Derivatives of Parametric Equations

A parametric curve describes $x$ and $y$ separately, each as a function of a third variable $t$ (often time). That lets you draw curves that loop, cross themselves, or double back, which no single function $y = f(x)$ can do. This page shows how to find slopes and concavity of these curves, using nothing more than the chain rule.

Key ideas

Parametric curves

A pair of equations

$$x = x(t), \qquad y = y(t)$$

traces out a curve: for each value of the parameter $t$, you get one point $(x(t), y(t))$. As $t$ increases, the point moves along the curve in a particular direction, called the orientation. Often you can’t (or don’t want to) solve for $y$ in terms of $x$, so we find slopes directly from $t$.

The slope dy/dx

By the chain rule, $\dfrac{dy}{dt} = \dfrac{dy}{dx} \cdot \dfrac{dx}{dt}$. Dividing both sides by $\dfrac{dx}{dt}$:

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad \text{provided } \frac{dx}{dt} \ne 0$$

The $y$-rate goes on top. The slope usually depends on $t$, so to find the slope at a point, plug in that point’s $t$-value.

Tangent lines

To write the tangent line at $t = t_0$:

1. Find the point: $x_0 = x(t_0)$, $y_0 = y(t_0)$.
2. Find the slope: $m = \dfrac{dy}{dx}$ at $t = t_0$.
3. Write $y - y_0 = m(x - x_0)$.

The tangent line is in $x$ and $y$, not $t$.

Horizontal and vertical tangents

Tangent	Condition
Horizontal	$\dfrac{dy}{dt} = 0$ and $\dfrac{dx}{dt} \ne 0$
Vertical	$\dfrac{dx}{dt} = 0$ and $\dfrac{dy}{dt} \ne 0$
Can’t tell yet	both are $0$; investigate further (for example, find $\displaystyle\lim_{t \to t_0} \frac{dy}{dx}$)

The second derivative

$\dfrac{d^2y}{dx^2}$ means “the derivative of $\dfrac{dy}{dx}$ with respect to $x$”. But $\dfrac{dy}{dx}$ is written in terms of $t$, so use the same trick again: differentiate with respect to $t$, then divide by $\dfrac{dx}{dt}$.

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

As usual, $\dfrac{d^2y}{dx^2} \gt 0$ means the curve is concave up, and $\dfrac{d^2y}{dx^2} \lt 0$ means concave down.

On the AP exam, parametric derivative questions can appear on both the calculator and no-calculator sections. Trig functions are always in radians (remember, $\pi$ radians $= 180^\circ$).

The curve $x = t^2$, $y = t^3 - 3t$ (Example 2): horizontal tangents at $(1, \pm 2)$, a vertical tangent at the origin, and a self-crossing at $(3, 0)$.

Worked examples

Example 1: Slope and tangent line

A curve is given by $x = t^2 + 1$, $y = t^3 - 2t$. Find $\dfrac{dy}{dx}$, and the equation of the tangent line at $t = 1$.

Solution. Differentiate each coordinate:

$$\frac{dx}{dt} = 2t, \qquad \frac{dy}{dt} = 3t^2 - 2, \qquad \frac{dy}{dx} = \frac{3t^2 - 2}{2t}$$

At $t = 1$: the point is $x = 2$, $y = 1 - 2 = -1$, and the slope is $\dfrac{3 - 2}{2} = \dfrac{1}{2}$. The tangent line is

$$y + 1 = \frac{1}{2}(x - 2)$$

Example 2: Horizontal and vertical tangents

Find all points where the curve $x = t^2$, $y = t^3 - 3t$ has a horizontal or vertical tangent.

Solution. $\dfrac{dx}{dt} = 2t$ and $\dfrac{dy}{dt} = 3t^2 - 3 = 3(t - 1)(t + 1)$.

Horizontal: $\dfrac{dy}{dt} = 0$ at $t = \pm 1$. There $\dfrac{dx}{dt} = \pm 2 \ne 0$, so both work.

• $t = 1$: the point $(1, 1 - 3) = (1, -2)$.
• $t = -1$: the point $(1, -1 + 3) = (1, 2)$.

Vertical: $\dfrac{dx}{dt} = 0$ at $t = 0$. There $\dfrac{dy}{dt} = -3 \ne 0$, so the tangent is vertical at $(0, 0)$.

Compare with the figure above: the curve flattens out at $(1, \pm 2)$ and stands straight up at the origin.

Example 3: Concavity

For the same curve, $x = t^2$, $y = t^3 - 3t$, find $\dfrac{d^2y}{dx^2}$ and decide where the curve is concave up.

