# Tagged Questions

33 views

### Polar coordinates in the cartesian plane.

${dy}/{dx} = {dy}/{d\theta}$ divided by $dx/d\theta$ where $x$ and $y$ are in the Cartesian plane and $\theta$ is in the polar plane and $x = r\cos( \theta), \ y = r \sin (\theta)$. If $dy/dx = 0$ ...
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### how to find slope of this polar curve: $r^2=\sin(2\theta)$.

Given $r^2=\sin(2\theta),\;$ how to find the slope of the tangent line at $x=0$ ? If the question were $r=\sin(2\theta)$, it would be o.k. but since it is $r^2=\sin(2\theta)$, I don't know how to ...
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### Polar coordinates: Slope of tangent

Would anyone mind telling me how to solve this problem? It seems strange as my answer is $-1$. Do I have to apply this formula, $(r'sinθ +rcosθ)/(r'cosθ -rsinθ)$ ?
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### Points on $(x^2 + y^2)^2 = 2x^2 - 2y^2$ with slope of $1$

Let the curve in the plane defined by the equation: $(x^2 + y^2)^2 = 2x^2 - 2y^2$ How can i graph the curve in the plane and determine the points of the curve where $\frac{dy}{dx} = 1$. My work: ...
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### Converting polar equation into cartesian equation to obtain derivatives

If I have a polar equation such as $r=1-2\cos(\theta)$ How would I convert this into an equation for $x$ and $y$ so that I can get $dx/d\theta$ and $dy/d\theta$ ?
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### Representation of differentials in Polar Coordinates

We define polar coordinates in $\mathbb{R}^{n}$\ $\{ 0\}$ by $x=ry$, where $r=|x|>0$ and $y \in \partial B(0,1)$ is a point on the unit sphere. In the coordinates, Lebesgue measure has the ...
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### gradient of an axis symmetric vector field in cylindical coordiantes

I am trying to calculate with a general approach the gradient of an axis symmetric vector field in cylindrical coordinates and then express it in cartesian coordinates. First I write my vector ...
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### Diff eq. transformation polar coordinates

I have $(x',y')=(x-y-x(x^2+y^2)+\frac{xy}{\sqrt{x^2+y^2}},x+y-y(x^2+y^2)-\frac{x^2}{\sqrt{x^2+y^2}} )$ Now I want to use polar coordinates $(x,y)=(r\cos(t),r\sin(t))$ to get ...
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### How do I calculate numerically a tensor in polar coordinates?

You can formulate the question also like this: What is the easiest way of calculating directed derivative of a function if its values are evaluated in a cartesian grid? a) fit a (spline) surface, ...
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### Simple partial differentiation $x = r\cos\theta$ and $y = r\sin\theta$

If \begin{align} x &= r\cos\theta,\\ y &= r\sin\theta, \end{align} find $$\dfrac{\partial^2\theta}{\partial{x}\partial{y}}.$$ How can I find this partial derivative? I need to prove ...
I have a missing $\frac{1}{r}\partial_r$ -term (notice the question mark) but cannot see why, could someone hint where I am doing mistake.