Calc 3 partial derivative review for PDE's class

Nick Bruno
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1. Homework Statement

I am suppose to use polar coordinate data to find derivatives, ie

x = r cos(theta)
y = r sin(theta)

r^2 = x^2 + y^2


2. Homework Equations

show dtheta/dy = cos(theta)/r
show dtheta/dx = -sin(theta)/r

in other words since i don't have the math script
find the equation for theta and take the derivatives
These are partial derivatives by the way (as you can tell by inspection)

3. The Attempt at a Solution

d theta / dy = cos(theta)/r

I separate and integrate

dtheta/cos(theta) = dy/sqrt(x^2+y^2)

ln(cos(theta)) = ln(sqrt(x^2+y^2))/(0.5(x^2+y^2)^-.5*2y) => per chain rule

ln(cos(theta))=ln(r)*r/ r*sin(theta)

ln(cos(theta)) = ln(r)/ sin(theta)

now what?

any help is very much appreciated. Thanks for looking. have a good one.
 
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how about trying implictly differentiating both sides of all your equations w.r.t x & y, then a little subtitution to get the form you want?

you'll need to consider
r = r(x,y)
theta = theta(x,y)
 
do you mind refreshing me on implicit differentiation?
 
Implicit differentiation is where you differentiate every term with respect to one of them without solving one term as a function of another.

For example

x2 + y2 = 1

I want to find dy/dx. Rather than calculate y as a function of x, I just differentiate both sides with respect to x. I get a dy/dx because of the chain rule

2x + 2y dy/dx = 0

dy/dx = -x/y

Similarly, you could differentiate your equations implicitly with respect to e.g. y in an attempt to find dtheta/dy
 
Thanks, this worked and was actually quite a clever way to solve the problem.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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