Second derivative of a parametric equation with trig functions

Question

I am solving a problem where I have to find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ given these parametric equations:

$ x = \cos t $
$ y = 3 \sin t $

I got $-3\operatorname{cot}t\;$ for the first derivative by finding $\frac{dy}{dt} / \frac{dx}{dt}$.

For the second derivative, I simply took the derivative of $-3\operatorname{cot}t$ to get $3\operatorname{csc}^2t$.

However, my teacher did this differently. He wrote that $\frac{d^2y}{dx^2} = \frac{-3(-\operatorname{csc}^2t)}{-\sin t} = -3\operatorname{csc}^3t$

I'd really appreciate it if someone could explain where I went wrong in my own solution.

Thanks $:)$

Answer 1

You took the derivative with respect to $t$. Your teacher took the derivative with respect to $x$. $$\frac{dy}{dx} = -3\cot t$$ $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{\frac{d}{dt}(-3\cot t)}{-\sin t} = \frac{3\csc^2 t}{-\sin t}$$

Answer 2

Note that $$\eqalign{ \def\L{\left} \def\R{\right} \def\b#1{\color{blue}{#1}} \def\c#1{\color{red}{#1}} &\b{9x^2 + y^2 = 9} \;=\; \L(9 \cos^2 t + 9\sin^2t\R) \\ }$$ This can be implicitly differentiated to yield $$\eqalign{ 0 &= 18x\:dx + 2y\:dy \\ \c{dy} &\c{= -\L(\frac{9x}{y}\R) dx} \\ \frac{dy}{dx} &= -\frac{9x}y \;=\; p \\ }$$ Further implicit differentiation yields $$\eqalign{ \def\qiq{\quad\implies\quad} dp &= -9\L(\frac{dx}{y}\R) + 9\L(\frac{x\,\c{dy}}{y^2}\R) \\ &= -9\L(\frac{dx}{y}\R) \c{-} 9\L(\frac{x\;\c{9x\,dx}}{y^2\,\c{y}}\R) \\ &= -9\L(\frac{\b{y^2 + 9x^2}}{y^3}\R)dx \\ &= -9\L(\frac{\b9}{y^3}\R)dx \\ \frac{dp}{dx} &= -\L(\frac{81}{y^3}\R) \;=\; \frac{d^2y}{dx^2} \\ \\ }$$ To recover your teacher's answer substitute $\:y=3\sin t$