Chapter 10 · Parametric Equations and Polar Coordinates

10.2 Calculus of Parametric Curves

Slopes, concavity, areas, lengths and surface areas computed straight from $x(t)$ and $y(t)$ — no need to solve for $y$ as a function of $x$.

In Section 10.1 we learned to describe curves with a parameter. Now we want to do calculus on them: find tangent lines, decide which way a curve bends, measure the area beneath it, and — most importantly — measure how long it is. For a cycloid, an ellipse or a projectile's path we could sometimes eliminate the parameter first, but the resulting Cartesian equations are often awkward (the cycloid's is hopeless) and they throw away the motion. It is far better to work with $x(t)$ and $y(t)$ directly.

Every formula in this section comes from one idea: a small change $dt$ in the parameter moves the point by $dx=x^\prime(t)\,dt$ horizontally and $dy=y^\prime(t)\,dt$ vertically. Slopes are ratios of these, areas are sums of $y\,dx$, and lengths are sums of $\sqrt{dx^2+dy^2}$. You already know the first of these from Calculus I, Section 2.5; the integral formulas generalise those of 6.1 and 6.4.

Slopes and Tangent Lines

📐 Recall — Slope of a parametric curve (Calculus I, Theorem 2.15)
If $x(t)$ and $y(t)$ are differentiable at $t$ and $x^\prime(t)\neq0$, then at the point $\big(x(t),y(t)\big)$ $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{y^\prime(t)}{x^\prime(t)}.$$

This is the chain rule $\dfrac{dy}{dt}=\dfrac{dy}{dx}\cdot\dfrac{dx}{dt}$ solved for $\dfrac{dy}{dx}$. The tangent line at parameter $t_0$ passes through $\big(x(t_0),y(t_0)\big)$ with this slope. (Revise the details in Section 2.5.)

The formula breaks down where $x^\prime(t)=0$, and that is exactly where the interesting geometry happens. Think of the point as a car on the curve with velocity $\big(x^\prime(t),y^\prime(t)\big)$: if it is moving purely sideways the tangent is horizontal; if purely up or down, vertical; and if it stops altogether, anything can happen.

📘 Definition 10.2 — Smooth parametric curve
A curve $x=x(t)$, $y=y(t)$, $a\le t\le b$, is smooth if $x^\prime$ and $y^\prime$ are continuous on $[a,b]$ and are never both zero there. A smooth curve has a well-defined tangent direction $\big(x^\prime(t),y^\prime(t)\big)$ at every point. A point where $x^\prime(t)=y^\prime(t)=0$ is a singular point; the curve may have a sharp cusp there, as the cycloid does where it meets the road.
🧭 Problem-solving strategy — Horizontal and vertical tangents
  1. Compute $x^\prime(t)$ and $y^\prime(t)$.
  2. Horizontal tangent: solve $y^\prime(t)=0$ and keep the solutions with $x^\prime(t)\neq0$.
  3. Vertical tangent: solve $x^\prime(t)=0$ and keep the solutions with $y^\prime(t)\neq0$.
  4. If both are zero at $t_0$, the test is inconclusive: study $\displaystyle\lim_{t\to t_0}\frac{y^\prime(t)}{x^\prime(t)}$ (often with a trigonometric identity or L'Hospital's rule) and sketch the curve near $t_0$.
  5. Convert each $t$ into the point $\big(x(t),y(t)\big)$ — the question is about points, not parameter values.
✏️ Example 1 — Tangents to a curve that crosses itself
Consider the curve $x=t^2$, $y=t^3-3t$, $t\in\mathbb R$.
  • a Find the tangent line at the point where $t=2$.
  • b Find the points with horizontal or vertical tangents.
  • c Show that the curve passes through $(3,0)$ twice, and find both tangent lines there.
Derivatives $x^\prime(t)=2t$, $y^\prime(t)=3t^2-3$, so $\dfrac{dy}{dx}=\dfrac{3t^2-3}{2t}$ for $t\neq0$.
(a) At $t=2$ the point is $(4,\,8-6)=(4,2)$ and the slope is $\dfrac{12-3}{4}=\dfrac94$. Tangent line: $$y-2=\frac94(x-4)\quad\Longleftrightarrow\quad y=\frac94x-7.$$
(b) Horizontal: $3t^2-3=0$ gives $t=\pm1$, where $x^\prime=\pm2\neq0$. The points are $t=1$: $(1,-2)$ and $t=-1$: $(1,2)$. Vertical: $2t=0$ gives $t=0$, where $y^\prime(0)=-3\neq0$. The point is $(0,0)$.
(c) $x=t^2=3$ when $t=\pm\sqrt3$, and $y=t\left(t^2-3\right)=0$ for both. So $(3,0)$ is reached at $t=\sqrt3$ and again at $t=-\sqrt3$. The slopes are $$\frac{3\cdot3-3}{2(\pm\sqrt3)}=\pm\frac{6}{2\sqrt3}=\pm\sqrt3,$$ giving two tangent lines $y=\sqrt3(x-3)$ and $y=-\sqrt3(x-3)$. A curve that crosses itself can have two tangents at one point — something no graph $y=f(x)$ can do.
✏️ Example 2 — Tangents and cusps of the cycloid
For the cycloid $x=\theta-\sin\theta$, $y=1-\cos\theta$ (radius $1$):
  • a find the tangent line at $\theta=\tfrac\pi3$;
  • b find where the tangent is horizontal, and describe what happens at $\theta=0,\pm2\pi,\dots$
Slope $x^\prime(\theta)=1-\cos\theta$, $y^\prime(\theta)=\sin\theta$, so where $\cos\theta\neq1$, $$\frac{dy}{dx}=\frac{\sin\theta}{1-\cos\theta}=\frac{2\sin\frac\theta2\cos\frac\theta2}{2\sin^2\frac\theta2}=\cot\frac\theta2.$$
(a) At $\theta=\tfrac\pi3$: the point is $\left(\tfrac\pi3-\tfrac{\sqrt3}{2},\ \tfrac12\right)\approx(0.181,\,0.5)$ and the slope is $\cot\tfrac\pi6=\sqrt3$. Tangent line: $$y-\tfrac12=\sqrt3\left(x-\tfrac\pi3+\tfrac{\sqrt3}{2}\right)\quad\Longleftrightarrow\quad y=\sqrt3\,x+2-\frac{\pi}{\sqrt3}.$$
(b) $y^\prime=\sin\theta=0$ with $x^\prime=1-\cos\theta\neq0$ means $\cos\theta=-1$: $\theta=\pi+2k\pi$, the tops of the arches $\big((2k+1)\pi,\,2\big)$. At $\theta=2k\pi$ both derivatives are $0$ — a singular point. There $\cot\frac\theta2\to+\infty$ as $\theta\to0^+$ and $\to-\infty$ as $\theta\to0^-$: the curve comes down vertically onto the road and leaves vertically upward. That is a cusp.
Interactive 1 — The Tangent Rider

