the chain rule on both the sides of the equation we will get the same result. If not then what is the reason for not applying the chain rule? Or is it that $v$ is not a function of $x$ but then $v$ is related to $x$. Where am I going wrong in a such a basic thing is puzzling me a lot. | Calculus of variations -- how does it make sense to vary the position and the velocity independently? In the calculus of variations, particularly Lagrangian mechanics, people often say we vary the position and the velocity independently. But velocity is the derivative of position, so how can you treat them as independent variables? | Ball motion with air resistance coupled differential equation for fourth-order Runge-Kutta I've created a MATLAB function for solving coupled differential equation with the fourth-order Runge-Kutta method based on what is provided (Simultaneous Equations of First Order). Here the function: function [y] = rk4_c(f, g, h, x, y, z, n) % Runge-Kutta % Implementation of the fourth-order method for coupled equations % h = dt % x is the time here for ii=1:(n-1) k1 = h * f(x(ii), y(ii), z(ii)); l1 = h * g(x(ii), y(ii), z(ii)); k2 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1); l2 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1); k3 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2); l3 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2); k4 = h * f(x(ii) + h, y(ii) + k3, z(ii) + l3); l4 = h * g(x(ii) + h, y(ii) + k3, z(ii) + l3); y(ii+1) = y(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4); z(ii+1) = z(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end I would like to use this function for solving motion equations for a ball with air resistance. So starting with the basic equations (I hope they are correct): $$md^2x/dt^2 = F_{D_x}$$ and $$md^2y/dt^2 = F_{D_y}-mg$$ with $$F_{D_x} = -Dvvcos(\theta)$$ and $$F_{D_y} = -Dvvsin(\theta)$$ where $D$ is a drag coefficient. From my understanding, and from what I read, both second-order equation can be converted to two coupled equations that could be used with my coupled Runge-Kutta method function. Can someone help me understand how to convert my equation to a set of coupled equations that could be used with my MATLAB function? EDIT I was able to use my rk4_c function, however it only works for solving the velocities. I'm now trying to see how I could solve for the positions x and y. I'm kind of learning about Runge-Kutta methods at the same time, so sorry for the questions that may sound trivials. % Runge Kutta code to solve projectile motion with quadratic drag % dVx/dt = -(D/m)*vx*sqrt(vx^2+vy^2) % dVy/dt = -(D/m)*vy*sqrt(vx^2+vy^2) - g clc clear all % Constant D = 0.24; % m = 2; % kg g = 9.80665; % m/s^2 % Define function handles fVx = @(t,vx,vy) -(D/m)*vx*sqrt(vx^2+vy^2); fVy = @(t,vx,vy) -(D/m)*vy*sqrt(vx^2+vy^2) - g; % Initial conditions v0 = 200; % m/s theta = 30*pi/180; % rad t(1) = 0; vx(1) = v0*cos(theta); vy(1) = v0*sin(theta); % Step size h = 0.01; % s tFinal = 2; N = ceil(tFinal/h); % RK4 simultaneous coupled loop for ii = 1:N % Update time t(ii+1) = t(ii) + h; % Update vx and vy k1 = h * fVx(t(ii), vx(ii), vy(ii)); l1 = h * fVy(t(ii), vx(ii), vy(ii)); k2 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1); l2 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1); k3 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2); l3 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2); k4 = h * fVx(t(ii) + h, vx(ii) + k3, vy(ii) + l3); l4 = h * fVy(t(ii) + h, vx(ii) + k3, vy(ii) + l3); vx(ii+1) = vx(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4); vy(ii+1) = vy(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end % Plot the solution figure(1) plot(t, vx, t,vy) xlabel('Time (s)') ylabel('Velocities (m/s)') legend('Vx', 'Vy') | eng_Latn | 18,746 |
Lagrange's equation implying Newton's 2nd law? The typical first application of Lagrange's equation is showing that it implies Newton's law for a particle whose Lagrangian is $L=\frac{1}{2}mv^2-V(x)$. Plugging this Lagrangian into Lagrange's equation, we have that $$\frac{\partial}{\partial x}\left(\frac{1}{2}mv^2-V(x)\right)=\frac{d}{dt}\frac{\partial}{\partial v}\left(\frac{1}{2}mv^2-V(x)\right)$$ Typically at this point books will conclude that, therefore, we must have Newton's 2nd law $$-\frac{dV}{dx}=m\frac{dv}{dt}.$$ However, how can we be sure of the following? $$\frac{\partial}{\partial x} \frac{1}{2}mv^2+\frac{d}{dt}\frac{\partial}{\partial v}V=0$$ The above result of Newton's law seems to depend on this, but I'm not sure how to go about showing that the last equation is true. | Calculus of variations -- how does it make sense to vary the position and the velocity independently? In the calculus of variations, particularly Lagrangian mechanics, people often say we vary the position and the velocity independently. But velocity is the derivative of position, so how can you treat them as independent variables? | How is the Schroedinger equation a wave equation? Wave equations take the form: $$\frac{ \partial^2 f} {\partial t^2} = c^2 \nabla ^2f$$ But the Schroedinger equation takes the form: $$i \hbar \frac{ \partial f} {\partial t} = - \frac{\hbar ^2}{2m}\nabla ^2f + U(x) f$$ The partials with respect to time are not the same order. How can Schroedinger's equation be regarded as a wave equation? And why are interference patterns (e.g in the double-slit experiment) so similar for water waves and quantum wavefunctions? | eng_Latn | 18,747 |
3rd order differential equations examples one example of 2nd order diff equation that I can think of is spring mass damper system. $ x''+c/mx'+k/mx = f $ What are examples of 3rd order differential equations? | Are there examples of third-(or higher)-order linear differential equations in physics or applied mathematics? The classical second-order linear ordinary differential equation is that named after Sturm and Liouville: formally, \begin{equation} (pu')'=ru. \end{equation} It arises naturally in many physical situations, for example through radial considerations of Schrödinger's equation. A higher-order "analogue" of it is the Orr$-$Sommerfeld equation, which, after relabelling the coefficients, may be written as \begin{equation} (\phi u'')''=\psi u. \end{equation} It also arises naturally, namely from certain simplifications applied to Navier$-$Stokes' (in)famous equation, and describes to great accuracy the cross-stream behaviour of channel fluid flow. I have been thinking long and hard about whether there are additional higher-order linear differential equations that emerge naturally from our mathematical models of the world. It even seems pretty much all of the linear partial differential equations (heat, wave, Schrödinger, etc.) are of second order. Does anybody know of higher-order examples? | How do I apply differential cryptanalysis to a block cipher? I have read a lot of summaries of block ciphers particularly with regards to the NIST competitions stating that reduced-round block ciphers are – for example – vulnerable to differential cryptanalysis. I have a general idea that the application of differential cryptanalysis is to look at the difference between inputs; makes that fairly clear. However, I could take any two inputs for any given block cipher and I am pretty certain I'd be staring at random differences. I am aware this is the idea of a well written block cipher; however, assuming a broken or vulnerable cipher (feel free to provide simple examples) how do I go about choosing differences to try? Are there any clues in algorithm design that would inform a decision on which values to choose? How does being vulnerable to differential cryptanalysis impact a cipher in the wild? If all I have are differences between known plain-texts and known keys as my analysis and a captured ciphertext as my data to exploit, what can I actually deduce? | eng_Latn | 18,748 |
Hi the question I have to solve is: $u_x + u_y + u = e^{x+2y} \ $ where $ u(x,0) = 0$ First, I tried to solve the question of the form: $$ u_x + u_y + u = 0$$ We know that $\dot x = 1 \ $ and $\dot y = 1 \ $. From that we can introduce s where $$x(s) = s +x_0$$ and $$y(s) = s$$. Then if $z = u(x(s), y(s))$ we have that $\dot z + z = 0$. By setting $s = y$ we have then that $x_0 = y-x$. Then we can get that $u(x,y) = e^{-y} \ g(y-x)$. Now, to get $\dot z + z = e^{x+2y}$ I am completely stuck | It is asked to solve the PDE $$u_x + u_y + u = e^{x+2y}$$ My attempt: We have that $$\frac{dx}{1}=\frac{dy}{1}=\frac{du}{e^{x+2y}-u} $$ $\Rightarrow \frac{dy}{dx}=1 \Rightarrow y=x+C_1 \iff C_1=y-x$ $\Rightarrow \frac{du}{dx}= e^{x+2y}-u \iff \dot{u}+u=e^{x+2y} \iff \\ \frac{d}{dx}(e^x u) = e^{2x+2y} \iff u = e^{-x} ( \frac{1}{2}e^{2x+2y} + C_2)$ As $u$ is constant along its characteristics: $$u(x,y)=\frac{e^{-x}}{2}(e^{2x+2y} + 2C_2(y-x))$$ where $C_2$ is any function of one variable. But wolfram gave the following answer $$u(x,y)=\frac{e^{-x}}{4}(e^{2x+2y} + 4C_2(y-x))$$ as you can check What am I doing wrong? | Towards the end of , how do I get $$ \frac{\sqrt{8z + 1} - 1}{2} < w + 1 $$ Here $w = x + y \geq 0, t = (w^2 + w)/2 , z = t + w, x,y \in \mathbb N_{\geq 0}$. Thanks. | eng_Latn | 18,749 |
