Math Core

The first derivative $\dfrac{dy}{dx}$ tells you which way a parametric curve is heading. The second derivative $\dfrac{d^2y}{dx^2}$ tells you how it bends: concave up or concave down. For parametric curves there is one extra step that trips up many students, and this lesson is about getting it right.

Why you can't just differentiate twice with respect to t

The second derivative $\dfrac{d^2y}{dx^2}$ is the derivative of $\dfrac{dy}{dx}$ with respect to $x$ . But for a parametric curve, $\dfrac{dy}{dx}$ comes out as an expression in $t$ . You can differentiate it with respect to $t$ easily, and that gives $\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)$ , which is a rate per unit of $t$ , not per unit of $x$ .

To convert, use the same trick as for the first derivative. If $w$ is any quantity that depends on $t$ , then

\frac{dw}{dx} = \frac{dw/dt}{dx/dt}.

Apply that with $w = \dfrac{dy}{dx}$ .

Second derivative of a parametric curve

If $x = x(t)$ and $y = y(t)$ are twice differentiable and $\dfrac{dx}{dt} \ne 0$ , then

\frac{d^2y}{dx^2} = \frac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}.

Three steps: find $\dfrac{dy}{dx}$ as a function of $t$ ; differentiate that with respect to $t$ ; divide by $\dfrac{dx}{dt}$ .

Common mistake

The most common mistake is stopping after step 2. $\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)$ is not the second derivative. You must divide by $\dfrac{dx}{dt}$ again. Another wrong shortcut is $\dfrac{d^2y/dt^2}{d^2x/dt^2}$ , which has no justification and usually gives the wrong value.

Working through the steps

Worked example: The looping curve from the last lesson

For $x = t^2$ , $y = t^3 - 3t$ , find $\dfrac{d^2y}{dx^2}$ and decide whether the curve is concave up or down at $t = 1$ and at $t = -1$ .

Step 1. $\dfrac{dy}{dx} = \dfrac{3t^2 - 3}{2t}$ . It helps to split this before differentiating:

\frac{dy}{dx} = \frac{3}{2}t - \frac{3}{2}t^{-1}.

Step 2. Differentiate with respect to $t$ :

\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{3}{2} + \frac{3}{2}t^{-2} = \frac{3t^2 + 3}{2t^2}.

Step 3. Divide by $\dfrac{dx}{dt} = 2t$ :

\frac{d^2y}{dx^2} = \frac{3t^2 + 3}{2t^2} \cdot \frac{1}{2t} = \frac{3t^2 + 3}{4t^3}.

At $t = 1$ : $\dfrac{6}{4} = \dfrac{3}{2} > 0$ , so the curve is concave up at $(1, -2)$ . At $t = -1$ : $\dfrac{6}{-4} = -\dfrac{3}{2}$ , so it is concave down at $(1, 2)$ .

That matches the picture: the point $(1, -2)$ is the bottom of the loop, which cups upward, and $(1, 2)$ is the top, which cups downward.

x = t², y = t³ − 3t for −2 ≤ t ≤ 2. The branch traced for t > 0, through (1, −2), is concave up; the branch for t < 0, through (1, 2), is concave down.Open in grapher →

Notice that the numerator $3t^2 + 3$ is always positive, so the sign of $\dfrac{d^2y}{dx^2}$ is the sign of $t^3$ . The whole branch with $t > 0$ is concave up and the whole branch with $t < 0$ is concave down.

Trig parametrizations

Worked example: A point on an ellipse

For $x = 2\cos t$ , $y = 3\sin t$ , find $\dfrac{d^2y}{dx^2}$ at $t = \dfrac{\pi}{6}$ .

Step 1. $\dfrac{dx}{dt} = -2\sin t$ and $\dfrac{dy}{dt} = 3\cos t$ , so $\dfrac{dy}{dx} = -\dfrac{3}{2}\cot t$ .

Step 2. $\dfrac{d}{dt}\left(-\dfrac{3}{2}\cot t\right) = \dfrac{3}{2}\csc^2 t$ .

Step 3. Divide by $-2\sin t$ :

\frac{d^2y}{dx^2} = \frac{\frac{3}{2}\csc^2 t}{-2\sin t} = -\frac{3}{4}\csc^3 t.

At $t = \dfrac{\pi}{6}$ , $\csc t = 2$ , so $\dfrac{d^2y}{dx^2} = -\dfrac{3}{4}(8) = -6$ . The point is on the top half of the ellipse, which is concave down, as expected.

Intervals of concavity

To find where a parametric curve is concave up or down, find where $\dfrac{d^2y}{dx^2}$ is positive or negative as a function of $t$ . Be careful about the sign of $\dfrac{dx}{dt}$ , because step 3 divides by it.

Worked example: Concavity on an interval

For $x = e^t$ , $y = t^2$ , find the values of $t$ for which the curve is concave up.

Step 1. $\dfrac{dy}{dx} = \dfrac{2t}{e^t} = 2te^{-t}$ .

Step 2. By the product rule, $\dfrac{d}{dt}\left(2te^{-t}\right) = 2e^{-t} - 2te^{-t} = (2 - 2t)e^{-t}$ .

Step 3. Divide by $\dfrac{dx}{dt} = e^t$ :

\frac{d^2y}{dx^2} = (2 - 2t)e^{-2t}.

Since $e^{-2t} > 0$ , the sign depends only on $2 - 2t$ . The curve is concave up for $t < 1$ and concave down for $t > 1$ . It has an inflection point at $t = 1$ , the point $(e, 1)$ .

Tip

If you can eliminate the parameter, you can check your answer. For $x = e^t$ , $y = e^{2t}$ you have $y = x^2$ , so $\dfrac{d^2y}{dx^2}$ should be $2$ everywhere. Working it parametrically: $\dfrac{dy}{dx} = \dfrac{2e^{2t}}{e^t} = 2e^t$ , then $\dfrac{d}{dt}(2e^t) = 2e^t$ , and dividing by $e^t$ gives $2$ . It matches.

Practice

Practice 1

For $x = t^2$ and $y = t^3$ , find $\dfrac{d^2y}{dx^2}$ at $t = 2$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

A student finds $\dfrac{d^2y}{dx^2}$ for $x = t^2$ , $y = t^3$ at $t = 1$ by computing $\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)$ and gets $\dfrac{3}{2}$ . What is the correct value?

Practice 3

For $x = t^2 + 1$ and $y = t^3 + 3t$ with $t > 0$ , find $\dfrac{d^2y}{dx^2}$ in terms of $t$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 4

For the curve in the previous problem ( $x = t^2 + 1$ , $y = t^3 + 3t$ , $t > 0$ ), on what interval is the curve concave down?

Practice 5

A curve is defined parametrically, and for all $t$ , $\dfrac{dx}{dt} = 2t + 1$ and $\dfrac{dy}{dx} = t^3$ . Find $\dfrac{d^2y}{dx^2}$ at $t = 1$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

For the unit circle $x = \cos t$ , $y = \sin t$ , find $\dfrac{d^2y}{dx^2}$ at $t = \dfrac{\pi}{3}$ . Give an exact answer or a decimal to three places.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

For $x = e^t$ and $y = e^{2t}$ , find $\dfrac{d^2y}{dx^2}$ at $t = \ln 3$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.