use the equations $r=\sqrt{x^2 +y^2}$ and $\theta=\arctan(\frac{y}{x})$. I was able to get the partial derivative of of $r$ with respect to $x$ and $y$ and the partial derivative of $\theta$ with respect to $x$ and $y$. I have spent hours trying to figure this problem out by looking at the polar form of the Laplace equation as a reference but everything I try does not get me to that point. If anyone is up to the challenge please help me out! I apologize for not putting in Latex format, I have not quite gotten the hang of it yet.
123331
I am stuck with an exercise that requires me to find the Laplacian $\Delta u=(D_x^2u+D_y^2u)$ of a 2d-function $u$ in polar coordinates (in the standard Euclidean plane).  I found the following article on the net, and tried to follow its logic, but I could not understand two steps:   at first, the representation of $D_x$ and $D_y$ in terms of $r, \theta$, at the bottom of page 2:   $D_y=\sin\theta D_r+\frac{\cos\theta}{r}D_\theta$ and  $D_x=\cos\theta D_r-\frac{\sin\theta}{r}D_\theta$ . When I draw a sketch of the plane with a circle and all the coordinates, I get that it should be $D_y=\sin\theta D_r+r\ \cos\theta D_\theta$, because the larger the radius is, the greater will be the impact of a change in $\theta$ on a change in $y$. What am I making wrong here?  And then, secondly, what looks like an easy multiplication, namely taking the square of the above terms (on the top of page 3 in the link):  $D_y^2=(\sin\theta D_r+\frac{\cos\theta}{r}D_\theta)(\sin\theta D_r+\frac{\cos\theta}{r}D_\theta)$  $=\sin^2\theta D_r^2+\frac{2\sin\theta \cos\theta}{r}D_\theta D_r+\frac{\sin^2\theta}{r^2}D_\theta^2 \\ -\frac{\cos\theta}{r^2}D_\theta + \frac{\cos^2\theta}{r}D_r - \frac{\cos\theta \sin\theta}{r^2}D_\theta$  and similarly for $D_x$.  I don't see from where the last three summands come, what I see is: $(a+b)(a+b)=a^2+2ab+b^2+c+d+e$ and I cannot see the context of $c,d,e.$  It would be great if you could explain those issues to me!
113749
This is an exercise from Kelley's book. Could someone help to show me a proof?   It seems very natural, and it is easy to prove by utilizing the arctan function in $\mathbb{R}^1$.  Thanks a lot.
101641
Consider the differential form:  $$\omega=-\frac y{x^2 + y^2}dx + \frac x{x^2 + y^2}dy$$  I believe that this form is not exact. i.e: There does not exist a $\psi$ such that $d\psi=\omega$.  In particular, this is supposed to be true since the line integral along a loop, $$\int_{C}\omega = 2\pi \neq 0$$  But, what about $$\psi=-\arctan(\frac x{y})$$  According to Maple,    As you can see, Maple says that $d\psi=\omega$. So what's the deal here? Is $\omega$ exact or not?  Thank You!
53169
It is given that $X,Y \overset{\text{i.i.d.}}{\sim} N(0,1)$  Show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}} \overset{\text{i.i.d.}}{\sim} N(0,\frac{1}{4})$  I was thinking of making polar transformations $X=r \cos \theta, Y=r \sin \theta$  Then I am getting stuck at ranges of $\theta$
74734
I'm trying to teach myself separation of variables and have been following some notes for the wave equation, but there's one part which really confuses me and I'm not exactly sure how it makes the step.  For the wave equation   $$u_{tt} - c^2 u_{xx} = 0$$   with length $l$ and fixed ends, $u(0,t)=u(l,t)=0$ we seek a solution in the form $$u(x,t)=v(x)q(t)$$  and substituting this into the equation gives $$\frac{1}{q(t)}\frac{d^2 q}{dt^2}=c^2 \frac{1}{v(x)}\frac{d^2 v}{dx^2}=-\omega ^2$$   and I understand all of this so far. But when solving the equation  $$q''+\omega ^2 q = 0$$ the general solution is $$q(t)=A \cos(\omega t +\alpha)$$ where $A$ and $\alpha$ are constants - which is the part which I'm not exactly sure on how it gets to.   Could someone explain this step to me? Thanks!