How would I show that $$f(x,y) = \frac{2xy}{x^2 + y^2}$$ is not differentiable at the origin? Is it enough to show that as the function tends to the origin along the paths $y = x$ and $y=2x$ that we get different limits and hence the function is not continuous? | $$f(x,y) = \left\{\begin{array}{cc} \frac{xy}{x^2+y^2} & (x,y)\neq(0,0) \\ f(x,y) = 0 & (x,y)=(0,0) \end{array}\right.$$ In order to verify if this function is differentiable, I tried to prove it by the theorem that says that if $\frac{∂f}{∂x}$ and $\frac{∂f}{∂y}$ exist and are continuous at the point $(x_0,y_0)$, then the function is differentiable at this point. So I did: $$\frac{\partial f}{\partial x}(0,0) = \lim_{h\to 0}\frac{f(0+h,0)-f(0,0)}{h} = 0$$ $$\frac{\partial f}{\partial y}(0,0) = \lim_{h\to 0}\frac{f(0,0+h)-f(0,0)}{h} = 0$$ so we have that the partial derivatives at point $(0,0)$ is $0$. Now, if we take the derivative at $(x,y)\neq (0,0)$ and then take the limit of it as $(x,y)\to(0,0)$, we can see if the derivatives are continuous or not. So here it is: $$\frac{\partial f}{\partial x}(x,y) = \frac{y(y^2-x^2)}{(x^2+y^2)}$$ but $$\lim_{(x,y)\to(0,0)} \frac{y(y^2-x^2)}{(x^2+y^2)} $$ does not exist (by wolfram alpha... but can anybody tell me an easy way to prove this limit does not exist? easier than taking the limit in different directions?), therefore the derivative is not continuous at $(0,0)$, so we can't say $f$ is differentiable at $(0,0)$, but for $(x,y)\neq (0,0)$ the function is continuous, as it is a quotient of continuous functions. So $f$ is at least differentiable at $(x,y)\neq (0,0)$. Now, to verify differentiability at $(0,0)$ I think we must use the limit definition of differentiablity: A function is differentiable at $(0,0)$ iff: $$\lim_{(h,k)\to (0,0)} \frac{f(0+h,0+k)-f(0,0)-\frac{\partial f}{\partial x}(0,0)-\frac{\partial f}{\partial y}(0,0)}{\|(h,k)\|} = 0$$ Let's calculate this limit: $$\lim_{(h,k)\to (0,0)} \frac{f(0+h,0+k)-f(0,0)-\frac{\partial f}{\partial x}(0,0)-\frac{\partial f}{\partial y}(0,0)}{\|(h,k)\|} = \\ \lim_{(h,k)\to (0,0)} \frac{\frac{hk}{h^2+k^2}}{\sqrt{h^2+k^2}} = \\ \lim_{(h,k)\to (0,0)} \frac{hk}{(h^2+k^2)\sqrt{h^2+k^2}}$$ which I think, it's a limit that does not exist, therefore the function isn't differentiable at $(0,0)$ | Consider this equation : $$\sqrt{\left( \frac{dy\cdot u\,dt}{L}\right)^2+(dy)^2}=v\,dt,$$ where $t$ varies from $0$ to $T$ , and $y$ varies from $0$ to $L$. Now how to proceed ? This equation arises out of following problem : A cat sitting in a field suddenly sees a standing dog. To save its life, the cat runs away in a straight line with speed $u$. Without any delay, the dog starts with running with constant speed $v>u$ to catch the cat. Initially, $v$ is perpendicular to $u$ and $L$ is the initial separation between the two. If the dog always changes its direction so that it is always heading directly at the cat, find the time the dog takes to catch the cat in terms of $v, u$ and $L$. See my solution below : Let initially dog be at $D$ and cat at $C$ and after time $dt$ they are at $D'$ and $C'$ respectively. Dog velocity is always pointing towards cat. Let $DA = dy, \;AD' = dx$ Let $CC'=udt,\;DD' = vdt$ as interval is very small so $DD'$ can be taken straight line. Also we have $\frac{DA}{DC}= \frac{AD'}{ CC'}$ using triangle property. $\frac{dy}{L}= \frac{dx}{udt}\\ dx = \frac{dy.udt}{L}$ $\sqrt{(dx)^2 + (dy)^2} = DD' = vdt \\ \sqrt{(\frac{dy.udt}{L})^2 + (dy)^2} = vdt $ Here $t$ varies from $0-T$, and $y$ varies from $0-L$. Now how to proceed? | eng_Latn | 18,750 |
I've been asked by a friend to help him solve this equation, but since we couldn't find the right answer, I thought about posting it here. Firstly, I thought about derivating both sides and get: $a=be^{bx}$ and from here we could find the answer pretty quickly, but then I noticed that what I did was incorrect because I can't derivate both sides - the functions aren't equal. | A certain physics problem I have been working on has turned into a math problem. Particularly, I want to find the solutions of some equation of the form $$f(x)=ae^{bx}+cx+d = 0$$ where $a, b, c,$ and $d$ are constant, real numbers that come from the physics problem and $x$ will be a real number. I do not know how to find the solutions of an equation of this form. In the special case where $c=0$, the solution is simply $$x = \frac{\ln(\frac{-d}{a})}{b}$$ but for other values of $c$, I am stumped. If there is no exact form for the roots, is there a relatively simple way to estimate the roots? | I am really satisfied that $(x \to a) ≠(x=a)$ and if that is not right , Then all the process of $Limits$ is dividing by zero and that is a crime. Since $(x \to a) + h = (x=a)$ , $h ≠ 0$,So Why does $f(x \to a) = f(x = a)$ ? NOTE:I am talking about continuous function. | eng_Latn | 18,751 |
For example, given the differential equation $\dfrac{\mathrm{d}y}{\mathrm{d}x}=y\tag*{}$ We do, $\dfrac{\mathrm{d}y}{y}=\mathrm{d}x\tag{*}$ At which point we go and integrate both sides. Is there a rigorous justification for being able to go to $(*)?$ | I see an equation like this: $$y\frac{\textrm{d}y}{\textrm{d}x} = e^x$$ and solve it by "separating variables" like this: $$y\textrm{d}y = e^x\textrm{d}x$$ $$\int y\textrm{d}y = \int e^x\textrm{d}x$$ $$y^2/2 = e^x + c$$ What am I doing when I solve an equation this way? Because $\textrm{d}y/\textrm{d}x$ actually means $$\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$$ they are not really separate entities I can multiply around algebraically. I can check the solution when I'm done this procedure, and I've never run into problems with it. Nonetheless, what is the justification behind it? What I thought of to do in this particular case is write $$\int y \frac{\textrm{d}y}{\textrm{d}x}\textrm{d}x = \int e^x\textrm{d}x$$ $$\int \frac{\textrm{d}}{\textrm{d}x}(y^2/2)\textrm{d}x = e^x + c$$ then by the fundamental theorem of calculus $$y^2/2 = e^x + c$$ Is this correct? Will such a procedure work every time I can find a way to separate variables? | The new Top-Bar does not show reputation changes from Area 51. | eng_Latn | 18,752 |
I'm trying to write a program that finds the salvage value of an item after entering the purchase price, annual depreciation, and years of service. Annual Depreciation= (Purchase Price-Salvage Value)/Years of Service When inputting 5000, 8, 5.25 I should return 4958.00000 | Why is it that scanf() needs the l in "%lf" when reading a double, when printf() can use "%f" regardless of whether its argument is a double or a float? Example code: double d; scanf("%lf", &d); printf("%f", d); | I see an equation like this: $$y\frac{\textrm{d}y}{\textrm{d}x} = e^x$$ and solve it by "separating variables" like this: $$y\textrm{d}y = e^x\textrm{d}x$$ $$\int y\textrm{d}y = \int e^x\textrm{d}x$$ $$y^2/2 = e^x + c$$ What am I doing when I solve an equation this way? Because $\textrm{d}y/\textrm{d}x$ actually means $$\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$$ they are not really separate entities I can multiply around algebraically. I can check the solution when I'm done this procedure, and I've never run into problems with it. Nonetheless, what is the justification behind it? What I thought of to do in this particular case is write $$\int y \frac{\textrm{d}y}{\textrm{d}x}\textrm{d}x = \int e^x\textrm{d}x$$ $$\int \frac{\textrm{d}}{\textrm{d}x}(y^2/2)\textrm{d}x = e^x + c$$ then by the fundamental theorem of calculus $$y^2/2 = e^x + c$$ Is this correct? Will such a procedure work every time I can find a way to separate variables? | eng_Latn | 18,753 |
In most textbooks of physics I've found this demonstration of work-kinetic energy theorem: $$\begin{align} W &= \int_{x_{1}}^{x_{2}} F(x)\ dx \tag{1}\\ &= \int_{x_{1}}^{x_{2}} m\cdot a\ dx \tag{2}\\ &= m\int_{x_{1}}^{x_{2}} \frac{dv}{dt}\ dx \tag{3}\\ &= m\int_{x_{1}}^{x_{2}} \frac{dv}{dx} \cdot \frac{dx}{dt}\ dx \tag{4}\\ &= m\int_{x_{1}}^{x_{2}} \frac{dv}{dx} \cdot v \ dx \tag{5}\\ &= m\int_{v_{1}}^{v_{2}} v \ dv \tag{6}\\ &= \frac{1}{2}mv_{2}^{2} - \frac{1}{2}mv_{1}^{2} \tag{7} \end{align}$$ I don't fully understand how they go from (5) to (6). It seems that they cancel the dx as if they were algebraic elements. I know from calculus that $\int f(g(x))\cdot g'(x)\ dx = F(g(x))$, where $f(x)=F'(x)$ because we are applying the chain rule reversed. And in order to do so, Leibniz's notation can help us. If we define $u=g(x)$ and the "differential" $du=g'(x)dx$ the former integral turns in $\int f(u)\ du = F(u)$. But du and dx doesn't exist, they are just a notation we use to realize more easily that we can find the antiderivatives applying the chain rule reversed. So, what are the true mathematical operation they are doing between (5) and (6)? | Referring to unidimensional motion, it is obvious that it doesn't always make sense to write the speed as a function of position. Seems to me that this is a necessary condition to derive formulas like: $$v^2=v_0 ^2 +2\int_{x_0}^{x}a\cdot dx$$ In fact, in the first step of the demonstration (the one I saw, but I think that this step is crucial) it's required to write $a=dv/dt=(dv/dx)(dx/dt)$, that doesn't make sense if $v$ isn't a function of $x$. When can one rigorously write $v=v(x)$? | $ \int_{-2}^{0} \frac{x}{\sqrt{e^x+(x+2)^2}} dx $ = -$ 2.887270... $ The function has no antiderivate and there's no symmetry here to help, if you solve this I would be thankful if you would for a highschooler ^^ | eng_Latn | 18,754 |
How could I solve legendre differential equation by power series method? | This is a very standard problem. Since it is impossibe to type this in text, I provide the following link to the solution from the Math World Web site:\nhttp://mathworld.wolfram.com/LegendreDifferentialEquation.html | if ur using the windows calculator, do \n25000/20000 -> ln -> / .08\nthat should give the answer | eng_Latn | 18,755 |
What is the name of the general method for solving first-order systems of differential equations of the form:? | from first equation put the value of b in second and then form a second degree differential equation in "a"...good luck! | Click ON\nClick PRGM\nSelect NEW tab and hit ENTER\n\nTo select DISP and PROMPT:\n Hit PRGM when creating or editing a program,\n Select Tab I/O and you will see you selection\n\nTo get the arrow, click STO-> just to the left of 1\n\n\nThis is the Program: (or the one I use)\n\n:Disp "AX^2+BX+C"\n\n:Prompt A\n:Prompt B\n:Prompt C\n:(-B+*square root*(B^2-4AC))/(2A)->D\n:(-B-*square root*(B^2-4AC))/(2A)->E\n:Disp D\n:Disp E\n\nTo run program, go to the main screen, click PRGM, select your program, hit ENTER, input values for A, B, and C, hit ENTER and you result will show your two possible answers. | eng_Latn | 18,756 |
How would you solve: 3^(x) = 2^(x-1) ??? | 3^x=2^x/2\n\n(1.5)^x=0.5\n\nx=-1.71 | dP/P(a-bP)=dt\n\nusing partial fraction to integrate the left side of the eq.\n\nSince 1/P(a-bP)=(1/a)/P + (b/a)/(a-bP)..then\n\ndP/[(1/aP)+(b/a)(1/(a-bP))]=dt ...... Integrate \n\n(1/a)Ln(P)-(1/a)Ln(a-bP)=t+c ...... c is a constant\n\n(1/a) Ln[P/(a-bP)]= t+c\nOR\nLn[P/(a-bP)]= at +c1 ....... c1=a*c\nOR\nP/(a-bP) =exp[at+c1]\nP/(a-bP) =exp(at)*exp(c1) ...... let exp(c1)=c2\nP/(a-bP) = c2 exp(at)\n\nBut P=f(t) then\n\n1/[c2 exp(at)] = (a-bP)/P\n-->\n(1/c2) exp(-at)=(a/P)-b .... let 1/c2 = c3\n-->\nc3 exp(-at)+b= a/P\n-->\nP(t)= a/[c3 exp(-at)+b]\nOR\n a\nP(t)= --------------- ..... where c3 is a constant\n c3 exp(-at) + b | eng_Latn | 18,757 |
How to extract expanded equations that use commands? | LaTeX macro expander | Unable to solve equation for a variable by any method | eng_Latn | 18,758 |
Why the method of separation of variables works? | Why separation of variables works in PDEs? | Unable to solve equation for a variable by any method | eng_Latn | 18,759 |
Help with complicated system of coupled differential equations with a parameter | DSolve gets stuck on system of differential equations with unassigned variable | A fiber bundle over Euclidean space is trivial. | eng_Latn | 18,760 |
Runge Kutta 4th order for system coupled equations-how to do? | Solving a system of ODEs with the Runge-Kutta method | A fiber bundle over Euclidean space is trivial. | eng_Latn | 18,761 |
Assign variable to solve | Assign the results from a Solve to variable(s) | Unable to solve equation for a variable by any method | eng_Latn | 18,762 |
Can we differentiate a differential equation to determine its order and degree? | Order and degree of a differential equation | Order and degree of a differential equation | eng_Latn | 18,763 |
Is there any connection between theta series and physics? I have heared that there is connection between number theory and physics. I specially ask that: Can anyone give a concrete example which shows this connection? I want to see an example which is related to (integral) quadratic forms or theta series. Let $f$ to be a quadratic form; i.e a homogeneous polynomial of degree two (in any number of variables), for example $4x_1^2 + 3x_1x_2 + 5x_1x_3 - 8x_3^2$. Also let's assume that all of the co-efficeints of $f$ are integral. Now let's fix a (integral) quadratic form $f$ in $m$ variables, and let $n$ varies in $\mathbb{N}$; let's define: $$ N(n) := N(f,n) = N\big(f(x_1, ..., x_m),n\big) = \# \{(x_1, ..., x_m) \in \mathbb{Z}^m : f(x_1, ..., x_m)=n \} ; $$ note that $N(n)$ is the number of integral solutions to the equation $f=n$; now by a associated to the quadratic form $f$ we mean the following: $$ \Theta(q) := \Theta_f(q) = \sum_{n\in \mathbb{N}_0}N(n)q^n =\sum_{(x_1, ..., x_m) \in \mathbb{Z}^m}q^{f(x_1, ..., x_m)} ; $$ also if we assume that $q=e^{2\pi i z}$; then there is theorem which states that $\Theta_f(q)$ is a modular form in $z$. I have asked my question again. | Number theory in Physics As a Graduate Mathematics student, my interest lies in Number theory. I am curious to know if Number theory has any connections or applications to physics. I have never even heard of any applications of Number theory to physics. I have heard Applications of linear algebra and analysis to many branches of physics, but not number theory. Waiting forward in receiving interesting answers! | Ball motion with air resistance coupled differential equation for fourth-order Runge-Kutta I've created a MATLAB function for solving coupled differential equation with the fourth-order Runge-Kutta method based on what is provided (Simultaneous Equations of First Order). Here the function: function [y] = rk4_c(f, g, h, x, y, z, n) % Runge-Kutta % Implementation of the fourth-order method for coupled equations % h = dt % x is the time here for ii=1:(n-1) k1 = h * f(x(ii), y(ii), z(ii)); l1 = h * g(x(ii), y(ii), z(ii)); k2 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1); l2 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1); k3 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2); l3 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2); k4 = h * f(x(ii) + h, y(ii) + k3, z(ii) + l3); l4 = h * g(x(ii) + h, y(ii) + k3, z(ii) + l3); y(ii+1) = y(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4); z(ii+1) = z(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end I would like to use this function for solving motion equations for a ball with air resistance. So starting with the basic equations (I hope they are correct): $$md^2x/dt^2 = F_{D_x}$$ and $$md^2y/dt^2 = F_{D_y}-mg$$ with $$F_{D_x} = -Dvvcos(\theta)$$ and $$F_{D_y} = -Dvvsin(\theta)$$ where $D$ is a drag coefficient. From my understanding, and from what I read, both second-order equation can be converted to two coupled equations that could be used with my coupled Runge-Kutta method function. Can someone help me understand how to convert my equation to a set of coupled equations that could be used with my MATLAB function? EDIT I was able to use my rk4_c function, however it only works for solving the velocities. I'm now trying to see how I could solve for the positions x and y. I'm kind of learning about Runge-Kutta methods at the same time, so sorry for the questions that may sound trivials. % Runge Kutta code to solve projectile motion with quadratic drag % dVx/dt = -(D/m)*vx*sqrt(vx^2+vy^2) % dVy/dt = -(D/m)*vy*sqrt(vx^2+vy^2) - g clc clear all % Constant D = 0.24; % m = 2; % kg g = 9.80665; % m/s^2 % Define function handles fVx = @(t,vx,vy) -(D/m)*vx*sqrt(vx^2+vy^2); fVy = @(t,vx,vy) -(D/m)*vy*sqrt(vx^2+vy^2) - g; % Initial conditions v0 = 200; % m/s theta = 30*pi/180; % rad t(1) = 0; vx(1) = v0*cos(theta); vy(1) = v0*sin(theta); % Step size h = 0.01; % s tFinal = 2; N = ceil(tFinal/h); % RK4 simultaneous coupled loop for ii = 1:N % Update time t(ii+1) = t(ii) + h; % Update vx and vy k1 = h * fVx(t(ii), vx(ii), vy(ii)); l1 = h * fVy(t(ii), vx(ii), vy(ii)); k2 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1); l2 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1); k3 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2); l3 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2); k4 = h * fVx(t(ii) + h, vx(ii) + k3, vy(ii) + l3); l4 = h * fVy(t(ii) + h, vx(ii) + k3, vy(ii) + l3); vx(ii+1) = vx(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4); vy(ii+1) = vy(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end % Plot the solution figure(1) plot(t, vx, t,vy) xlabel('Time (s)') ylabel('Velocities (m/s)') legend('Vx', 'Vy') | eng_Latn | 18,764 |
Help me with differential equation of harmonic motion. I asked this question before, but I don't quite understand it. So the question is A particle undergoes simple harmonic motion. Initially its displacement is 1 , velocity 1 and acceleration is -12. and find out displacement and acceleration when the velocity is square root of 8. Someone says that harmonic equation takes the form y(t)=Acos(Wt+@) But, I can not solve this equation by taking derivative and putting in the values// Someone help me plz | Differential equation for Harmonic Motion Particle undergoes simple harmonic motion. Initially Its displacement is $1$, velocity $1$ and acceleration is $-12$ Compute displacement and acceleration when the velocity is square root of $8$. I am not familiar with this kind of application question. So, its displacement is $1$, does that mean that $y=1$ when $x=0$ ? | How to determine the direction of a wave propagation? In the textbook, it said a wave in the form $y(x, t) = A\cos(\omega t + \beta x + \varphi)$ propagates along negative $x$ direction and $y(x, t) = A\cos(\omega t - \beta x + \varphi)$ propagates along positive $x$ direction. This statement looks really confusing because when it says the wave is propagating along $\pm$ x direction, to my understanding, we can drop the time term and ignore the initial phase $\varphi$ while analyzing the direction, i.e. $y(x, 0) = A\cos(\pm\beta x)$, however, because of the symmetry of the cosine function, $\cos(\beta x)\equiv \cos(-\beta x)$, so how can we determine the direction of propagation from that? I know my reasoning must be incorrect but I don't know how to determine the direction. So if we don't go over the math, how to figure out the direction of propagation from the physical point of view? Why $-\beta x$ corresponding to the propagation on positive x direction but not the opposite? | eng_Latn | 18,765 |
