Help with Partial Differentialtion

[cshum00]

Homework Statement

I am given $$x^2 + y^2 = r^2$$Need to find $$\frac{\partial y}{\partial x}=-\frac{1}{r}sin(\frac{v}{r}t)$$

2. Revelant equations
x parametriztion of circle is:
$$x=rcos(\frac{v}{r}t)$$Trigonometric Identity
$$sin^2t + cos^2t=1$$
$$r^2sin^2t=r^2-r^2cos^2t$$
$$1 + \frac{cos^2t}{sin^2t} = \frac{1}{sin^2t}$$

The Attempt at a Solution

so $$y = (r^2 - x^2)^\frac{1}{2}$$$$\frac{\partial y}{\partial x}=\frac{1}{2}(r^2 - x^2)^\frac{-1}{2}(-2x)$$
simplify: $$\frac{\partial y}{\partial x}=-x(r^2 - x^2)^\frac{-1}{2}$$$$\frac{\partial ^2y}{\partial x^2}=-(r^2 - x^2)^\frac{-1}{2}+\frac{-1}{2}(r^2 - x^2)^\frac{-3}{2}(-2x)(x)$$
Simplify: $$\frac{\partial ^2y}{\partial x^2} = -\frac{1}{(r^2 - x^2)^\frac{1}{2}}-\frac{x^2}{(r^2 - x^2)^\frac{3}{2}}$$Substitute parametrization to second partial derivative:
$$\frac{\partial ^2y}{\partial x^2} = -\frac{1}{(r^2 - r^2cos^2(\frac{v}{r}t))^\frac{1}{2}}-\frac{r^2cos^2(\frac{v}{r}t)}{(r^2 - r^2cos^2(\frac{v}{r}t))^\frac{3}{2}}$$
$$\frac{\partial ^2y}{\partial x^2} = -\frac{1}{rsin(\frac{v}{r}t)}-\frac{r^2cos^2(\frac{v}{r}t)}{r^3sin^3(\frac{v}{r}t)}$$
$$\frac{\partial ^2y}{\partial x^2} = -\frac{1}{rsin(\frac{v}{r}t)}(1 + \frac{r^2cos^2(\frac{v}{r}t)}{r^2sin^2(\frac{v}{r}t)})$$
$$\frac{\partial ^2y}{\partial x^2} = -\frac{1}{rsin(\frac{v}{r}t)}\frac{1}{sin^2(\frac{v}{r}t)}$$

For some reason i ended up with $$\frac{1}{sin^2(\frac{v}{r}t)}$$ instead of $$sin^2(\frac{v}{r}t)$$. Did i do any intermediate steps wrong?

 

[Dick]

I don't think you are doing anything wrong. I get y''=(-r^2)/y^3 which seems to agree with you final answer using implicit differentiation.