Alternative Formula for Second Derivative of Parametric Equations

Question

So, if you have $y(t)$ and $x(t)$ and you're trying to find the $~\frac{dy}{dx}~$, you can do $~\frac{dy/dt}{dx/dt}~$.

So to find the second derivative I did $~\frac{d^2y/dt^2}{d^2x/dt^2}~$ because I did not yet know the correct formula $~\frac{\frac{d}{dt}(dy/dx)}{dx/dt}~$.

My method got me a different (wrong) answer but I don't understand why it doesn't work.

Why doesn't it work?

-"It" being $~\frac{d^2y/dt^2}{d^2x/dt^2}~$ to find the second derivative.

BTW: I was doing this question on Khan Academy

Did you see this link on the Khanacademy page you shared? https://math.stackexchange.com/questions/49734/taking-the-second-derivative-of-a-parametric-curve — Zach Favakeh, Jul 08 '19 at 18:25
Even if you think of the $dt$’s as formally “canceling” in $(dy/dt)/(dx/dt)$, this doesn’t work: if you “cancel” the $dt^2$s against each other in $(d^2y/dt^2)/(d^2x/dt^2)$, you end up with $d^2y/d^2x$ instead of $d^2y/dx^2$. — amd, Jul 08 '19 at 19:28

score 1 · Answer 1 · answered Jul 09 '19 at 04:00

$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$$

${}$

$$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{dy/dt}{dx/dt}\right)=\frac{d}{dt}\left(\frac{dy/dt}{dx/dt}\right)\cdot\frac{dt}{dx}$$

$$= \frac{\frac{dx}{dt}\cdot\frac{d^2y}{dt^2}-\frac{dy}{dt}\cdot\frac{d^2x}{dt^2}}{\left(\frac{dx}{dt}\right)^2}\cdot\frac{dt}{dx}$$

$$=\left\{\frac{dx}{dt}\cdot\frac{d^2y}{dt^2}-\frac{dy}{dt}\cdot\frac{d^2x}{dt^2}\right\}\cdot{\left(\frac{dt}{dx}\right)^3}$$

Now using this formula you can proceed to solve your problem.

Alternative Formula for Second Derivative of Parametric Equations

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