A complex ordinary differential equation I’m trying to see what instruments I could use to analyse the following ODE in the complex plane: $ \dot{z} = \exp(it)\cdot \bar{z}$. Where $z$ is a function of real valued time. ${{}}$ | Complex differential equation How do I solve $$x'=e^{it}\overline{x}?$$ This is a complex differential equation, but I don't see how to solve it. Edit: the original ODE is given by $$(x', y')=\begin{pmatrix}\cos t& \sin t\\ \sin t&-\cos t\end{pmatrix}(x,y)$$ I want to show that solutions of this ODE do not remain bounded for all $t$, and the idea was to solve the complex equation. | $\int_0^{\pi/2}\ln(\sin(x))$? From this paper: Shouldn't $du$ be $dt$? And also how do you get from that line to the final result if $du$ is not $dt$? | eng_Latn | 18,766 |
Calculating the inverse for a Rijndael s-box I want to calculate a Rijndael S-box but something went wrong. How can I calculate the inverse? In the following is my approach: given: $f(x)=x^8+x^4+x^3+x+1 \rightarrow$ binary: 100011011 0x12 = 10010 calculation: ggT(100011011,10010): (x^8+x^4+x^3+x+1)/(x^4+x)=x^4+x+1 -(x^8+x^5) __________ x^5+x^4+x^3+x+1 -(x^5+x^2) __________ x^4+x^3+x^2+x+1 -(x^4+x) __________ x^3+x^2+1 => 1101=100011011+10010*10011 ggT(10010,1101): (x^4+x)/(x^3+x^2+1)=x+1 -(x^4+x^3+x) __________ x^3 -(x^3+x^2+1) ____________ x^2+1 => 101=10010+1101*11 ggT(1101,101): (x^3+x^2+1)/(x^2+1)=x+1 -(x^3+x) __________ x^2+x+1 -(x^2+1) _________ x => 10=1101+101*11 ggT(101,10): (x^2+1)/(x)=x -(x^2) __________ 1 => 1=101+10*10 1 = 101+10*10 = (10010+1101*11)+10(1101+101*11) = (10010+11(100011011+10010*10011))+10((100011011+10010*10011)+11(10010+1101*11))) = (10010+11(100011011+10010*10011))+10((100011011+10010*10011)+11(10010+(100011011+10010*10011)*11))) a = 100011011 b = 10010 = (b+11(a+10011b))+10((a+10011b)+11(b+11(a+10011b))) = b+11a+11*10011b+10(a+10011b+11(b+11a+11*10011b)) = b+11a+11*10011b+10(a+10011b+11b+11*11a+11*11*10011b) = b+11a+11*10011b+10a+10*10011b+10*11+10*11*11a+10*11*1110011b = b(1+11*10011+10*10011+10*11+10*11*11*10011) = b(110111100) -> 9 bit but i need 8 bit | Multiplicative inverse in $\operatorname{GF}(2^8)$? I know how to do multiplication over ${\rm GF}(2^8)$: uint8_t gmul(uint8_t a, uint8_t b) { uint8_t p=0; uint8_t carry; int i; for(i=0;i<8;i++) { if(b & 1) p ^=a; carry = a & 0x80; a = a<<1; if(carry) a^=0x1b; b = b>>1; } return p; } So, I tried to create a ${\rm GF}(2^8)$ multiplication table using this code. I've given below the values in the 3rd row of the table, but I don't think they're correct: 0 1 2 3 4 5 6 7 8 9 A B C D E F 0 1 2 0 2 4 6 8 A C E 10 12 14 16 18 1A 1C 1E 3 . . E F I don't know what went wrong. I built the table by multiplying the values in the first row with those in the first column. E.g. in the third row, I multiplied 2 × 0, 2 × 1, …, 2 × E, 2 × F. How can I create a multiplication table for arithmetic in ${\rm GF}(2^8)$? Also, how can I find the multiplicative inverse of a number in ${\rm GF}(2^8)$? For example, how can I determine that the inverse of 95 is 8A? I tried to do this using the multiplication table above, but when I took 9th row and the 5th column in the multiplication table I got 2D, not 8A. | Ball motion with air resistance coupled differential equation for fourth-order Runge-Kutta I've created a MATLAB function for solving coupled differential equation with the fourth-order Runge-Kutta method based on what is provided (Simultaneous Equations of First Order). Here the function: function [y] = rk4_c(f, g, h, x, y, z, n) % Runge-Kutta % Implementation of the fourth-order method for coupled equations % h = dt % x is the time here for ii=1:(n-1) k1 = h * f(x(ii), y(ii), z(ii)); l1 = h * g(x(ii), y(ii), z(ii)); k2 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1); l2 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1); k3 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2); l3 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2); k4 = h * f(x(ii) + h, y(ii) + k3, z(ii) + l3); l4 = h * g(x(ii) + h, y(ii) + k3, z(ii) + l3); y(ii+1) = y(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4); z(ii+1) = z(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end I would like to use this function for solving motion equations for a ball with air resistance. So starting with the basic equations (I hope they are correct): $$md^2x/dt^2 = F_{D_x}$$ and $$md^2y/dt^2 = F_{D_y}-mg$$ with $$F_{D_x} = -Dvvcos(\theta)$$ and $$F_{D_y} = -Dvvsin(\theta)$$ where $D$ is a drag coefficient. From my understanding, and from what I read, both second-order equation can be converted to two coupled equations that could be used with my coupled Runge-Kutta method function. Can someone help me understand how to convert my equation to a set of coupled equations that could be used with my MATLAB function? EDIT I was able to use my rk4_c function, however it only works for solving the velocities. I'm now trying to see how I could solve for the positions x and y. I'm kind of learning about Runge-Kutta methods at the same time, so sorry for the questions that may sound trivials. % Runge Kutta code to solve projectile motion with quadratic drag % dVx/dt = -(D/m)*vx*sqrt(vx^2+vy^2) % dVy/dt = -(D/m)*vy*sqrt(vx^2+vy^2) - g clc clear all % Constant D = 0.24; % m = 2; % kg g = 9.80665; % m/s^2 % Define function handles fVx = @(t,vx,vy) -(D/m)*vx*sqrt(vx^2+vy^2); fVy = @(t,vx,vy) -(D/m)*vy*sqrt(vx^2+vy^2) - g; % Initial conditions v0 = 200; % m/s theta = 30*pi/180; % rad t(1) = 0; vx(1) = v0*cos(theta); vy(1) = v0*sin(theta); % Step size h = 0.01; % s tFinal = 2; N = ceil(tFinal/h); % RK4 simultaneous coupled loop for ii = 1:N % Update time t(ii+1) = t(ii) + h; % Update vx and vy k1 = h * fVx(t(ii), vx(ii), vy(ii)); l1 = h * fVy(t(ii), vx(ii), vy(ii)); k2 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1); l2 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1); k3 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2); l3 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2); k4 = h * fVx(t(ii) + h, vx(ii) + k3, vy(ii) + l3); l4 = h * fVy(t(ii) + h, vx(ii) + k3, vy(ii) + l3); vx(ii+1) = vx(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4); vy(ii+1) = vy(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end % Plot the solution figure(1) plot(t, vx, t,vy) xlabel('Time (s)') ylabel('Velocities (m/s)') legend('Vx', 'Vy') | yue_Hant | 18,767 |
Solve $2tx'(t)-x(t)=\ln x'(t)$ Solve $2tx'(t)-x(t)=\ln \left[x'(t)\right]$ That would be an easy Clairaut's equation if $tx'(t)$ wasn't multiplied by $2$. But unfortunately it is, and I have no idea what to do here. | Second-order non-linear ODE $2tx'-x=lnx'$ I differentiated both sides with respect to x: $x'+2tx''=\frac {x''}{x'}$ Substituting $p=x'$, $p+2tp'=\frac{p'}{p}$ But I have no clue what can I do from here on. EDIT: $t$ is the non-dependent variable. | Kepler problem in time: how do two gravitationally attracted particles move? Two particles with initial positions and velocities $r_1,v_1$ and $r_2,v_2$ are interacting by the inverse square law (with G=1), so that $$ {d^2r_1\over dt^2} = - { m_2(r_1-r_2)\over |r_1-r_2|^3} $$ $$ {d^2r_2\over dt^2} = - { m_1(r_2-r_1)\over |r_1-r_2|^3} $$ (the inverse square law along the line of separation). What is the complete solution of these differential equations? What is the position of the two objects as a function of time? After reading a lot on Wikipedia, I've come to the definition of center of mass and relative coordinates: $$R(t) ~=~ \frac{m_1 r_1 + m_2 r_2}{m_1+m_2}$$ $$\ddot{r}(t) r(t)^2 ~=~ (m_1+m_2)G$$ Where $R$ is the center of mass, and $r$ is the displacement between the particles... Is this correct? How do I proceed to solve the differential equation? | eng_Latn | 18,768 |
