2.5 Parametric Differentiation
Curves traced by a moving point — the slope is the ratio of the two rates, $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$.
Parametric equations describe curves that are not necessarily functions. A parameter — often $t$, thought of as time — is an independent variable on which both $x$ and $y$ depend. As $t$ increases, the point $\big(x(t),y(t)\big)$ traces out a path in the plane.
If $x$ and $y$ are continuous functions of $t$ on an interval $I$, the equations $x=x(t)$ and $y=y(t)$ are called parametric equations and $t$ is the parameter. The set of points obtained as $t$ varies over $I$ is the graph of the parametric equations, called a parametric curve or plane curve.
For example, the position of a baseball after it leaves a pitcher's hand can be written as $\big(x(t),y(t)\big)$. Calculus should then tell us the slope of its path, its speed, and more.
Derivatives of Parametric Equations
Consider the plane curve $x(t)=2t+3$, $y(t)=3t-4$, $-2\le t\le3$. It is a line segment from $(-1,-10)$ to $(9,5)$. Eliminating the parameter, $t=\dfrac{x-3}{2}$, so $$y=3\left(\frac{x-3}{2}\right)-4=\frac{3x}{2}-\frac{17}{2},$$ whose slope is $\dfrac{dy}{dx}=\dfrac32$. Now notice $x^\prime(t)=2$ and $y^\prime(t)=3$, and $\dfrac{dy/dt}{dx/dt}=\dfrac32$ as well. That is no coincidence.
📐 Theorem 2.15 — Derivative of parametric equations
Consider the plane curve defined by $x=x(t)$ and $y=y(t)$. Suppose that $x^\prime(t)$ and $y^\prime(t)$ exist, and assume that $x^\prime(t)\neq0$. Then
$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{y^\prime(t)}{x^\prime(t)}.$$
This is the chain rule rearranged: $\dfrac{dy}{dt}=\dfrac{dy}{dx}\cdot\dfrac{dx}{dt}$. Where $x^\prime(t)=0$ but $y^\prime(t)\neq0$, $\dfrac{dy}{dx}$ is undefined (the point is moving straight up or down); where $y^\prime(t)=0$ but $x^\prime(t)\neq0$, $\dfrac{dy}{dx}=0$ (the path is momentarily level).
✏️ Example 1 — Finding the derivative of a parametric curve
Calculate $\dfrac{dy}{dx}$ for each plane curve.
- a $x(t)=t^2-3$, $y(t)=2t-1$
- b $x(t)=2t+1$, $y(t)=t^3-3t+4$
- c $x(t)=5\cos t$, $y(t)=5\sin t$
💡 Show solution▶
(a) $x^\prime(t)=2t$ and $y^\prime(t)=2$, so $$\frac{dy}{dx}=\frac{2}{2t}=\frac1t.$$
(b) $x^\prime(t)=2$ and $y^\prime(t)=3t^2-3$, so $$\frac{dy}{dx}=\frac{3t^2-3}{2}.$$
(c) $x^\prime(t)=-5\sin t$ and $y^\prime(t)=5\cos t$, so $$\frac{dy}{dx}=\frac{5\cos t}{-5\sin t}=-\cot t.$$ This curve is the circle of radius $5$; at $t=0$ the slope is undefined because $x^\prime(0)=0$ — the point is moving straight up.
Second-Order Derivatives
The second derivative is the derivative of the first derivative. Since $\dfrac{dy}{dx}$ is itself a function of $t$, apply Theorem 2.15 to it — replace $y$ by $\dfrac{dy}{dx}$:
$$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{(d/dt)(dy/dx)}{dx/dt}.$$⚠️ A common slip
The second derivative is not $\dfrac{y^{\prime\prime}(t)}{x^{\prime\prime}(t)}$. Differentiate $\dfrac{dy}{dx}$ with respect to $t$, then divide by $\dfrac{dx}{dt}$ once more.
✏️ Example 2 — Finding a second derivative
Calculate $\dfrac{d^2y}{dx^2}$ for the plane curve $x(t)=t^2-3$, $y(t)=2t-1$, $-3\le t\le4$.