The red tangent line rides along the curve in the direction of the velocity $\big(x^\prime(t),y^\prime(t)\big)$. Green squares mark horizontal tangents ($y^\prime=0$, $x^\prime\neq0$), orange diamonds vertical tangents ($x^\prime=0$, $y^\prime\neq0$), and red crosses singular points where both vanish. Press Next special point to jump between them.

$t$—
$(x,y)$—
$\big(x^\prime(t),\,y^\prime(t)\big)$—
$dy/dx$—
tangent is…—

Second Derivatives and Concavity

Because $\dfrac{dy}{dx}$ is itself a function of $t$, we differentiate it the same way — rate of change with respect to $t$, divided by $\dfrac{dx}{dt}$ (see Calculus I, 2.5): $$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}},\qquad\text{provided }x^\prime(t)\neq0.$$ As for graphs, the curve is concave up where $\dfrac{d^2y}{dx^2}\gt0$ and concave down where $\dfrac{d^2y}{dx^2}\lt0$.

⚠️ Two common slips

1. $\dfrac{d^2y}{dx^2}$ is not $\dfrac{y^{\prime\prime}(t)}{x^{\prime\prime}(t)}$. For the line $x=t^2$, $y=2t^2$ that ratio is $2$, yet a straight line has $\dfrac{d^2y}{dx^2}=0$.

2. Do not forget the final division by $\dfrac{dx}{dt}$. Differentiating $\dfrac{dy}{dx}$ with respect to $t$ only gives a rate per unit of $t$, not per unit of $x$.

✏️ Example 3 — Concavity of a parametric curve
For $x=t^2$, $y=t^3-3t$ (Example 1), find $\dfrac{d^2y}{dx^2}$ and determine where the curve is concave up and concave down.
Step 1 Simplify first: $\dfrac{dy}{dx}=\dfrac{3t^2-3}{2t}=\dfrac32t-\dfrac32t^{-1}$.
Step 2 Differentiate with respect to $t$: $\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)=\dfrac32+\dfrac{3}{2t^2}=\dfrac{3t^2+3}{2t^2}$.
Step 3 Divide by $\dfrac{dx}{dt}=2t$: $$\frac{d^2y}{dx^2}=\frac{3t^2+3}{2t^2}\cdot\frac{1}{2t}=\frac{3\left(t^2+1\right)}{4t^3}.$$
Meaning The numerator is always positive, so the sign is the sign of $t$: concave up for $t\gt0$ (the lower-right branch through $(1,-2)$) and concave down for $t\lt0$ (the upper branch through $(1,2)$). Check at $t=2$: $\dfrac{3\cdot5}{4\cdot8}=\dfrac{15}{32}\gt0$, and the curve near $(4,2)$ does lie above its tangent $y=\tfrac94x-7$.

