Concept preview

What you will be able to do

1. 01Compute the second derivative of a parametric curve with $\dfrac{d^2y}{dx^2}=\dfrac{\frac{d}{dt}\!\left(\frac{dy}{dx}\right)}{dx/dt}$ — differentiating the slope with respect to $t$ and then dividing by $\dfrac{dx}{dt}$ a second time.
2. 02Explain why the naive shortcut $\dfrac{d^2y/dt^2}{d^2x/dt^2}$ is wrong, and diagnose answers produced by it or by dividing by $\left(dx/dt\right)^2$.
3. 03Use the sign of $\dfrac{d^2y}{dx^2}$ to determine the concavity of a parametric curve at a point (concave up when positive, concave down when negative).

The second derivative $\dfrac{d^2y}{dx^2}$ means "the derivative of $\dfrac{dy}{dx}$ with respect to $x$." We already have $\dfrac{dy}{dx}$ as a function of $t$ from 9.1. To differentiate any quantity that lives in terms of $t$ with respect to $x$, we use the very same trick that gave us the first derivative: divide its $t$-derivative by $\dfrac{dx}{dt}$. Replacing "$y$" in the first-derivative formula with "$\dfrac{dy}{dx}$" yields $\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\!\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}.$ Read it as a two-step recipe: (1) take the slope $\dfrac{dy}{dx}$ you already found and differentiate it with respect to $t$; (2) divide that by $\dfrac{dx}{dt}$ AGAIN.