💡 Show solution▶
Step 1 From Example 1(a), $\dfrac{dy}{dx}=\dfrac{2}{2t}=\dfrac1t$.
Step 2 Differentiate with respect to $t$ and divide by $\dfrac{dx}{dt}=2t$: $$\frac{d^2y}{dx^2}=\frac{(d/dt)(1/t)}{2t}=\frac{-t^{-2}}{2t}=-\frac{1}{2t^3}.$$
Meaning For $t>0$ the curve bends down; for $t<0$ it bends up — exactly what the sideways parabola $x=\tfrac{(y+1)^2}{4}-3$ does above and below its vertex.
Exercises
I
Each set of parametric equations represents a line. Without eliminating the parameter, find the slope of each line.
- 1 $x=3+t$, $y=1-t$
- 2 $x=8+2t$, $y=1$
- 3 $x=4-3t$, $y=-2+6t$
- 4 $x=-5t+7$, $y=3t-1$
Watch the rates
$t$—
$(x, y)$—
rates—
$dy/dx$—
💡 Show solutions▶
1 $\dfrac{dy}{dx}=\dfrac{-1}{1}=-1$.
2 $\dfrac{dy}{dx}=\dfrac{0}{2}=0$ — a horizontal line.
3 $\dfrac{dy}{dx}=\dfrac{6}{-3}=-2$.
4 $\dfrac{dy}{dx}=\dfrac{3}{-5}=-\dfrac35$.
Why For a line, $x^\prime$ and $y^\prime$ are constants, so their ratio — the slope — never changes.
II
Find $\dfrac{d^2y}{dx^2}$.
- 1 $x=t^4-1$, $y=t-t^2$
- 2 $x=\sin(\pi t)$, $y=\cos(\pi t)$
- 3 $x=e^{-t}$, $y=te^{2t}$
- 4 $x=t^2-4t$, $y=2t^3-6t$
See the curves
$t$—
$(x, y)$—
rates—
$dy/dx$—
💡 Show solutions▶
1 $\dfrac{dy}{dx}=\dfrac{1-2t}{4t^3}=\dfrac14t^{-3}-\dfrac12t^{-2}$. Then $\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)=-\dfrac34t^{-4}+t^{-3}=\dfrac{4t-3}{4t^4}$, and dividing by $x^\prime=4t^3$: $$\frac{d^2y}{dx^2}=\frac{4t-3}{16t^7}.$$
2 $\dfrac{dy}{dx}=\dfrac{-\pi\sin\pi t}{\pi\cos\pi t}=-\tan(\pi t)$. Then $\dfrac{d}{dt}=-\pi\sec^2(\pi t)$, and dividing by $x^\prime=\pi\cos(\pi t)$: $$\frac{d^2y}{dx^2}=-\sec^3(\pi t).$$
3 $y^\prime=e^{2t}(1+2t)$ and $x^\prime=-e^{-t}$, so $\dfrac{dy}{dx}=-e^{3t}(1+2t)$. Then $\dfrac{d}{dt}=-e^{3t}(3+6t+2)=-e^{3t}(6t+5)$, and dividing by $-e^{-t}$: $$\frac{d^2y}{dx^2}=e^{4t}(6t+5).$$
4 $\dfrac{dy}{dx}=\dfrac{6t^2-6}{2t-4}=\dfrac{3\left(t^2-1\right)}{t-2}$. Quotient rule: $\dfrac{d}{dt}=\dfrac{3\left[2t(t-2)-\left(t^2-1\right)\right]}{(t-2)^2}=\dfrac{3\left(t^2-4t+1\right)}{(t-2)^2}$. Dividing by $x^\prime=2(t-2)$: $$\frac{d^2y}{dx^2}=\frac{3\left(t^2-4t+1\right)}{2(t-2)^3}.$$
III
Find $\dfrac{dy}{dx}$ at the value of the parameter.