Area under a Parametric Curve

For a graph, the area under $y=F(x)$ from $x=\alpha$ to $x=\beta$ is $\int_\alpha^\beta y\,dx$. If the same curve is traced by $x=x(t)$, $y=y(t)$, substitute $x=x(t)$, $dx=x^\prime(t)\,dt$: the limits $x=\alpha$, $x=\beta$ become the parameter values $t=a$, $t=b$, and the integral becomes $\int_a^b y(t)\,x^\prime(t)\,dt$. This works even when we cannot write $y$ as a function of $x$ explicitly.

📐 Theorem 10.3 — Area under a parametric curve
Suppose that $x^\prime$ is continuous on $[a,b]$ with $x^\prime(t)\ge0$ (equal to $0$ at no more than finitely many points), so that the curve $x=x(t)$, $y=y(t)$ is traced exactly once from left to right, and that $y$ is continuous with $y(t)\ge0$ on $[a,b]$. Then the area of the region between the curve and the $x$-axis is $$A=\int_a^b y(t)\,x^\prime(t)\,dt.$$ If instead the curve is traced from right to left ($x^\prime(t)\le0$), the area is $\displaystyle A=-\int_a^b y(t)\,x^\prime(t)\,dt=\int_b^a y(t)\,x^\prime(t)\,dt$.
✏️ Example 4 — The area under one arch of a cycloid
Find the area under one arch of the cycloid $x=r(\theta-\sin\theta)$, $y=r(1-\cos\theta)$, $0\le\theta\le2\pi$, and compare it with the area of the rolling circle.
Hypotheses $x^\prime(\theta)=r(1-\cos\theta)\ge0$, zero only at $\theta=0,2\pi$; and $y\ge0$. So Theorem 10.3 applies.
Set up $$A=\int_0^{2\pi}r(1-\cos\theta)\cdot r(1-\cos\theta)\,d\theta=r^2\int_0^{2\pi}\left(1-2\cos\theta+\cos^2\theta\right)d\theta.$$
Integrate Using $\cos^2\theta=\tfrac12(1+\cos2\theta)$ (Section 5.1): $$A=r^2\Big[\theta-2\sin\theta+\tfrac12\theta+\tfrac14\sin2\theta\Big]_0^{2\pi}=r^2\left(2\pi+\pi\right)=3\pi r^2.$$
Compare The area under one arch is exactly three times the area $\pi r^2$ of the wheel that drew it — a result first found in the 1600s, long before calculus was fully developed. Check with $r=1$: a midpoint sum with $1000$ strips gives $9.42478\approx3\pi$. ✓

Arc Length and Speed

To measure the length of a curve, approximate it by a polygon. Split $[a,b]$ into $n$ pieces with points $t_0\lt t_1\lt\dots\lt t_n$ and join the points $P_k=\big(x(t_k),y(t_k)\big)$ by straight segments. By the Mean Value Theorem, $x(t_k)-x(t_{k-1})=x^\prime(t_k^*)\,\Delta t$ and $y(t_k)-y(t_{k-1})=y^\prime(t_k^{**})\,\Delta t$ for some $t_k^*,t_k^{**}$ in the $k$-th piece, so $$|P_{k-1}P_k|=\sqrt{\big(x^\prime(t_k^*)\big)^2+\big(y^\prime(t_k^{**})\big)^2}\;\Delta t.$$ As $n\to\infty$ (and $x^\prime,y^\prime$ are continuous), the total length of the polygon approaches an integral. Interactive 2 lets you watch this happen.