Solution. From Example 2,

$$\frac{dy}{dx} = \frac{3t^2 - 3}{2t} = \frac{3}{2}\left( t - \frac{1}{t} \right)$$

Differentiate with respect to $t$, then divide by $\dfrac{dx}{dt} = 2t$:

$$\begin{aligned} \frac{d}{dt}\left( \frac{dy}{dx} \right) &= \frac{3}{2}\left( 1 + \frac{1}{t^2} \right) = \frac{3(t^2 + 1)}{2t^2} \\ \frac{d^2y}{dx^2} &= \frac{3(t^2 + 1)}{2t^2} \div 2t = \frac{3(t^2 + 1)}{4t^3} \end{aligned}$$

The numerator is always positive, so the sign matches $t^3$: concave up for $t \gt 0$ and concave down for $t \lt 0$. For example, at $t = 2$, $\dfrac{d^2y}{dx^2} = \dfrac{15}{32} \gt 0$.

Example 4: A trig curve (radians)

An ellipse is given by $x = 2\cos t$, $y = 3\sin t$. Find the tangent line at $t = \dfrac{\pi}{4}$, and find $\dfrac{d^2y}{dx^2}$ there.

Solution.

$$\frac{dy}{dx} = \frac{3\cos t}{-2\sin t} = -\frac{3}{2}\cot t$$

At $t = \dfrac{\pi}{4}$: $\cot\dfrac{\pi}{4} = 1$, so the slope is $-\dfrac{3}{2}$. The point is $\left( 2 \cdot \dfrac{\sqrt{2}}{2},\ 3 \cdot \dfrac{\sqrt{2}}{2} \right) = \left( \sqrt{2}, \dfrac{3\sqrt{2}}{2} \right)$, so the tangent line is

$$y - \frac{3\sqrt{2}}{2} = -\frac{3}{2}\left( x - \sqrt{2} \right)$$

Second derivative:

$$\frac{d}{dt}\left( -\frac{3}{2}\cot t \right) = \frac{3}{2}\csc^2 t, \qquad \frac{d^2y}{dx^2} = \frac{\tfrac{3}{2}\csc^2 t}{-2\sin t} = -\frac{3}{4}\csc^3 t$$

At $t = \dfrac{\pi}{4}$, $\csc t = \sqrt{2}$, so $\dfrac{d^2y}{dx^2} = -\dfrac{3}{4}\left( 2\sqrt{2} \right) = -\dfrac{3\sqrt{2}}{2}$. The curve is concave down there, which makes sense: it’s on the top half of the ellipse.

Common mistakes

Dividing in the wrong order. The slope is $\dfrac{dy/dt}{dx/dt}$, with $y$ on top, just like “rise over run”. Writing $\dfrac{dx/dt}{dy/dt}$ gives the reciprocal.

Forgetting to divide by dx/dt for the second derivative. $\dfrac{d}{dt}\left( \dfrac{dy}{dx} \right)$ alone is not $\dfrac{d^2y}{dx^2}$. You must divide by $\dfrac{dx}{dt}$ once more. This is the most common error on AP parametric questions.

Using the ratio of second derivatives. $\dfrac{d^2y}{dx^2}$ is not $\dfrac{d^2y/dt^2}{d^2x/dt^2}$. In Example 4 that shortcut would give $\dfrac{-3\sin t}{-2\cos t}$, which is positive at $t = \dfrac{\pi}{4}$: the wrong concavity.

Not checking the other derivative. For a horizontal tangent you need $\dfrac{dy}{dt} = 0$ and $\dfrac{dx}{dt} \ne 0$. If both are zero, the point needs more work: it could be a cusp or a corner, not a horizontal tangent.

Writing the tangent line with t in it. Substitute $t_0$ to get an actual point $(x_0, y_0)$ and an actual number for the slope. The final line uses only $x$ and $y$.

Calculator in degree mode. AP calculus uses radians. A degree-mode calculator silently gives wrong slopes for trig curves.

Practice

1. (Warm-up) For $x = 3t - 1$, $y = t^2 + 4$, find $\dfrac{dy}{dx}$ and its value at $t = 3$.

Solution

$$\frac{dy}{dx} = \frac{2t}{3}, \qquad \text{at } t = 3: \ \frac{6}{3} = 2$$

2. (Warm-up) For $x = t^3$, $y = t^2 - t$, find the slope of the curve at $t = 2$.

Solution

$$\frac{dy}{dx} = \frac{2t - 1}{3t^2}, \qquad \text{at } t = 2: \ \frac{3}{12} = \frac{1}{4}$$

3. (Warm-up) For $x = 2t + 1$, $y = 4t - 3$, show that the slope is the same for every $t$. What kind of curve is this?

Solution

$\dfrac{dy}{dx} = \dfrac{4}{2} = 2$ for every $t$, so the curve is a line with slope $2$.

Check by eliminating $t$: $t = \dfrac{x - 1}{2}$, so $y = 4 \cdot \dfrac{x - 1}{2} - 3 = 2x - 5$. That’s a line with slope $2$.