- 1 $x=\cos t$, $y=\sin t$, $t=\tfrac{3\pi}{4}$
- 2 $x=\sqrt t$, $y=2t+4$, $t=9$
- 3 $x=4\cos(2\pi s)$, $y=3\sin(2\pi s)$, $s=-\tfrac14$
Jump to the given parameter
$t$—
$(x, y)$—
rates—
$dy/dx$—
💡 Show solutions▶
1 $\dfrac{dy}{dx}=\dfrac{\cos t}{-\sin t}=-\cot t$; at $t=\tfrac{3\pi}{4}$, $\cot\tfrac{3\pi}{4}=-1$, so $\dfrac{dy}{dx}=1$.
2 $\dfrac{dy}{dx}=\dfrac{2}{1/(2\sqrt t)}=4\sqrt t$; at $t=9$ this is $12$.
3 $\dfrac{dy}{dx}=\dfrac{6\pi\cos(2\pi s)}{-8\pi\sin(2\pi s)}=-\dfrac34\cot(2\pi s)$; at $s=-\tfrac14$, $\cot\left(-\tfrac\pi2\right)=0$, so $\dfrac{dy}{dx}=0$ — the bottom of the ellipse, at $(0,-3)$.
IV
Find $\dfrac{d^2y}{dx^2}$ at the given point without eliminating the parameter.
- 1 $x=\tfrac12t^2$, $y=\tfrac13t^3$, $t=2$
- 2 $x=\sqrt t$, $y=2t+4$, $t=1$
💡 Show solutions▶
1 $\dfrac{dy}{dx}=\dfrac{t^2}{t}=t$, so $\dfrac{d^2y}{dx^2}=\dfrac{1}{x^\prime}=\dfrac1t$, which at $t=2$ is $\dfrac12$.
2 $\dfrac{dy}{dx}=4\sqrt t$, so $\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)=\dfrac{2}{\sqrt t}$; dividing by $x^\prime=\dfrac{1}{2\sqrt t}$ gives $\dfrac{d^2y}{dx^2}=4$ for every $t$ — including $t=1$. (Indeed $y=2x^2+4$.)
V
Find $\dfrac{dy}{dx}$ for each parametric curve, and simplify.
- 1 $x=t^3-3t$, $y=t^2+2t$
- 2 $x=4\cos t$, $y=2\sin t$ (an ellipse)
- 3 $x=e^{2t}$, $y=e^{-t}$
- 4 $x=t-\sin t$, $y=1-\cos t$ (a cycloid)
- 5 $x=\ln t$, $y=t^2$, $t>0$
Watch the slope change
$t$—
$(x, y)$—
rates—
$dy/dx$—
💡 Show solutions▶
1 $x^\prime=3t^2-3$ and $y^\prime=2t+2$, so $$\frac{dy}{dx}=\frac{2(t+1)}{3(t-1)(t+1)}=\frac{2}{3(t-1)},\quad t\neq\pm1.$$ At $t=1$ the slope is undefined ($x^\prime=0$, $y^\prime=4$); at $t=-1$ both rates are $0$, so the formula gives no information there.
2 $x^\prime=-4\sin t$ and $y^\prime=2\cos t$, so $$\frac{dy}{dx}=\frac{2\cos t}{-4\sin t}=-\frac12\cot t.$$
3 $x^\prime=2e^{2t}$ and $y^\prime=-e^{-t}$, so $$\frac{dy}{dx}=\frac{-e^{-t}}{2e^{2t}}=-\frac12e^{-3t}.$$ It is negative for every $t$: the curve always falls from left to right.
4 $x^\prime=1-\cos t$ and $y^\prime=\sin t$, so $$\frac{dy}{dx}=\frac{\sin t}{1-\cos t},\quad t\neq2k\pi.$$ At $t=\pi$ (the top of the arch) this is $0$.
5 $x^\prime=\dfrac1t$ and $y^\prime=2t$, so $$\frac{dy}{dx}=\frac{2t}{1/t}=2t^2.$$ (Check: $t=e^x$, so $y=e^{2x}$ and $\dfrac{dy}{dx}=2e^{2x}=2t^2$.)
VI
Find $\dfrac{dy}{dx}$ at the given parameter, or find $\dfrac{d^2y}{dx^2}$ and decide the concavity as asked.
