Showing $\frac{d^2y}{dx^2}=\frac{r^2d\theta^3 +2d\theta dr^2- rd^2 rd\theta}{(\cos\theta dr-r\sin\theta d\theta)^3}$

Question

$\frac{d^2y}{dx^2}=\frac{r^2d\theta^3 +2d\theta dr^2- rd^2 rd\theta}{(\cos\theta dr-r\sin\theta d\theta)^3}$ given that $y=f(x)$, $x=r\cos\theta$, $y=r\sin\theta$ and $\theta$ is independent.

$x=r\cos\theta$

$dx=\cos\theta dr-r\sin\theta d\theta$

$y=r\sin\theta$

$dy=r\cos\theta d\theta+ \sin\theta dr$

$d^2 y=\cos\theta d\theta dr -r\sin\theta d\theta^2+r\cos\theta d^2\theta+\cos\theta d\theta dr+\sin\theta d^2 r=\cos\theta d\theta dr -r\sin\theta d\theta^2+\cos\theta d\theta dr+\sin\theta d^2 r$

$\frac{d^2y}{dx^2}$

$=\frac{(\cos\theta d\theta dr -r\sin\theta d\theta^2+\cos\theta d\theta dr+\sin\theta d^2 r)(\cos\theta dr-r\sin\theta d\theta)}{(\cos\theta dr-r\sin\theta d\theta)^3}$

$=\frac{\cos^2\theta d\theta dr^2-r\sin\theta\cos\theta d\theta^2 dr-r\sin\theta\cos\theta d\theta^2 dr +r^2 \sin^2 \theta d\theta^3+\cos^2 d\theta dr^2-r\sin\theta \cos\theta d\theta ^2 dr+\sin\theta \cos\theta d^2 r dr -r\sin\theta d\theta d^2 r}{(\cos\theta dr-r\sin\theta d\theta)^3}$

$=\frac{2\cos^2\theta d\theta dr^2-3r\sin\theta\cos\theta d\theta^2 dr +r^2 \sin^2 \theta d\theta^3+\sin\theta \cos\theta d^2 r dr -r\sin\theta d\theta d^2 r}{(\cos\theta dr-r\sin\theta d\theta)^3}$

$=\frac{r^2d\theta^3 +2d\theta dr^2- rd^2 rd\theta -2\sin^2 \theta d\theta dr^2 +r \cos^2 \theta d\theta d^2 r-3r\sin\theta\cos\theta d\theta^2 dr+\sin\theta \cos\theta d^2 r dr}{(\cos\theta dr-r\sin\theta d\theta)^3}$

I got stuck here but I assume we have to use the fact that $r^2 =(r\cos\theta)^2 +(r\sin\theta)^2$ and factorize similar terms?

Answer 1

You should start from $y=f(x)$. Putting $x=r\cos\theta,y=r\sin\theta$ in $y=f(x)$ gives $$ r\sin\theta=f(r\cos\theta). $$ So after differentiating both sides, you have $$ \sin\theta dr+r\cos\theta d\theta=f'(r\cos\theta)(\cos\theta dr-r\sin\theta d\theta) \tag{1}$$ and hence $$ f'(r\cos\theta)=\frac{\sin\theta dr+r\cos\theta d\theta}{\cos\theta dr-r\sin\theta d\theta}. $$ Differentiating both sides of (1), you have $$ \sin\theta d^2r+2\cos\theta d\theta dr-r\sin\theta d\theta^2+r\cos\theta d^2\theta=f''(r\cos\theta)(\cos\theta dr-r\sin\theta d\theta)^2+f'(r\cos\theta)(\cos\theta d^2r-2\sin\theta d\theta dr-r\cos\theta d\theta^2-r\sin\theta d^2\theta) $$ and hence \begin{eqnarray*} f''(r\cos\theta)&=&\frac{\sin\theta d^2r+2\cos\theta d\theta dr-r\sin\theta d\theta^2+r\cos\theta d^2\theta-f'(r\cos\theta)(\cos\theta d^2r-2\sin\theta d\theta dr-r\cos\theta d\theta^2-r\sin\theta d^2\theta)}{(\cos\theta dr-r\sin\theta d\theta)^2}\\ &=&\frac{\sin\theta d^2r+2\cos\theta d\theta dr-r\sin\theta d\theta^2+r\cos\theta d^2\theta-\frac{\sin\theta dr+r\cos\theta d\theta}{\cos\theta dr-r\sin\theta d\theta}(\cos\theta d^2r-2\sin\theta d\theta dr-r\cos\theta d\theta^2-r\sin\theta d^2\theta)}{(\cos\theta dr-r\sin\theta d\theta)^2}\\ &=&\frac{(\sin\theta d^2r+2\cos\theta d\theta dr-r\sin\theta d\theta^2+r\cos\theta d^2\theta)(\cos\theta dr-r\sin\theta d\theta)-(\sin\theta dr+r\cos\theta d\theta)(\cos\theta d^2r-2\sin\theta d\theta dr-r\cos\theta d\theta^2-r\sin\theta d^2\theta)}{(\cos\theta dr-r\sin\theta d\theta)^3}. \end{eqnarray*} After simplification, you have $$ f''(r\cos\theta)=\frac{r^2d\theta^3 +2d\theta dr^2- rd^2 rd\theta}{(\cos\theta dr-r\sin\theta d\theta)^3}. $$ But $$ f''(r\cos\theta)=f''(x)=\frac{d^2y}{dx^2} $$ and hence $$ \frac{d^2y}{dx^2}=\frac{r^2d\theta^3 +2d\theta dr^2- rd^2 rd\theta}{(\cos\theta dr-r\sin\theta d\theta)^3}. $$