📐 Theorem 10.4 — Arc length of a parametric curve
Suppose $x^\prime$ and $y^\prime$ are continuous on $[a,b]$ and the curve $x=x(t)$, $y=y(t)$ is traversed exactly once as $t$ increases from $a$ to $b$. Then the length of the curve is $$L=\int_a^b\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt=\int_a^b\sqrt{x^\prime(t)^2+y^\prime(t)^2}\,dt.$$ If $t$ is time, the integrand is the speed $\dfrac{ds}{dt}=\sqrt{x^\prime(t)^2+y^\prime(t)^2}$, and the integral is the distance travelled along the path. For a graph ($x=t$, $y=f(t)$) it reduces to $L=\int_a^b\sqrt{1+f^\prime(x)^2}\,dx$ from Section 6.4.
⚠️ “Exactly once” matters
The integral measures the distance the moving point travels. If the curve is traced twice, the integral counts it twice: for $x=\cos t$, $y=\sin t$, $0\le t\le4\pi$, the speed is $1$ and $\int_0^{4\pi}1\,dt=4\pi$ — but the circle has length $2\pi$. Likewise $x=\sin t$, $y=\sin^2t$ on $[0,2\pi]$ runs back and forth along a short arc; the integral gives the distance travelled, which is twice the arc's length (the arc from $(-1,1)$ to $(1,1)$ is covered once in each direction).
✏️ Example 5 — The circumference of a circle and the length of a cycloid arch
  • a Use Theorem 10.4 to find the circumference of the circle $x=r\cos t$, $y=r\sin t$, $0\le t\le2\pi$.
  • b Find the length of one arch of the cycloid $x=r(\theta-\sin\theta)$, $y=r(1-\cos\theta)$, $0\le\theta\le2\pi$.
(a) $x^\prime=-r\sin t$, $y^\prime=r\cos t$, so $\sqrt{x^{\prime2}+y^{\prime2}}=\sqrt{r^2\left(\sin^2t+\cos^2t\right)}=r$. The circle is traced once, and $$L=\int_0^{2\pi}r\,dt=2\pi r.$$ Reassuring: the formula reproduces the circumference.
(b) Set up $x^\prime=r(1-\cos\theta)$, $y^\prime=r\sin\theta$, so $$x^{\prime2}+y^{\prime2}=r^2\left(1-2\cos\theta+\cos^2\theta+\sin^2\theta\right)=2r^2(1-\cos\theta)=4r^2\sin^2\frac\theta2,$$ using the half-angle identity $1-\cos\theta=2\sin^2\frac\theta2$.
(b) Integrate For $0\le\theta\le2\pi$, $\sin\frac\theta2\ge0$, so the square root is $2r\sin\frac\theta2$ (no absolute value needed): $$L=\int_0^{2\pi}2r\sin\frac\theta2\,d\theta=\Big[-4r\cos\frac\theta2\Big]_0^{2\pi}=-4r(-1)+4r(1)=8r.$$
Meaning One arch is exactly $8r$ long, while the wheel moves forward only $2\pi r\approx6.28r$. A pebble stuck in a tyre travels $\tfrac{8}{2\pi}=\tfrac4\pi\approx1.27$ times as far as the bicycle! Interactive 2 shows inscribed polygons creeping up to $8$ when $r=1$.
Interactive 2 — Polygons Closing In on the Arc Length

Join $n+1$ equally spaced parameter points by straight segments. The polygon is always a little shorter than the curve (a straight segment is the shortest path), but as $n$ grows its length rises towards the exact value $\int_a^b\sqrt{x^{\prime2}+y^{\prime2}}\,dt$.

4
segments $n$—
polygon length—
exact length $L$—
$L-$ polygon—

Surface Area of Revolution

Rotate a small piece of the curve, of length $ds=\sqrt{x^{\prime2}+y^{\prime2}}\,dt$ and at height $y$, about the $x$-axis. It sweeps out a thin band of radius $y$, whose area is approximately (circumference) × (width) $=2\pi y\,ds$. Adding these bands, exactly as for graphs in Section 6.4, leads to the following.

📘 Definition 10.3 — Surface area of revolution for a parametric curve
Let $x^\prime$ and $y^\prime$ be continuous on $[a,b]$, and suppose the curve $x=x(t)$, $y=y(t)$ is traversed exactly once as $t$ increases from $a$ to $b$.
  • If $y(t)\ge0$ on $[a,b]$, the area of the surface obtained by rotating the curve about the $x$-axis is $$S=2\pi\int_a^b y(t)\sqrt{x^\prime(t)^2+y^\prime(t)^2}\,dt.$$
  • If $x(t)\ge0$ on $[a,b]$, the area of the surface obtained by rotating it about the $y$-axis is $$S=2\pi\int_a^b x(t)\sqrt{x^\prime(t)^2+y^\prime(t)^2}\,dt.$$
✏️ Example 6 — A sphere, and a cycloid arch spun about the road
  • a Show that the surface area of a sphere of radius $r$ is $4\pi r^2$ by rotating the semicircle $x=r\cos t$, $y=r\sin t$, $0\le t\le\pi$, about the $x$-axis.
  • b Find the area of the surface formed by rotating one arch of the cycloid $x=\theta-\sin\theta$, $y=1-\cos\theta$ about the $x$-axis.
(a) Here $y=r\sin t\ge0$ on $[0,\pi]$, the semicircle is traced once, and the speed is $r$ (Example 5a): $$S=2\pi\int_0^{\pi}r\sin t\cdot r\,dt=2\pi r^2\Big[-\cos t\Big]_0^{\pi}=2\pi r^2(1+1)=4\pi r^2.$$
(b) Set up From Example 5b with $r=1$, the speed is $2\sin\frac\theta2$, and $y=1-\cos\theta=2\sin^2\frac\theta2$. So $$S=2\pi\int_0^{2\pi}2\sin^2\frac\theta2\cdot2\sin\frac\theta2\,d\theta=8\pi\int_0^{2\pi}\sin^3\frac\theta2\,d\theta.$$
(b) Integrate Let $u=\frac\theta2$, $d\theta=2\,du$: $\;S=16\pi\displaystyle\int_0^{\pi}\sin^3u\,du$. With $\sin^3u=\left(1-\cos^2u\right)\sin u$ and $w=\cos u$ (Section 5.1): $$\int_0^\pi\sin^3u\,du=\Big[-\cos u+\tfrac13\cos^3u\Big]_0^{\pi}=\left(1-\tfrac13\right)-\left(-1+\tfrac13\right)=\tfrac43.$$ Hence $S=16\pi\cdot\tfrac43=\dfrac{64\pi}{3}\approx67.02$ square units.
Check A numerical integration of $2\pi(1-\cos\theta)\sqrt{(1-\cos\theta)^2+\sin^2\theta}$ over $[0,2\pi]$ gives $67.0206$. ✓