Null Reference Exception in Switch Case I'm getting an annoying NullReferenceException error. Can anyone help? This is the function, and below you can see how it is called: Error message: NullReferenceException: Object reference not set to an instance of an object BotController.Start () (at Assets/Scripts/BotController.cs:64) void Start() { InvokeRepeating("RandomMove", 0f, waitTime); } function: void RandomMove() { float behaviour = Random.Range(0,3); switch(behaviour) { //0: move left, 1: stop, 2: move right // case 0: rb.velocity = new Vector2(-moveSpeed, rb.velocity.y); movingRight = false; break; case 1: rb.velocity = new Vector2(0f, 0f); break; case 2: rb.velocity = new Vector2(moveSpeed, rb.velocity.y); movingRight = true; break; } return; } the console shows the error on this line: case 0: rb.velocity = new Vector2(-moveSpeed, rb.velocity.y); Edit: I've just changed the function to this: void RandomMove() { float behaviour = Random.Range(0,3); switch(behaviour) { //0: move left, 1: stop, 2: move right // case 0: movingRight = false; break; case 1: rb.velocity = new Vector2(0f, 0f); break; case 2: movingRight = true; break; } return; } and I've put an initial speed inside the Start() function like this: rb.velocity = new Vector2(moveSpeed, rb.velocity.y); Now it shows the error exactly on this line. I think something's up with either Rigidbody or the vectors. I'm using the same rigidbody inside the same script with no issues. | NullReferenceException in Unity Since many users are facing the NullReferenceException: Object reference not set to an instance of an object error in Unity, I thought that it would be a good idea to gather from multiple source some explanation and ways to fix this error. Symptoms I am getting the error below appearing in my console, what does it mean and how do I fix it? NullReferenceException: Object reference not set to an instance of an object | Ball motion with air resistance coupled differential equation for fourth-order Runge-Kutta I've created a MATLAB function for solving coupled differential equation with the fourth-order Runge-Kutta method based on what is provided (Simultaneous Equations of First Order). Here the function: function [y] = rk4_c(f, g, h, x, y, z, n) % Runge-Kutta % Implementation of the fourth-order method for coupled equations % h = dt % x is the time here for ii=1:(n-1) k1 = h * f(x(ii), y(ii), z(ii)); l1 = h * g(x(ii), y(ii), z(ii)); k2 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1); l2 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1); k3 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2); l3 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2); k4 = h * f(x(ii) + h, y(ii) + k3, z(ii) + l3); l4 = h * g(x(ii) + h, y(ii) + k3, z(ii) + l3); y(ii+1) = y(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4); z(ii+1) = z(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end I would like to use this function for solving motion equations for a ball with air resistance. So starting with the basic equations (I hope they are correct): $$md^2x/dt^2 = F_{D_x}$$ and $$md^2y/dt^2 = F_{D_y}-mg$$ with $$F_{D_x} = -Dvvcos(\theta)$$ and $$F_{D_y} = -Dvvsin(\theta)$$ where $D$ is a drag coefficient. From my understanding, and from what I read, both second-order equation can be converted to two coupled equations that could be used with my coupled Runge-Kutta method function. Can someone help me understand how to convert my equation to a set of coupled equations that could be used with my MATLAB function? EDIT I was able to use my rk4_c function, however it only works for solving the velocities. I'm now trying to see how I could solve for the positions x and y. I'm kind of learning about Runge-Kutta methods at the same time, so sorry for the questions that may sound trivials. % Runge Kutta code to solve projectile motion with quadratic drag % dVx/dt = -(D/m)*vx*sqrt(vx^2+vy^2) % dVy/dt = -(D/m)*vy*sqrt(vx^2+vy^2) - g clc clear all % Constant D = 0.24; % m = 2; % kg g = 9.80665; % m/s^2 % Define function handles fVx = @(t,vx,vy) -(D/m)*vx*sqrt(vx^2+vy^2); fVy = @(t,vx,vy) -(D/m)*vy*sqrt(vx^2+vy^2) - g; % Initial conditions v0 = 200; % m/s theta = 30*pi/180; % rad t(1) = 0; vx(1) = v0*cos(theta); vy(1) = v0*sin(theta); % Step size h = 0.01; % s tFinal = 2; N = ceil(tFinal/h); % RK4 simultaneous coupled loop for ii = 1:N % Update time t(ii+1) = t(ii) + h; % Update vx and vy k1 = h * fVx(t(ii), vx(ii), vy(ii)); l1 = h * fVy(t(ii), vx(ii), vy(ii)); k2 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1); l2 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1); k3 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2); l3 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2); k4 = h * fVx(t(ii) + h, vx(ii) + k3, vy(ii) + l3); l4 = h * fVy(t(ii) + h, vx(ii) + k3, vy(ii) + l3); vx(ii+1) = vx(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4); vy(ii+1) = vy(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end % Plot the solution figure(1) plot(t, vx, t,vy) xlabel('Time (s)') ylabel('Velocities (m/s)') legend('Vx', 'Vy') | eng_Latn | 18,769 |
local variable initialized in delphi? During a review process of an older programm code the following question arised: All local variables in a method are initialized right after begin. Usually local variables are not initialized. But we have a procedure where all variables are initialized to 0. Does anybody has an idea how this could happen? Example: type TPrices = array[0..10, 0..5] of Integer; procedure DoSomething(); var mPrices : TPrices; mValue : Integer; begin if (mPrices[0,0] = 0) then MessageDlg('Zero', mtInformation, [mbOK], 0); if (mValue = 0) then MessageDlg('Zero Integer', mtInformation, [mbOK], 0); end; | Are delphi variables initialized with a value by default? I'm new to Delphi, and I've been running some tests to see what object variables and stack variables are initialized to by default: TInstanceVariables = class fBoolean: boolean; // always starts off as false fInteger: integer; // always starts off as zero fObject: TObject; // always starts off as nil end; This is the behaviour I'm used to from other languages, but I'm wondering if it's safe to rely on it in Delphi? For example, I'm wondering if it might depend on a compiler setting, or perhaps work differently on different machines. Is it normal to rely on default initialized values for objects, or do you explicitly set all instance variables in the constructor? As for stack (procedure-level) variables, my tests are showing that unitialized booleans are true, unitialized integers are 2129993264, and uninialized objects are just invalid pointers (i.e. not nil). I'm guessing the norm is to always set procedure-level variables before accessing them? | Specific system of differential equations I have the following system of equations: \begin{eqnarray}\frac{dx}{dt} = x(1 - x^2 - y^2) \\ \frac{dy}{dt} = y(4 - x^2 - y^2) \end{eqnarray} I want to prove that if a solutions starts (at time $t = 0$) in the region $$ G = \{(x, y) \in \mathbb{R}^2 : x > 0, y > 0 \text{ and } 1 < x^2 + y^2 < 4\} $$ that it remains there for all $t \geq 0$. I have already proven that, from $G$, a solution cannot cross the $x$-axis where $0 < x < 2$ or the $y$-axis where $0 < y < 2$ (since those line segments are orbits), but the pieces of the circles with radius 1 and 2 (where $x > 0, y > 0$) are the lines I haven't done yet. In this question we can focus on just the circle with radius 1 since the proof for the other circle is completely analogous to it. In my book, there is a similar problem where they show that if a solution crosses a certain line, that solution attains its maximum or minimum value for $x$ and $y$, and then they derive a contradiction. However, in this case, for $n \geq 1$ the $n$-th derivative of $x$ equals 0 on the circle with radius 1. (If the second derivative were negative, $x$ would attain its maximum on the circle, a contradiction since $\frac{dx}{dt} < 0$ in $G$, so $x$ is decreasing.) I don't see why a solution can't cross the circle at all. On the circle, $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} > 0$, so after crossing the circle, the solution would go back up to $G$. Is the circle itself an orbit? EDIT I think I have found the answer. Using polar coordinates, I found $$ \frac{dr}{dt} = r - r^3 + 3r \sin^2(\theta) $$ Now, $$ \frac{dr}{dt} > 0 \Leftrightarrow 0 < r - r^3 + 3r \sin^2(\theta) \leq r - r^3 + 3r = 4r - r^3 $$ since $0 \leq \sin^2(\theta) \leq 1$. Note that $0 < r \leq 1$ satisfies $4r - r^3 > 0$, and thus $\frac{dr}{dt} > 0$. Therefore, if a solution goes from $G$ to the edge of the unit circle, $r$ must, on the edge, either be decreasing or constant (i.e., its derivative is less than or equal to 0), contradicting the fact that the derivative is positive. Can someone tell me if this is correct? | eng_Latn | 18,770 |