- 1 $x=t^2+1$, $y=t^3-t$: $\dfrac{dy}{dx}$ at $t=2$
- 2 $x=t-\sin t$, $y=1-\cos t$: $\dfrac{dy}{dx}$ at $t=\tfrac{\pi}{3}$
- 3 $x=3\cos t$, $y=3\sin t$: $\dfrac{d^2y}{dx^2}$, and the concavity at $t=\tfrac{\pi}{4}$
- 4 $x=t^2$, $y=t^3-3t$: $\dfrac{d^2y}{dx^2}$, and the concavity at $t=-1$ and at $t=1$
- 5 $x=e^t$, $y=te^t$: $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$; then eliminate the parameter to check both
Jump to the given t
$t$—
$(x, y)$—
rates—
$dy/dx$—
💡 Show solutions▶
1 $\dfrac{dy}{dx}=\dfrac{3t^2-1}{2t}$; at $t=2$ this is $\dfrac{12-1}{4}=\dfrac{11}{4}$.
2 $\dfrac{dy}{dx}=\dfrac{\sin t}{1-\cos t}$; at $t=\tfrac\pi3$, $\sin\tfrac\pi3=\tfrac{\sqrt3}{2}$ and $1-\cos\tfrac\pi3=\tfrac12$, so $\dfrac{dy}{dx}=\sqrt3$.
3 $\dfrac{dy}{dx}=\dfrac{3\cos t}{-3\sin t}=-\cot t$. Then $\dfrac{d}{dt}(-\cot t)=\csc^2t$, and dividing by $x^\prime=-3\sin t$: $$\frac{d^2y}{dx^2}=-\frac{\csc^3t}{3}.$$ At $t=\tfrac\pi4$, $\csc\tfrac\pi4=\sqrt2$, so $\dfrac{d^2y}{dx^2}=-\dfrac{2\sqrt2}{3}<0$: concave down (the upper half of the circle).
4 $\dfrac{dy}{dx}=\dfrac{3t^2-3}{2t}=\dfrac32t-\dfrac{3}{2t}$. Then $\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)=\dfrac32+\dfrac{3}{2t^2}=\dfrac{3\left(t^2+1\right)}{2t^2}$, and dividing by $x^\prime=2t$: $$\frac{d^2y}{dx^2}=\frac{3\left(t^2+1\right)}{4t^3}.$$ At $t=-1$ it equals $-\tfrac32$ (concave down); at $t=1$ it equals $\tfrac32$ (concave up). The sign follows the sign of $t$.
5 $x^\prime=e^t$, $y^\prime=e^t(1+t)$, so $\dfrac{dy}{dx}=1+t$. Then $\dfrac{d^2y}{dx^2}=\dfrac{1}{e^t}=e^{-t}>0$: concave up everywhere. Check: $t=\ln x$, so $y=x\ln x$, $y^\prime=\ln x+1=t+1$ ✓ and $y^{\prime\prime}=\dfrac1x=e^{-t}$ ✓.
VII
Checking, applying and thinking.
- 1 For $x=t+1$, $y=t^2-2t$, find $\dfrac{dy}{dx}$ in terms of $t$. Then eliminate the parameter, differentiate $y$ as a function of $x$, and show the two answers agree. Where is $\dfrac{dy}{dx}=0$?
- 2 For $x=3t$, $y=9t^2-2$, find $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ from the parametric formulas, and confirm them by eliminating $t$.
- 3 Projectile. A ball is thrown with speed $v_0=64$ ft/s at angle $\theta=30^\circ$, so its path is $x=v_0t\cos\theta=32\sqrt3\,t$, $y=v_0t\sin\theta-16t^2=32t-16t^2$ (feet, seconds). (a) Find $\dfrac{dy}{dx}$ as a function of $t$. (b) What is the slope of the path at launch? (c) When is the path level ($\dfrac{dy}{dx}=0$), and where is the ball then? (d) Find the slope when the ball lands, and the sign of $\dfrac{d^2y}{dx^2}$.
- 4 Conceptual. A student says that for $x=t^2$, $y=t^3$ ($t>0$) the second derivative is $\dfrac{y^{\prime\prime}(t)}{x^{\prime\prime}(t)}$. Compare this with the correct $\dfrac{d^2y}{dx^2}$ at $t=1$.