Exercises

I
Slopes and tangent lines.
  • 1 Find the tangent line to $x=3t+1$, $y=t^2-4$ at $t=3$.
  • 2 Find the tangent line to $x=t^2$, $y=t^3$ at $t=2$.
  • 3 Find the tangent line to the ellipse $x=2\cos t$, $y=3\sin t$ at $t=\tfrac\pi4$.
  • 4 Find the tangent line to $x=e^t$, $y=t^2$ at $t=1$.
  • 5 Find the tangent line to the cycloid $x=\theta-\sin\theta$, $y=1-\cos\theta$ at $\theta=\tfrac\pi2$.
  • 6 Find all points with horizontal or vertical tangents on $x=t^3-3t$, $y=t^2-4$.
  • 7 Find all points with horizontal or vertical tangents on $x=2+3\cos t$, $y=-1+2\sin t$, $0\le t\lt2\pi$.
  • 8 Find all points with horizontal or vertical tangents on $x=t^2-2t$, $y=t^3-12t$.
  • 9 For the astroid $x=\cos^3t$, $y=\sin^3t$, show that $\dfrac{dy}{dx}=-\tan t$ where it is defined, find the tangent line at $t=\tfrac\pi4$, and describe the curve at $t=0$.
  • 10 Show that the curve $x=t^2$, $y=t^3-4t$ passes through $(4,0)$ twice and find both tangent lines there.
1 $\dfrac{dy}{dx}=\dfrac{2t}{3}$. At $t=3$: point $(10,5)$, slope $2$. Tangent: $y=2x-15$.
2 $\dfrac{dy}{dx}=\dfrac{3t^2}{2t}=\dfrac{3t}{2}$. At $t=2$: point $(4,8)$, slope $3$. Tangent: $y=3x-4$.
3 $\dfrac{dy}{dx}=\dfrac{3\cos t}{-2\sin t}=-\tfrac32\cot t$. At $t=\tfrac\pi4$: point $\left(\sqrt2,\tfrac{3\sqrt2}{2}\right)$, slope $-\tfrac32$. Tangent: $y=-\tfrac32x+3\sqrt2$.
4 $\dfrac{dy}{dx}=\dfrac{2t}{e^t}$. At $t=1$: point $(e,1)$, slope $\tfrac2e$. Tangent: $y-1=\tfrac2e(x-e)$, i.e. $y=\tfrac2ex-1$.
5 $\dfrac{dy}{dx}=\dfrac{\sin\theta}{1-\cos\theta}$; at $\theta=\tfrac\pi2$ this is $\tfrac11=1$. Point $\left(\tfrac\pi2-1,\,1\right)$. Tangent: $y=x+2-\tfrac\pi2$.
6 $x^\prime=3t^2-3$, $y^\prime=2t$. Horizontal: $t=0$ (where $x^\prime=-3\neq0$): point $(0,-4)$. Vertical: $t=\pm1$ (where $y^\prime=\pm2\neq0$): $t=1$ gives $(-2,-3)$, $t=-1$ gives $(2,-3)$.
7 $x^\prime=-3\sin t$, $y^\prime=2\cos t$. Horizontal: $\cos t=0$, $t=\tfrac\pi2,\tfrac{3\pi}2$: points $(2,1)$ and $(2,-3)$. Vertical: $\sin t=0$, $t=0,\pi$: points $(5,-1)$ and $(-1,-1)$. (The top, bottom, right and left of the ellipse.)
8 $x^\prime=2t-2$, $y^\prime=3t^2-12$. Horizontal: $t=\pm2$: $t=2$ gives $(0,-16)$, $t=-2$ gives $(8,16)$. Vertical: $t=1$ (where $y^\prime=-9\neq0$): $(-1,-11)$.
9 $x^\prime=-3\cos^2t\sin t$, $y^\prime=3\sin^2t\cos t$, so $\dfrac{dy}{dx}=\dfrac{3\sin^2t\cos t}{-3\cos^2t\sin t}=-\tan t$ when $\sin t\cos t\neq0$. At $t=\tfrac\pi4$: point $\left(\tfrac{\sqrt2}{4},\tfrac{\sqrt2}{4}\right)$, slope $-1$, tangent $y=-x+\tfrac{\sqrt2}{2}$. At $t=0$ both derivatives vanish (a singular point at $(1,0)$); $-\tan t\to0$, so the two halves of the curve arrive tangent to the $x$-axis from above and below and meet in a sharp cusp.
10 $x=4$ at $t=\pm2$, and $y=t\left(t^2-4\right)=0$ for both. $\dfrac{dy}{dx}=\dfrac{3t^2-4}{2t}$: at $t=2$ it is $\tfrac84=2$; at $t=-2$ it is $-2$. Tangent lines: $y=2x-8$ and $y=-2x+8$.
II
Find $\dfrac{d^2y}{dx^2}$ and determine the values of the parameter for which the curve is concave up.
  • 1 $x=t^2$, $y=t^3$, $t\neq0$
  • 2 $x=2t+1$, $y=t^3-3t$
  • 3 $x=\cos t$, $y=\sin t$, $0\lt t\lt2\pi$, $t\neq\pi$
  • 4 $x=e^t$, $y=te^{-t}$
  • 5 The cycloid $x=r(\theta-\sin\theta)$, $y=r(1-\cos\theta)$, $0\lt\theta\lt2\pi$
1 $\dfrac{dy}{dx}=\dfrac{3t}{2}$; $\dfrac{d}{dt}=\dfrac32$; divide by $2t$: $\dfrac{d^2y}{dx^2}=\dfrac{3}{4t}$. Concave up for $t\gt0$.