39229
Verify that: $\arcsin \theta+\arccos \theta=\frac{\pi}{2}.$ (1)   How can one verify (1) when it is not generally true? We can rewrite (1) as      Verify that if $\sin u = \cos v$, then $u+v=\frac{\pi}{2}$.   What about $\sin \frac{2\pi}{3}$ and $\cos \frac{\pi}{6}$? $\sin \frac{2\pi}{3} = \cos \frac{\pi}{6}$, but $\frac{2\pi}{3}+\frac{\pi}{6} \neq \frac{\pi}{2}$.We may try to verify (1) as:  Let $\arcsin \theta=u$ and $\arccos \theta=v$, then $\sin u=\cos v = \theta.$ We can write   $$\sin(u+v)=\sin u \cos v+ \cos u \sin v,$$  which becomes   $$\sin(u+v)=\theta^2+ \cos u \sin v.$$  Now, $\cos u=\pm \sqrt{1-\theta^2}$, and $\sin v=\pm \sqrt{1-\theta^2}$. For $\cos u= \sqrt{1-\theta^2}$ and $\sin v= \sqrt{1-\theta^2}$, we have  $$\sin(u+v)= \theta^2+(1-\theta^2)=1,$$  $$\therefore u+v=\arcsin \theta + \arccos \theta = \arcsin(1)=\frac{\pi}{2}.$$ But $\arcsin(1)$ is also equal to $\frac{\pi}{2}+2n\pi, n \in \mathbb{Z}.$ What about that? And what about the cases when $\cos u = -\sqrt{1-\theta^2}$ and $\sin u = \sqrt{1-\theta^2}$ and the other way? Then $\sin(u+v)=2\theta^2-1=1\ \mbox{iff}\ \theta =\pm 1.$   I don't know where I am making a mistake, what assumptions are wrong, or where my reasoning goes wrong. Please help.
83729
let's consider $f: \mathbb R^2 \to \mathbb R, \quad (x,y) \mapsto xy$  We calcualte the partial-derivatives:  $\partial_x f=y$   $\partial_y f=y$  Now let's sitch to polar-coordinates, we get:  $f: (0,\infty)\times\mathbb R \to \mathbb R, \quad (r,\varphi) \mapsto r^2\sin \varphi \cos \varphi$  We calcualte the partial-derivatives:  $\partial_r f = 2r\sin\varphi\cos\varphi$   $\partial_\varphi f = r^2(-\cos^2\varphi+\sin^2\varphi)$  Now, If I calcualted the derivation in polra-coordinates, can I somehwo go back to eucledean coordinates? Just using $x=r\cos\varphi, y=r\sin\varphi$ won't do it, since if we take the partial-derivatives in polar coordinates we would get:  $2r\sin\varphi\cos\varphi=2y\cos\varphi$  $r^2\sin\varphi - r^2\cos^2\varphi = x^2-y^2$  I first expected to get $x$ and $y$ but that's bullshit since we aren't deriving in the same directions. So how could I now switch properly to eucleadean coordinates?
62341
This is a worked out example in my book, but I am having a little trouble understanding it:  Consider the system of equations:  $$x'=y+x(1-x^2-y^2)$$ $$y'=-x+y(1-x^2-y^2)$$  The orbits and limit sets of this example can be easily determined by using polar coordinates. (My question: what is the motivation for that thinking? What should clue me in to thinking that I should use polar coordinates?)  The polar coordinate satisfies $r^2=x^2+y^2$ so by differentiating with respect to t and using the differential equations we get:  $r\cdot r'=x\cdot x'+y\cdot y'$ (I am unclear about how the book even got this first equation from $r^2=x^2+y^2$)   $=x\cdot y+x^2(1-r^2)-x\cdot y+y^2(1-r^2)$ Substitute in $x'$ and $y'$ and then multiple out and replace with $r$, I get this step  $=r^2(1-r^2)$ cancel terms, I get this step too  $r'=r(1-r)$  similarly, the angle variable $\theta$ satisfies $\tan\theta=\frac yx$, so the derivative with respect to $t$ yields $\sec^2(\theta)\theta'=x^{-2}[x^2+xy(1-r^2)-y^2-xy(1-r^2)]=-\frac{r^2}{x^2}$ so $\theta=1$  Thus the solution goes clockwise around the origin at unit angular speed.  I don't understand the $\theta$ step at all or how they reached the conclusion of clockwise around the origin with unit angular speed..  But then it just jumps to saying "the origin is a fixed point, so α(0)=ω(0)={0} but I have no idea how they reached this conclusion..