How do you apply the 4th order Runge Kutta method of numerical integration to the problem of projectile motion through atmosphere? How do you apply the 4th order Runge Kutta method of numerical integration to the problem of projectile motion through atmosphere? | Ball motion with air resistance coupled differential equation for fourth-order Runge-Kutta I've created a MATLAB function for solving coupled differential equation with the fourth-order Runge-Kutta method based on what is provided (Simultaneous Equations of First Order). Here the function: function [y] = rk4_c(f, g, h, x, y, z, n) % Runge-Kutta % Implementation of the fourth-order method for coupled equations % h = dt % x is the time here for ii=1:(n-1) k1 = h * f(x(ii), y(ii), z(ii)); l1 = h * g(x(ii), y(ii), z(ii)); k2 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1); l2 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1); k3 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2); l3 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2); k4 = h * f(x(ii) + h, y(ii) + k3, z(ii) + l3); l4 = h * g(x(ii) + h, y(ii) + k3, z(ii) + l3); y(ii+1) = y(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4); z(ii+1) = z(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end I would like to use this function for solving motion equations for a ball with air resistance. So starting with the basic equations (I hope they are correct): $$md^2x/dt^2 = F_{D_x}$$ and $$md^2y/dt^2 = F_{D_y}-mg$$ with $$F_{D_x} = -Dvvcos(\theta)$$ and $$F_{D_y} = -Dvvsin(\theta)$$ where $D$ is a drag coefficient. From my understanding, and from what I read, both second-order equation can be converted to two coupled equations that could be used with my coupled Runge-Kutta method function. Can someone help me understand how to convert my equation to a set of coupled equations that could be used with my MATLAB function? EDIT I was able to use my rk4_c function, however it only works for solving the velocities. I'm now trying to see how I could solve for the positions x and y. I'm kind of learning about Runge-Kutta methods at the same time, so sorry for the questions that may sound trivials. % Runge Kutta code to solve projectile motion with quadratic drag % dVx/dt = -(D/m)*vx*sqrt(vx^2+vy^2) % dVy/dt = -(D/m)*vy*sqrt(vx^2+vy^2) - g clc clear all % Constant D = 0.24; % m = 2; % kg g = 9.80665; % m/s^2 % Define function handles fVx = @(t,vx,vy) -(D/m)*vx*sqrt(vx^2+vy^2); fVy = @(t,vx,vy) -(D/m)*vy*sqrt(vx^2+vy^2) - g; % Initial conditions v0 = 200; % m/s theta = 30*pi/180; % rad t(1) = 0; vx(1) = v0*cos(theta); vy(1) = v0*sin(theta); % Step size h = 0.01; % s tFinal = 2; N = ceil(tFinal/h); % RK4 simultaneous coupled loop for ii = 1:N % Update time t(ii+1) = t(ii) + h; % Update vx and vy k1 = h * fVx(t(ii), vx(ii), vy(ii)); l1 = h * fVy(t(ii), vx(ii), vy(ii)); k2 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1); l2 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1); k3 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2); l3 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2); k4 = h * fVx(t(ii) + h, vx(ii) + k3, vy(ii) + l3); l4 = h * fVy(t(ii) + h, vx(ii) + k3, vy(ii) + l3); vx(ii+1) = vx(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4); vy(ii+1) = vy(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end % Plot the solution figure(1) plot(t, vx, t,vy) xlabel('Time (s)') ylabel('Velocities (m/s)') legend('Vx', 'Vy') | Mathematical physics text with plenty of applications I'm looking for texts on mathematical physics. I've seen various other threads, but the texts recommended in those threads were mathematical methods of theoretical physics texts, that is to say those texts focused on the math methods useful for physicists and assumed the applications of those methods would be covered in some other text. What I'm looking for is a book which introduces some math and then looks at its applications (back and forth) rather than a text which just goes over some math. (Example of books which are NOT what I'm looking for: Mathematics for physicists by Dennery, Math methods for physicists by Weber and Arfken etc) I've taken courses in multi-variable calc, linear algebra, real analysis, and I'm currently taking a course in abstract algebra. But so far I've only seen applications of multi-variable to physics. I'm looking for books which look at the applications of the other subjects mentioned above to physics, of course books which cover subjects beyond those mentioned are also welcome. So far I'm liking the flavor of the text "A Course in Modern Mathematical Physics" by Szekeres, but I'd prefer a text which got to the physics side of things more quickly (in the Szekeres text the first eight chapters are strictly on pure math). After some amazon searching it looks like Theoretical Physics by Joos, Mathematical Physics by Henzel, and Physical Mathematics by Cahill may be good bets? | eng_Latn | 18,771 |
Tracking the path of two point masses through time under the influence of gravity and with no initial velocities Given the point mass $p_1$ at $(0,0)$ with mass $m_1$ and the point mass at $p_2$ at $(r,0)$ with mass $m_2$ how would you find the positions of $p_1$ or $p_2$ at any time. My first thought was to first solve the problem of what their positions were after some small period of time $\Delta t$. The force by gravity on $p_1$ is $F_1=\frac{Gm_1m_2}{r^2}$ and from the equation $F=ma$ substituting $\frac{v}{t}$ in for $a$ and the $\frac{d}{t}$ for $v$ the result should be $F=\frac{md}{t^2}$ we know the force on $p_1$ and its mass, as well as the time so substituting them results in $\frac{Gm_1m_2}{r^2}=\frac{m_1 d}{\Delta t^2}$ some quick algabra tells me that $p_1$ should have move a distance of $d_1=\frac{Gm_2\Delta t^2}{r^2}$. Taking similar steps brings me to find that $p_2$ travels a distance of $d_2=-\frac{Gm_1\Delta t^2}{r^2}$. From there the same process could be repeated to find the position of of $p_1$ and $p_2$ at the time $2\Delta t$ except by using $r-\frac{G(m_1+m_2)\Delta t^2}{r^2}$ for the new distance between the points. In general if $R(t)$ is the distance between $p_1$ and $p_2$ and if $D_1(t)$ is the distance that $p_1$ moves over the time $\Delta t$ at time the time $t$ then $D_1(t)=\frac{Gm_2\Delta t^2}{R(t)^2}$ and the corresponding function $p_2$ is $D_2(t)=\frac{Gm_1\Delta t^2}{R(t)^2}$. With those two equations $R(t)=\lim_{h\to\infty }\sum_{n=0}^h D_1(\frac{nt}{h})+D_2(\frac{nt}{h})$, but since $D_1$ and $D_2$ are defined that makes are self referential, is there any way out of that? Would the definition $R(t)=\int_0^t\frac{G(m_1+m_2)}{R(u)^2}du$ be equivalent or useful? Most of this work is back of the envelope stuff combined with only really a half understanding of both calculus and Newtonian Mechanics, so any pointers or advice would be greatly appreciated. | Kepler problem in time: how do two gravitationally attracted particles move? Two particles with initial positions and velocities $r_1,v_1$ and $r_2,v_2$ are interacting by the inverse square law (with G=1), so that $$ {d^2r_1\over dt^2} = - { m_2(r_1-r_2)\over |r_1-r_2|^3} $$ $$ {d^2r_2\over dt^2} = - { m_1(r_2-r_1)\over |r_1-r_2|^3} $$ (the inverse square law along the line of separation). What is the complete solution of these differential equations? What is the position of the two objects as a function of time? After reading a lot on Wikipedia, I've come to the definition of center of mass and relative coordinates: $$R(t) ~=~ \frac{m_1 r_1 + m_2 r_2}{m_1+m_2}$$ $$\ddot{r}(t) r(t)^2 ~=~ (m_1+m_2)G$$ Where $R$ is the center of mass, and $r$ is the displacement between the particles... Is this correct? How do I proceed to solve the differential equation? | How can I prove that $\frac{n^2}{x_1+x_2+\dots+x_n} \le \frac{1}{x_1}+ \frac{1}{x_2} +\dots+ \frac{1}{x_n}$? How can I prove that $\frac{n^2}{x_1+x_2+\dots+x_n} \le \frac{1}{x_1}+ \frac{1}{x_2} +\dots+ \frac{1}{x_n}$? im trying to use AM-GM $\sqrt[n]{ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}} \le \sum_{k=1}^n \frac{{\frac{1}{x_1} +\frac{1}{x_2}+ \frac{1}{x_3}+ ..\frac{1}{x_n}}}{n}$ $ln\sqrt[n]{ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}} \le ln \frac{1}{n} \sum_{k=1}^n {\frac{1}{x_1} +\frac{1}{x_2}+ \frac{1}{x_3}+ ..\frac{1}{x_n}}$ $ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}\le \sum_{k=1}^n {\frac{1}{x_1} +\frac{1}{x_2}+ \frac{1}{x_3}+ ..\frac{1}{x_n}}$ im not sure is this right or not, however i dont know how to include the $n^2$ in the nominator? is there also alternative proof using jensen inequality? | eng_Latn | 18,772 |
Condition on a & b so that the two tangents drawn to the parabola $y^2=4ax$ from a point are normals to the parabola $x^2=4by$ Find the condition on a & b so that the two tangents drawn to the parabola $y^2=4ax$ from a point are normals to the parabola $x^2=4by$ I tried finding the joint equation of tangents to the parabola $y^2=4ax$ like $SS_1=T^2$.Then I tried to apply the condition of normality to the joint pair of straight lines.But its becoming too huge.Please let me know if you can think of a shorter and more elegant method to prove $a^2>8b^2$.Thanks. | Find the condition on $a$ and $b$ so that the two tangents drawn to the parabola $y^2=4ax$ from a point are normals to the parabola $x^2=4by$ Find the condition on $a$ and $b$ so that the two tangents drawn to the parabola $y^2=4ax$ from a point are normals to the parabola $x^2=4by.$ The required condition is $a^2>8b^2$.I dont know how to prove it.I tried. Let $(h,k)$ be the point from where tangents are drawn to the parabola $y^2=4ax$.Let $m_1,m_2$ be the slopes of the tangents.Equations of tangents are $y=m_1x+\frac{a}{m_1}$ and $y=m_2x+\frac{a}{m_2}$ and equations of normal to the parabola $x^2=4by$ is $y=m_1x-2bm_1-bm^3_1$ and $y=m_2x-2bm_2-bm^3_2$. As $y=m_1x+\frac{a}{m_1}$ and $y=m_1x-2bm_1-bm^3_1$ are the same lines.Therefore,$\frac{a}{m_1}=-2bm_1-bm^3_1$ $bm^4+2bm^2+a=0$ And I am stuck.I cannot get the desired condition.Please help me. | Specific system of differential equations I have the following system of equations: \begin{eqnarray}\frac{dx}{dt} = x(1 - x^2 - y^2) \\ \frac{dy}{dt} = y(4 - x^2 - y^2) \end{eqnarray} I want to prove that if a solutions starts (at time $t = 0$) in the region $$ G = \{(x, y) \in \mathbb{R}^2 : x > 0, y > 0 \text{ and } 1 < x^2 + y^2 < 4\} $$ that it remains there for all $t \geq 0$. I have already proven that, from $G$, a solution cannot cross the $x$-axis where $0 < x < 2$ or the $y$-axis where $0 < y < 2$ (since those line segments are orbits), but the pieces of the circles with radius 1 and 2 (where $x > 0, y > 0$) are the lines I haven't done yet. In this question we can focus on just the circle with radius 1 since the proof for the other circle is completely analogous to it. In my book, there is a similar problem where they show that if a solution crosses a certain line, that solution attains its maximum or minimum value for $x$ and $y$, and then they derive a contradiction. However, in this case, for $n \geq 1$ the $n$-th derivative of $x$ equals 0 on the circle with radius 1. (If the second derivative were negative, $x$ would attain its maximum on the circle, a contradiction since $\frac{dx}{dt} < 0$ in $G$, so $x$ is decreasing.) I don't see why a solution can't cross the circle at all. On the circle, $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} > 0$, so after crossing the circle, the solution would go back up to $G$. Is the circle itself an orbit? EDIT I think I have found the answer. Using polar coordinates, I found $$ \frac{dr}{dt} = r - r^3 + 3r \sin^2(\theta) $$ Now, $$ \frac{dr}{dt} > 0 \Leftrightarrow 0 < r - r^3 + 3r \sin^2(\theta) \leq r - r^3 + 3r = 4r - r^3 $$ since $0 \leq \sin^2(\theta) \leq 1$. Note that $0 < r \leq 1$ satisfies $4r - r^3 > 0$, and thus $\frac{dr}{dt} > 0$. Therefore, if a solution goes from $G$ to the edge of the unit circle, $r$ must, on the edge, either be decreasing or constant (i.e., its derivative is less than or equal to 0), contradicting the fact that the derivative is positive. Can someone tell me if this is correct? | eng_Latn | 18,773 |
What is the derivative of t/1+t^2 and then how do I know what its local max and local mins are? | derivitive is \n\n1 + 2t\n\nmaximum value is when this value equals 1. Therefore, the max value occurs when t = 0.\n\nThe minimum value is when this value equals 0. Therefore, the min value occurs when t = -1/2. | try sosmath.com they helped me before. if they don't have the answer you are looking for they have a list of links to tons of other math sites. \n\nand there probably is someone on your campus that can help as well. One on one help is always best...\n\nGood luck! | eng_Latn | 18,774 |
Susan's 10 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30° above the floor. The tension is a constant 34 N and the coefficient of friction is 0.20. Use work and energy to find Paul's speed after being pulled 3.4 m.\n\nThank you | plz ask in physics category | I think the velocity expression is meant to be in terms of t (rather than x)\nv(t) = ln(t+3) -cost*e^(t/2 -1).\n\nIf the particle VELOCITY is decreasing, it means that the time derivative of v(t), which is the acceleration, a(t), is negative. But if the particle SPEED is decreasing, it means that the acceleration is the opposite sign of the velocity. So if you really mean speed, you are looking for times when a(t)*v(t) < 0.\n\nThe derivatives are as follows\nd/dt(ln(t+3)) = 1/(t+3)\nd/dt(e^((t/2)-1)) = 1/2*e^((t/2)-1)\nd/dt(cos(t)) = -sin(t)\nso\na(t) = 1/(t+3) -cos(t)*1/2*e^((t/2)-1) + sin(t)*e^((t/2)-1)\n\nThen find t, such that a(t) < 0 (velocity decreasing)\nor a(t)*v(t) < 0 (speed decreasing).\n\nHope this helps!\n\n(Note if the velocity really is given in terms of x (rather than t) you will be solving a second order differential equation, which is doable, but I don't think that was the point of the question). | eng_Latn | 18,775 |
Solve the SDE $dX_t=\alpha \,dt + \sigma X_t \,dB_t$, $X_0=x_0$ | Solving SDE: $dX(t) = udt + \sigma X(t)dB(t)$ | Totally disconnected space that is not $T_2$ | eng_Latn | 18,776 |
How to handle triangle in PDE to do separation of variables? | Using separation of variables to solve the wave equation | Why separation of variables works in PDEs? | eng_Latn | 18,777 |
How to find complex solutions of equations? | Solving $(z+1)^5 = z^5$ | Unable to solve equation for a variable by any method | eng_Latn | 18,778 |
Writing an equation with two domains? | How to write a function (piecewise) with bracket outside? | Prove that equation has exactly 2 solutions | eng_Latn | 18,779 |
How to understand the results of solve:ifun? | Accessing Reduce from DSolve | Can't transfer purchases from iPhone to iTunes in iOS 9 | eng_Latn | 18,780 |
Analytical solve for biharmonic equation | 2D inhomogeneous biharmonic equation | 2D inhomogeneous biharmonic equation | eng_Latn | 18,781 |
Perform a Scaling Substitution | In calculus, a useful technique for integrating is u-substitution. Changing variables allows us to obtain an integral that is easier to evaluate.
| You have a rusty knife but need a sharp one. But you have no whetstone or other tools at hand. | eng_Latn | 18,782 |
On solving ode/pde with Neural Networks | What should I do when my neural network doesn't learn? | Weak solutions to the Neumann's problem (Evans PDE) | eng_Latn | 18,783 |
Solving differential equations with sums (power series) | Solving an ODE in power series | Finite Sum of Power? | eng_Latn | 18,784 |
I can prove that $dx dy = r.dr d\theta$ by drawing a circle and calculating the area of a small square in the polar coordinates, but when I try proving it using the equations below, I fail to prove it. What is my mistake? $x = r\cos(\theta) => dx = \cos(\theta).dr - \sin(\theta)r.d\theta$ $y = r\sin(\theta)=>dy=\sin(\theta).dr + \cos(\theta)r.d\theta$ $=> dxdy = r(\cos^2(\theta)-\sin^2(\theta))drd\theta=r\cos(2\theta).drd\theta$ I actually saw a similar question in this site: My problem was that I didn't understand why $drd\theta = -d\theta dr$ | In polar coordinate how we can get $dx\;dy=r\;dr\;d\theta$? with these parameters: $r=\sqrt{x^2+y^2}$ $x=r\cos\theta$ $y=r\sin\theta$ Tanks. | In polar coordinate how we can get $dx\;dy=r\;dr\;d\theta$? with these parameters: $r=\sqrt{x^2+y^2}$ $x=r\cos\theta$ $y=r\sin\theta$ Tanks. | eng_Latn | 18,785 |
Find the Particular Integral of the following ODE using the variation of parameters? | How to solve this ode $xy''-(1+x)y'+y=x^2$? | Integral of periodic function over the length of the period is the same everywhere | eng_Latn | 18,786 |
dA in polar coordinates using total differentials | Explain $\iint \mathrm dx\,\mathrm dy = \iint r \,\mathrm \,d\alpha\,\mathrm dr$ | Boundary condition with spatial derivative is ignored by NDSolve | eng_Latn | 18,787 |
NDSolve with NIntegrate | How to plot and solve the numerical solution of a integro-differential equation | Can NDSolve handle discontinuous data? | eng_Latn | 18,788 |
Simple Ordinary Differential Equation | Initial value problem $t\frac{dx}{dt}=x+\sqrt{t^2+x^2}$ | No group of order 400 is simple | eng_Latn | 18,789 |
Solve Differential Equations Using Laplace Transforms | The Laplace transform is an integral transform that is widely used to solve linear differential equations with constant coefficients. When such a differential equation is transformed into Laplace space, the result is an algebraic equation, which is much easier to solve. | You'll learn to create Sine Waves on Cylinders by Method 1, which is taken from a recent article, as revised for this project. | eng_Latn | 18,790 |
Second derivative using ND | Numerical partial derivative | Derivative of function solved from NDSolve can't be plotted | eng_Latn | 18,791 |
What is a unsaturated solution? What are some examples? | What are examples of unsaturated solutions? | What is an isotonic solution? What are some examples? | eng_Latn | 18,792 |
When considering the Doppler shift, the 'canonical equation' is $$f=\frac{c+vr}{c+vs}f_0$$ However, this equation seems to run into trouble in the following situation: A light source inside water is moving at a speed $v$ towards a receiver outside the water. Can we modify the above equation to deal with a transition between two media? Or is there a completely different formula that applies here? | When considering the Doppler shift, the 'canonical equation' is $$f=\frac{c+vr}{c+vs}f_0$$ However, this equation seems to run into trouble in the following situation: A light source inside water is moving at a speed $v$ towards a receiver outside the water. Can we modify the above equation to deal with a transition between two media? Or is there a completely different formula that applies here? | When considering the Doppler shift, the 'canonical equation' is $$f=\frac{c+vr}{c+vs}f_0$$ However, this equation seems to run into trouble in the following situation: A light source inside water is moving at a speed $v$ towards a receiver outside the water. Can we modify the above equation to deal with a transition between two media? Or is there a completely different formula that applies here? | eng_Latn | 18,793 |