2 $\dfrac{dy}{dx}=\dfrac{3t^2-3}{2}$; $\dfrac{d}{dt}=3t$; divide by $2$: $\dfrac{d^2y}{dx^2}=\dfrac{3t}{2}$. Concave up for $t\gt0$.
3 $\dfrac{dy}{dx}=-\cot t$; $\dfrac{d}{dt}=\csc^2t$; divide by $-\sin t$: $\dfrac{d^2y}{dx^2}=-\csc^3t=-\dfrac{1}{\sin^3t}$. Concave up when $\sin t\lt0$: $\pi\lt t\lt2\pi$ (the lower half of the circle) — as expected.
4 $\dfrac{dy}{dx}=\dfrac{(1-t)e^{-t}}{e^t}=(1-t)e^{-2t}$; $\dfrac{d}{dt}=-e^{-2t}-2(1-t)e^{-2t}=(2t-3)e^{-2t}$; divide by $e^t$: $\dfrac{d^2y}{dx^2}=(2t-3)e^{-3t}$. Concave up for $t\gt\tfrac32$.
5 $\dfrac{dy}{dx}=\dfrac{\sin\theta}{1-\cos\theta}$; $\dfrac{d}{d\theta}=\dfrac{\cos\theta(1-\cos\theta)-\sin^2\theta}{(1-\cos\theta)^2}=\dfrac{\cos\theta-1}{(1-\cos\theta)^2}=\dfrac{-1}{1-\cos\theta}$; divide by $r(1-\cos\theta)$: $\dfrac{d^2y}{dx^2}=-\dfrac{1}{r(1-\cos\theta)^2}\lt0$. Never concave up: every arch is concave down.
III
Areas. Check the hypotheses of Theorem 10.3 (direction of tracing, sign of $y$) before integrating.
  • 1 Find the area under $x=t^2$, $y=t+1$, $0\le t\le2$, and check your answer by eliminating the parameter.
  • 2 Find the area enclosed by the ellipse $x=a\cos t$, $y=b\sin t$ ($a,b\gt0$).
  • 3 Find the area under $x=e^t$, $y=e^{-t}$, $0\le t\le1$.
  • 4 Find the area enclosed by the astroid $x=\cos^3t$, $y=\sin^3t$. (Use symmetry, and $\displaystyle\int_0^{\pi/2}\sin^4t\cos^2t\,dt=\tfrac{\pi}{32}$.)
  • 5 Applied. The entrance of a tunnel is shaped like one arch of a cycloid traced by a wheel of radius $2$ m. Find its width, its height and the area of the opening.
1 $x^\prime=2t\ge0$, $y\gt0$: $A=\displaystyle\int_0^2(t+1)\,2t\,dt=\Big[\tfrac23t^3+t^2\Big]_0^2=\tfrac{16}{3}+4=\tfrac{28}{3}$. Check: $y=\sqrt x+1$ on $[0,4]$, $\displaystyle\int_0^4\left(\sqrt x+1\right)dx=\tfrac23\cdot8+4=\tfrac{28}3$. ✓
2 The upper half ($0\le t\le\pi$) is traced right to left, so its area is $-\displaystyle\int_0^\pi b\sin t\,(-a\sin t)\,dt=ab\int_0^\pi\sin^2t\,dt=\tfrac{\pi ab}{2}$. By symmetry the whole ellipse encloses $\pi ab$ (a circle when $a=b=r$: $\pi r^2$ ✓).
3 $x^\prime=e^t\gt0$, $y\gt0$: $A=\displaystyle\int_0^1e^{-t}e^t\,dt=1$. (It is the area under $y=\tfrac1x$ from $1$ to $e$, which is $\ln e=1$. ✓)
4 The first-quadrant arc ($0\le t\le\tfrac\pi2$) runs from $(1,0)$ to $(0,1)$, right to left, with $x^\prime=-3\cos^2t\sin t$. Quarter area $=-\displaystyle\int_0^{\pi/2}\sin^3t\left(-3\cos^2t\sin t\right)dt=3\int_0^{\pi/2}\sin^4t\cos^2t\,dt=\tfrac{3\pi}{32}$. Total: $4\cdot\tfrac{3\pi}{32}=\tfrac{3\pi}{8}\approx1.178$.
5 Width $=2\pi r=4\pi\approx12.57$ m, height $=2r=4$ m, area $=3\pi r^2=12\pi\approx37.70$ m² (Example 4).
IV
Arc length, speed and surface area.
  • 1 Find the length of $x=3t^2$, $y=2t^3$, $0\le t\le1$.
  • 2 Find the length of the spiral arc $x=e^t\cos t$, $y=e^t\sin t$, $0\le t\le1$.
  • 3 Find the length of $x=1+2t$, $y=3-t$, $0\le t\le4$, and check with the distance formula.
  • 4 Find the total length of the astroid $x=\cos^3t$, $y=\sin^3t$, $0\le t\le2\pi$.
  • 5 Conceptual. For $x=\cos t$, $y=\sin t$, $0\le t\le4\pi$, the arc-length integral equals $4\pi$. What is the length of the curve, and why do they differ?
  • 6 A particle moves on $x=t^2$, $y=t^3$ (metres, seconds). Find its speed at $t=1$ and the distance it travels for $0\le t\le1$.
  • 7 Applied. For the football of Section 10.1, Example 5 ($x=10\sqrt3\,t$, $y=10t-4.9t^2$), find its speed at the top of its flight and at landing, and the length of its path. (Use $\displaystyle\int_0^c\sqrt{u^2+a^2}\,du=\tfrac c2\sqrt{c^2+a^2}+\tfrac{a^2}{2}\ln\frac{c+\sqrt{c^2+a^2}}{a}$.)
  • 8 Rotate the segment $x=3t$, $y=4t$, $0\le t\le1$, about the $x$-axis and find the surface area. Check with the cone formula $\pi r\ell$.
  • 9 Rotate $x=t^2$, $y=2t$, $0\le t\le1$, about the $x$-axis and find the surface area.
  • 10 Applied. A pebble is stuck in the tread of a bicycle tyre of radius $0.3$ m. How far does the pebble travel while the wheel makes one revolution, and how far does the bicycle move?
Check: accumulate the integral numerically as $t$ runs, and compare with your answer
$t$—
integral from $a$ to $t$—
numerical total—
exact answer—
1 $x^\prime=6t$, $y^\prime=6t^2$, speed $=\sqrt{36t^2+36t^4}=6t\sqrt{1+t^2}$ for $t\ge0$. $L=\displaystyle\int_0^16t\sqrt{1+t^2}\,dt=\Big[2\left(1+t^2\right)^{3/2}\Big]_0^1=4\sqrt2-2\approx3.657$.
2 $x^\prime=e^t(\cos t-\sin t)$, $y^\prime=e^t(\sin t+\cos t)$; $x^{\prime2}+y^{\prime2}=e^{2t}\cdot2\left(\cos^2t+\sin^2t\right)=2e^{2t}$. $L=\displaystyle\int_0^1\sqrt2\,e^t\,dt=\sqrt2(e-1)\approx2.430$.
3 Speed $=\sqrt{4+1}=\sqrt5$, so $L=4\sqrt5\approx8.944$. Distance from $(1,3)$ to $(9,-1)$: $\sqrt{64+16}=\sqrt{80}=4\sqrt5$. ✓
4 $x^{\prime2}+y^{\prime2}=9\cos^4t\sin^2t+9\sin^4t\cos^2t=9\sin^2t\cos^2t$, so the speed is $3|\sin t\cos t|$. By symmetry, $L=4\displaystyle\int_0^{\pi/2}3\sin t\cos t\,dt=4\cdot\tfrac32=6$.
5 The curve is the unit circle, of length $2\pi$. Theorem 10.4 requires the curve to be traversed exactly once; on $[0,4\pi]$ it is traversed twice, so the integral gives the distance travelled, $4\pi$.
6 Speed $=\sqrt{(2t)^2+\left(3t^2\right)^2}$; at $t=1$: $\sqrt{4+9}=\sqrt{13}\approx3.61$ m/s. Distance $=\displaystyle\int_0^1t\sqrt{4+9t^2}\,dt=\Big[\tfrac{1}{27}\left(4+9t^2\right)^{3/2}\Big]_0^1=\dfrac{13\sqrt{13}-8}{27}\approx1.440$ m.
7 Speed $=\sqrt{300+(10-9.8t)^2}$. At the top ($t=\tfrac{50}{49}$): $\sqrt{300}=10\sqrt3\approx17.3$ m/s. At landing ($t=\tfrac{100}{49}$): $\sqrt{300+100}=20$ m/s (the launch speed, by symmetry). Path length: with $u=10-9.8t$, $L=\dfrac{1}{9.8}\displaystyle\int_{-10}^{10}\sqrt{u^2+300}\,du=\dfrac{2}{9.8}\left(5\cdot20+150\ln\dfrac{10+20}{\sqrt{300}}\right)=\dfrac{2}{9.8}\left(100+75\ln3\right)\approx37.2$ m — about $1.9$ m more than the $35.35$ m range.
8 Speed $=5$, $y=4t\ge0$: $S=2\pi\displaystyle\int_0^14t\cdot5\,dt=20\pi\approx62.83$. The cone has radius $r=4$ and slant height $\ell=5$: $\pi\cdot4\cdot5=20\pi$. ✓
9 $x^\prime=2t$, $y^\prime=2$, speed $=2\sqrt{t^2+1}$, $y=2t\ge0$: $S=2\pi\displaystyle\int_0^12t\cdot2\sqrt{t^2+1}\,dt=8\pi\Big[\tfrac13\left(t^2+1\right)^{3/2}\Big]_0^1=\dfrac{8\pi}{3}\left(2\sqrt2-1\right)\approx15.32$.
10 The pebble traces one arch of a cycloid with $r=0.3$ m, so it travels $8r=2.4$ m, while the bicycle moves $2\pi r\approx1.885$ m. (The ratio $\tfrac4\pi\approx1.27$ does not depend on the size of the wheel.)
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How long is an ellipse? There is no elementary formula for its perimeter — how good are the shortcuts that engineers and astronomers use?