65657
What is the gradient of $f(x, y, z) = \sqrt{x^2+y^2+z^2}$?  I know that $\nabla f = \left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right\rangle$ or equivalently $\nabla f = \frac{\partial f}{\partial x}\mathbf{i} + \frac{\partial f}{\partial y}\mathbf{j} + \frac{\partial f}{\partial z}\mathbf{k}$.  Eventually, I am going to evaluate this gradient at a point and then dot it with some given unit vector to find a directional derivative. I think maybe what is troubling me is the radical in $f$. Will I have to use the chain rule in order to find the three partial derivatives?  I tried this:  $$\frac{\partial f}{\partial x} = \frac{1}{2}(x^2 + y^2 + z^2)^{-\frac{1}{2}}2x$$ $$= \frac{x}{\sqrt{x^2 + y^2 + z^2}}$$  That result looks ugly to me. However, if I did that partial derivative correctly, then I think I know what I am doing.
118512
I can not get the correct answer.   $$\int \frac {dx}{\sqrt {x^2 + 16}}$$  $x = 4 \tan \theta$, $dx = 4\sec^2 \theta$  $$\int \frac {dx}{\sqrt {16 \sec^2 \theta}}$$  $$\int \frac {4 \sec^ 2 \theta}{\sqrt {16 \sec^2 \theta}}$$  $$\int \frac {4 \sec^ 2 \theta}{4 \sec \theta}$$  $$\int \sec \theta$$  $$\ln| \sec \theta + \tan \theta|$$  Then I solve for $\theta$:  $x = 4 \tan \theta$  $x/4 = \tan \theta$  $\arctan (\frac{x}{4}) = \theta$  $$\ln| \sec (\arctan (\tfrac{x}{4})) + \tan (\arctan (\tfrac{x}{4}))|$$  $$\ln| \sec (\arctan (\tfrac{x}{4})) + \tfrac{x}{4}))| + c$$  This is wrong and I do not know why.
1937
How to prove for $x&gt;0$, $\dfrac{x}{1+x^2}&lt;\arctan x&lt;x$?   I started with saying $0&lt;\arctan x&lt;\frac{\pi}{2}$  when $x&gt;0$  I'm just not sure how to proceed with this proof, it works when I take random  values but I cant see how to prove that it works for any $x$.
86582
Can someone please post a detailed step-by-step procedure. Given the circle with a radius a, what is the differential equation of the circle.
45540
Express $\sqrt{3}\sin\theta - \cos\theta$ as: $a\cos (\theta + \alpha) $  Can someone please explain to me how to go about doing this?
44147
We need to prove that $$\dfrac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$$  I have tried and it gets confusing.
17812
I'm not sure on how to find the gradient in polar coordinates. The thing that troubles me the most is how to find the unit vectors $\hat{r}$ and $\hat{\theta}$. My approach for the rest is expressing the partial derivatives in respect of $r$ and $\theta$ using the chain rule.   How can I get around solving this problem?
33229
Find the value of $$\cos ^2\theta+\cos^2 (\theta+1^{\circ})+\cos^2(\theta+2^{\circ})+......+\cos^2(\theta+179^{\circ})$$  Can anyone teach me where to start with? I've no idea.
91561
The cylinder is given by the equation $x^2 + (y-\frac{a}{2})^2 = (\frac{a}{2})^2$.  The region of the cylinder is given by the limits $0 \le \theta \le \pi$, $0 \le r \le a\sin \theta$ in polar coordinates.  We need to only calculate the surface from a hemisphere and multiply it by two. By implicit functions we have:  $$A=2\iint\frac{\sqrt{\left(\frac{\partial F}{\partial x}\right)^2 + \left(\frac{\partial F}{\partial y}\right)^2 + \left(\frac{\partial F}{\partial z}\right)^2}}{\left|\frac{\partial F}{\partial z} \right|} dA$$  where $F$ is the equation of the sphere.  Plugging in the expressions and simplifying ($z \ge 0)$, we get:  $$A=2a\iint\frac{1}{\sqrt{a^2 - x^2 - y^2}} dxdy$$  Converting to polar coordinates, we have:  $$A = 2a \int_{0}^\pi \int_{0}^{a\sin(\theta)} \frac{r}{\sqrt{a^2 - r^2}} drd\theta$$  Calculating this I get $2\pi a^2$. The answer is $(2\pi - 4)a^2$. Where am I  going wrong?