I'm trying to make a ping sweep using bash: #!/bin/bash for x in seq 74 254 do ping -w 4 192.168.1.$x done According to what I've read on the Internet this should be looping from 74 to 254 but outputs only results from 74 and 254. What am I doing wrong ? | How do I iterate over a range of numbers in Bash when the range is given by a variable? I know I can do this (called "sequence expression" in the Bash ): for i in {1..5}; do echo $i; done Which gives: 1 2 3 4 5 Yet, how can I replace either of the range endpoints with a variable? This doesn't work: END=5 for i in {1..$END}; do echo $i; done Which prints: {1..5} | (A kind soul at suggested I post here as well, sorry if out of bounds.) I'm trying to programmatically model a damped harmonic spring for use in mobile UI animations (physics mathematics isn't my background, please pardon any misconceptions). Having derived the parameters for the general case equation, I can iteratively calculate values until I reach a suitable threshold, though because this is bound to "simple" trigonometric and $e^{x}$ functions on the CPU, the 4000-some-odd steps can cause about 0.25 seconds lag on slow devices while it calculates. I'd like to speed this up using my platform's super-optimized vector and BLAS/LAPACK variants. The requirement for doing this is precalculating the number of steps necessary to reach my threshold value. In the underdamped case, where the roots of the characteristic function of the differential equation are non-real, I can use algebraic tricks to get my values: $$x(t) = c_{1}e^{r_{1}}\cos(i_{1}t) + c_{2}e^{r_{2}}\sin(i_{2}t)$$ (Given $r_{1}$, $i_{1}$, $r_{2}$, and $i_{2}$ are the real and irrational components of my two roots, respectively.) Knowing that $r_{1} = r_{2}$ and $i_{1} = -i_{2}$, I can simplify to: $$x(t) = c_{1}e^{r_{1}}\cos(i_{1}t)$$ And get my desired value of $t$ for my threshold $a$: $$t = \arccos(a / c_{1} / e^{r_{1}}) / i_{1}$$ When the roots are real, the equation looks a lot simpler: $$x(t) = c_{1}e^{r_{1}} + c_{2}e^{r_{2}}$$ However, I don't have my trig functions floating around to help me solve it (even if I did, the irrational components being 0 would cause problems, of course). Take the concrete example on pages 3-4 of (my bible during this process), since they at least solve cleanly: $$x(t) = 1.5e^{-t} - 0.5e^{-3t}$$ I know how I would solve for $t$ to get my for when $x(t) = a$ on paper, by setting $x=e^{t}$, solving, and back substituting, but I don't have that luxury here. I can make a few assumptions: the roots and constants are all real. I'm always going to be looking for the smallest, first, positive value of $t$. Obviously, the iterative solution is the simplest for this case, but in the end that would involve more steps and therefore be slower no matter what my other optimizations would be. How, then, would I go about solving for my threshold value algorithmically in this (supposedly) simplified case? Addendum The underdamped solution presents an extra requirement. The motion curve will oscillate back and forth a few times across the endpoint. Therefore, "first and lowest" $t$ requirement is not necessarily true. In my current, iterative code, the threshold value is both checked against the distance from the current $x(t)$ to the endpoint, as well as to the distance from the previous $x(t)$ as well to allow for a number of oscillations. This might make a more efficient solution nearly impossible. | eng_Latn | 18,794 |
I've studied simple harmonic motion and the relation that force $\mathbf F$ is proportional to $-x$ and time period of motion. But if in some other motion, $\mathbf F$ is proportional to some other power of $x$ say $- \mathbf F$ proportional to $-x^3$, and we just doubled the amplitude of motion, then how would the time period change? | How would I find the period of an oscillator with the following force equation? $$F(x)=-cx^3$$ I've already found the potential energy equation by integrating over distance: $$U(x)={cx^4 \over 4}.$$ Now I have to find a function for the period (in terms of $A$, the amplitude, $m$, and $c$), but I'm stuck on how to approach the problem. I can set up a differential equation: $$m{d^2x(t) \over dt^2}=-cx^3,$$ $$d^2x(t)=-{cx^3 \over m}dt^2.$$ But I am not sure how to solve this. Wolfram Alpha gives a particularly nasty solution involving the hypergeometric function, so I don't think the solution involves differential equations. But I don't have any other leads. How would I find the period $T$ of this oscillator? | Both (if you take a list of values, the distribution of the most significant digit is rougly proportional to the logarithm of the digit) and (given a corpus of natural language utterances, the frequency of any word is roughly inversely proportional to its rank in the frequency table) are not theorems in a mathematical sense, but they work quite good in the real life. Does anyone have an idea why this happens? (see also ) | eng_Latn | 18,795 |
Applying uncertainty principle to energy states Often for this I have heard, the longer the lifetime of the energy state, the uncertainty in the energy state decreases as a result of heisenberg's uncertainty principle. However doesn't that look at uncertainty in time rather than the actual time? | What is $\Delta t$ in the time-energy uncertainty principle? In non-relativistic QM, the $\Delta E$ in the time-energy uncertainty principle is the limiting standard deviation of the set of energy measurements of $n$ identically prepared systems as $n$ goes to infinity. What does the $\Delta t$ mean, since $t$ is not even an observable? | The formal solution of the time-dependent Schrödinger equation Consider the time-dependent (or some equation in Schrödinger form) written down as $$ \tag 1 i \partial_{0} \Psi ~=~ \hat{ H}~ \Psi . $$ Usually, one likes to write that it has a formal solution of the form $$ \tag 2 \Psi (t) ~=~ \exp\left[-i \int \limits_{0}^{t} \hat{ H}(t^{\prime}) ~\mathrm dt^{\prime}\right]\Psi (0). $$ However, this form for the solution of $(1)$ is actually built by the method of successive approximations which actually returns a solution of the form $$ \tag 3 \Psi (t) ~=~ \hat{\mathrm T} \exp\left[-i \int \limits_{0}^{t} \hat{H}(t^{\prime})~\mathrm dt^{\prime}\right]\Psi (0), \qquad t>0, $$ where $\hat{\mathrm T}$ is the time-ordering operator. It seems that $(3)$ doesn't coincide with $(2)$, but formally $(2)$ seems to be perfectly fine: it satisfies $(1)$ and the initial conditions. So where is the mistake? | eng_Latn | 18,796 |
Why is it bad to use Pearson distance in K-means clustering? I have implemented this algorithm in MATLAB and when I produce plots I notice that using Euclidean distance, I usually get presented with a clear pattern (sum of squares decreases with the number of iterations). However, when I run the algorithm using the modified Pearson correlation distance (1 - r, where r is the Pearson correlation coefficient), sometimes I would see no trend at all. In fact, on some occasions the sum of squares seems to increase with the number of iterations. Is there a reason for this? Also I know that Euclidean distance is the preferred metric for this algorithm, but does it have any drawbacks? | Why does k-means clustering algorithm use only Euclidean distance metric? Is there a specific purpose in terms of efficiency or functionality why the k-means algorithm does not use for example cosine (dis)similarity as a distance metric, but can only use the Euclidean norm? In general, will K-means method comply and be correct when other distances than Euclidean are considered or used? [Addition by @ttnphns. The question is two-fold. "(Non)Euclidean distance" may concern distance between two data points or distance between a data point and a cluster centre. Both ways have been attempted to address in the answers so far.] | find an approximate solution, up to the order of epsilon The question is to find an approximate solution, up to the order of epsilon of following problem. $$y'' + y+\epsilon y^3 = 0$$ $$y(0) = a$$ $$y'(0) = 0$$ I tried to solve the given problem using perturbation theory. $$y(t) = y_0(t) + \epsilon y_1(t) + \epsilon^2 y_2(t) + \cdots$$ $$t = w(\epsilon)t = (1 + \epsilon w_1 + \epsilon^2 w_2 + \cdots )t$$ However, i failed to find an appropriate approximate solution... help me!! | eng_Latn | 18,797 |
I tried to solve but could not develop the left part of the equation Thanks | a) Give a combinatorial proof that for every $n \geq r \geq 1$ that: $$\sum_{i = 0}^{n} \binom{i}{r - 1} = \binom{n + 1}{r}$$ And use (a) to concoct a formula for $1^2 + 2^2 + ... + n^2$ | Using the method of characteristics, find a solution to Burgers' equation \begin{cases} u_t+\left(\frac{u^2}{2} \right)_x =0 & \text{in }\mathbb{R}\times(0,\infty) \\ \qquad \qquad \, \, u=g & \text{on } \mathbb{R} \times\{t=0\} \end{cases} with the initial conditions $$g(x)=\begin{cases} 0 & \text{if }x < 0 \\ 1 & \text{if }0 \le x \le 1 \\ 0 & \text{if }x > 1 \end{cases}$$ First, I realized that the equation $u_t+\left(\frac{u^2}{2} \right)_x =0$ is equivalent to this form: $$u_t+uu_x =0$$ Then should I generally follow the method of solution as outlined in the answer of ? Note that this is not a duplicate question of that page. Rather I want to know if that page can be used for my problem, even though my ICs are different. By the way, the solution printed in my book (PDE Evans, 2nd edition, page 142) is $$u(x,t) = \begin{cases} 0 & \text{if } x < 0 \\ \frac xt & \text{if } 0 < x < t \\ 1 & \text{if } t < x < 1 + \frac t2 \\ 0 & \text{if } x > 1 + \frac t2 \tag{$0 \le t \le 2$} \end{cases}$$ | eng_Latn | 18,798 |
Differential equation problem. Integrating the logistic equation. | Modelling population with $\frac{dP}{dt}=P(\beta - \delta P)$ | Integral of periodic function over the length of the period is the same everywhere | eng_Latn | 18,799 |
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