Why it matters. Elliptical shapes appear in satellite and planetary orbits, running tracks, gear and pipe cross-sections, and stadium roofs; their perimeters set orbit path lengths, material quantities and costs.

What to do.

  1. For $x=a\cos t$, $y=b\sin t$, show with Theorem 10.4 that $L=\displaystyle\int_0^{2\pi}\sqrt{a^2\sin^2t+b^2\cos^2t}\,dt$, and explain why the substitution tricks of Chapter 5 fail.
  2. Evaluate $L$ numerically with Simpson's rule (Section 5.4) for $(a,b)=(3,2)$ and $(3,1)$; double $n$ until four decimal places stop changing.
  3. Compare with two shortcuts: the “average radius” formula $2\pi\sqrt{\tfrac{a^2+b^2}{2}}$ and Ramanujan's formula $\pi\left[3(a+b)-\sqrt{(3a+b)(a+3b)}\right]$. Record the percentage errors.
  4. Apply the best formula to Earth's orbit ($a\approx149.6$ million km, eccentricity $0.0167$, $b=a\sqrt{1-e^2}$) and estimate Earth's average orbital speed in km/s.

Tools: Spreadsheet or a short program; Interactive 2 for a visual check.

Hand in: A one-page note with a table of exact (numerical) and approximate perimeters, the percentage errors, and a recommendation of which shortcut to use when.

Going further. Find the perimeter of a real elliptical structure in your country (a stadium, a roundabout or a park) from a map, and estimate the cost of fencing it.

Instructor note: for $(3,2)$, $L\approx15.8654$ (Ramanujan $15.8654$; average-radius formula $16.0190$, about $1\%$ high); for $(3,1)$, $L\approx13.3649$ (Ramanujan $13.3644$, average-radius $14.0496$, about $5\%$ high). Earth's orbit is about $9.40\times10^8$ km long, giving an average speed of roughly $29.8$ km/s.

🧪 Self-check quiz

0/5 correct

Q1. The curve $x=t^2$, $y=t^3-3t$ has horizontal tangents where

Q2. The length of one arch of the cycloid traced by a wheel of radius $r$ is

Q3. For $x=\cos t$, $y=\sin t$, $0\le t\le4\pi$, the integral $\int_0^{4\pi}\sqrt{x^{\prime2}+y^{\prime2}}\,dt=4\pi$. This is twice the circumference because

Q4. For a parametric curve, $\dfrac{d^2y}{dx^2}$ equals

Q5. A point moves on $x=3\cos2t$, $y=3\sin2t$. Its speed is

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