4. (Core) Find all points where $x = t^2 + 2t$, $y = t^3 - 12t$ has a horizontal tangent or a vertical tangent.

Solution

$\dfrac{dx}{dt} = 2t + 2$ and $\dfrac{dy}{dt} = 3t^2 - 12 = 3(t - 2)(t + 2)$.

Horizontal: $\dfrac{dy}{dt} = 0$ at $t = \pm 2$.

• $t = 2$: $\dfrac{dx}{dt} = 6 \ne 0$. Point $(4 + 4,\ 8 - 24) = (8, -16)$.
• $t = -2$: $\dfrac{dx}{dt} = -2 \ne 0$. Point $(4 - 4,\ -8 + 24) = (0, 16)$.

Vertical: $\dfrac{dx}{dt} = 0$ at $t = -1$, where $\dfrac{dy}{dt} = 3 - 12 = -9 \ne 0$. Point $(1 - 2,\ -1 + 12) = (-1, 11)$.

5. (Core) Find the tangent line to $x = e^t$, $y = te^t$ at $t = 0$.

Solution

$$\frac{dy}{dx} = \frac{e^t + te^t}{e^t} = 1 + t$$

At $t = 0$: slope $1$, point $(e^0, 0) = (1, 0)$. Tangent line: $y = x - 1$.

6. (Core) For $x = t^2$, $y = t^3 + 3t$ with $t \gt 0$, find $\dfrac{d^2y}{dx^2}$. For which $t \gt 0$ is the curve concave up?

Solution

$$\frac{dy}{dx} = \frac{3t^2 + 3}{2t} = \frac{3}{2}\left( t + \frac{1}{t} \right), \qquad \frac{d}{dt}\left( \frac{dy}{dx} \right) = \frac{3}{2}\left( 1 - \frac{1}{t^2} \right)$$

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

$$\frac{d^2y}{dx^2} = \frac{3(t^2 - 1)}{4t^3}$$

For $t \gt 0$ the denominator is positive, so the sign matches $t^2 - 1$: concave up for $t \gt 1$ (and concave down for $0 \lt t \lt 1$).

7. (Core) The curve $x = t - \sin t$, $y = 1 - \cos t$ is called a cycloid (radians).

• (a) Find the tangent line at $t = \dfrac{\pi}{2}$.
• (b) Find the point where the tangent is horizontal for $0 \lt t \lt 2\pi$.

Solution

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

(a) At $t = \dfrac{\pi}{2}$: slope $\dfrac{1}{1 - 0} = 1$, point $\left( \dfrac{\pi}{2} - 1,\ 1 \right)$. Tangent line: $y - 1 = x - \left( \dfrac{\pi}{2} - 1 \right)$, or $y = x + 2 - \dfrac{\pi}{2}$.

(b) $\sin t = 0$ at $t = \pi$ in this interval, and there $\dfrac{dx}{dt} = 1 - (-1) = 2 \ne 0$. The horizontal tangent is at $(\pi - 0,\ 1 - (-1)) = (\pi, 2)$, the top of the arch.

8. (Challenge) For $x = t^2$, $y = t^3$, show that $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ are both $0$ at $t = 0$. Then find $\displaystyle\lim_{t \to 0} \frac{dy}{dx}$ and describe the tangent direction at the origin.

Solution

$\dfrac{dx}{dt} = 2t$ and $\dfrac{dy}{dt} = 3t^2$ are both $0$ at $t = 0$, so the table gives no answer. For $t \ne 0$,

$$\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}, \qquad \lim_{t \to 0} \frac{3t}{2} = 0$$

The slopes on both sides approach $0$, so the curve comes into the origin tangent to the $x$-axis. (The curve has a sharp point there, called a cusp: the branch for $t \gt 0$ is above the $x$-axis and the branch for $t \lt 0$ is below, both touching it at the origin.)

9. (Challenge) The curve $x = t^2$, $y = t^3 - 3t$ from Example 2 crosses itself at $(3, 0)$. Find the two values of $t$ that give this point, and the equations of both tangent lines there.

Solution

$y = 0$ means $t(t^2 - 3) = 0$, so $t = 0$ or $t = \pm\sqrt{3}$. We need $x = t^2 = 3$, so $t = \pm\sqrt{3}$.

$$\frac{dy}{dx} = \frac{3t^2 - 3}{2t} = \frac{6}{\pm 2\sqrt{3}} = \pm\sqrt{3}$$

At $t = \sqrt{3}$ the slope is $\sqrt{3}$: $y = \sqrt{3}(x - 3)$. At $t = -\sqrt{3}$ the slope is $-\sqrt{3}$: $y = -\sqrt{3}(x - 3)$.

Two different tangent lines at one point: the curve passes through $(3, 0)$ twice, heading in different directions each time.