💡 Show solutions▶
1 $\dfrac{dy}{dx}=\dfrac{2t-2}{1}=2t-2$. Eliminating: $t=x-1$, so $y=(x-1)^2-2(x-1)=x^2-4x+3$ and $\dfrac{dy}{dx}=2x-4=2(t+1)-4=2t-2$ ✓. It is $0$ at $t=1$, the point $(2,-1)$ — the vertex of the parabola.
2 $\dfrac{dy}{dx}=\dfrac{18t}{3}=6t$ and $\dfrac{d^2y}{dx^2}=\dfrac{6}{3}=2$. Eliminating: $t=\tfrac x3$, so $y=x^2-2$, $y^\prime=2x=6t$ ✓ and $y^{\prime\prime}=2$ ✓.
3 (a) $x^\prime=32\sqrt3$, $y^\prime=32-32t$, so $\dfrac{dy}{dx}=\dfrac{32(1-t)}{32\sqrt3}=\dfrac{1-t}{\sqrt3}$. (b) At $t=0$: $\dfrac{1}{\sqrt3}=\tan30^\circ$ — the path leaves at the launch angle. (c) $\dfrac{dy}{dx}=0$ when $t=1$ s; then $x=32\sqrt3\approx55.4$ ft and $y=32-16=16$ ft, the highest point. (d) $y=0$ again at $t=2$ s, where $\dfrac{dy}{dx}=-\dfrac{1}{\sqrt3}$. Also $\dfrac{d^2y}{dx^2}=\dfrac{-1/\sqrt3}{32\sqrt3}=-\dfrac{1}{96}<0$: the path is concave down throughout.
4 The student's formula gives $\dfrac{6t}{2}=3t$, which is $3$ at $t=1$. Correctly, $\dfrac{dy}{dx}=\dfrac{3t^2}{2t}=\dfrac32t$, so $\dfrac{d^2y}{dx^2}=\dfrac{3/2}{2t}=\dfrac{3}{4t}$, which is $\dfrac34$ at $t=1$. (Check: $y=x^{3/2}$, $y^{\prime\prime}=\tfrac34x^{-1/2}=\tfrac{3}{4t}$ ✓.) Dividing second derivatives is not the chain rule.
🔬 Research–Teaching NexusResearch-based · optional, about 30 minutes
Research question. Where along the path of a thrown ball is the slope $\frac{dy}{dx}$ zero, and when does the ball reach its top?
Why it matters. Ball tracking in sports broadcasting and missile or drone guidance describe motion with parametric equations $x(t)$, $y(t)$.
What to do.
- Film a ball thrown across the camera view, with a metre stick in the frame for scale.
- Use Tracker (free) or step frame-by-frame to record $x$ and $y$ at about 10 times.
- Fit $x(t)=pt+q$ and $y(t)=at^2+bt+c$ in a spreadsheet or Desmos.
- Compute $\frac{dy}{dx}=\frac{y'(t)}{x'(t)}$ and find the time and position of the top ($y'(t)=0$). Compare with the video.
Tools: Phone video, Tracker or spreadsheet, Desmos
Hand in: A half-page note: your fitted equations, the predicted and observed top, and one source of error.
Going further. Anywhere: check whether your value of $2a$ is close to $-g\approx-9.8$ m/s$^2$.
Instructor note: Students should notice that $x'(t)$ is almost constant, so the top of the path is where $y'(t)=0$.
🧪 Self-check quiz
0/5 correct
Q1. For $x=t^2-3$, $y=2t-1$, $\dfrac{dy}{dx}$ is:
Q2. For a parametric curve, $\dfrac{dy}{dx}$ is undefined at a value of $t$ where:
Q3. For the circle $x=5\cos t$, $y=5\sin t$, $\dfrac{dy}{dx}=$
Q4. $\dfrac{d^2y}{dx^2}$ for a parametric curve equals:
Q5. For $x=2t$, $y=t^3$, the value of $\dfrac{dy}{dx}$ at $t=-1$ is: