Averages, Arc Length, Speed and Surface Area

5.1 The average value of a function

Suppose a cable is suspended between two poles, so that each end of the cable is $3$ metres above the ground and the lowest point of the cable is $2$ metres above the ground. What is the average height of the cable above the ground? Clearly the answer is somewhere between $2$ and $3$ metres, but 'somewhere between' is not an answer. Recall from MATH1131 that a suspended cable is always the graph of a function of the form

$$f(x)=\frac{1}{c}\cosh (cx)$$

over some interval $[a,b]$, where $c$ is a constant depending on the tension in and the mass of the cable, and where the coordinate system has been chosen conveniently. So the question becomes: what is the average value of $f$ on $[a,b]$?

Averaging finitely many numbers is easy, but a function takes uncountably many values on $[a,b]$, so we have to say what we even mean. Divide $[a,b]$ into $n$ subintervals of equal length, sample the height of the graph in the $k$​th subinterval by choosing a point $c_k$ in it, and take the ordinary arithmetic mean of the sampled heights:

$$a_n=\frac{1}{n}(f(c_1)+f(c_2)+\cdots +f(c_n))=\frac{1}{n}\sum _{k=1}^{n}f(c_k).$$

As $n$ increases, $a_n$ should get closer and closer to what we intuitively call the average height of the graph. Now comes the trick that turns this into an integral: multiply and divide by $b-a$,

$$\begin{array}{rl}{a}_n& =\frac{1}{n}\sum _{k=1}^{n}f(c_k)\\ & =\frac{1}{n}\sum _{k=1}^{n}f(c_k)\cdot \frac{b-a}{b-a}\\ & =\frac{1}{b-a}\sum _{k=1}^{n}f(c_k)\frac{b-a}{n}.\end{array}$$

Notice what the last sum is: $f(c_k)$ is the height and $\frac{b-a}{n}$ is the width of the $k$​th rectangle in the usual picture, so it is a Riemann sum for $f$ on $[a,b]$. Hence, provided $f$ is Riemann integrable,

$$\underset{n\to \infty }{\lim }{a}_n=\frac{1}{b-a}\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f(c_k)\frac{b-a}{n}=\frac{1}{b-a}{\int }_a^{b}f(x)dx.$$

This limit is what we take as the definition.

Rearranging this formula gives a second way to read it: $\bar{f}$ is the unique constant with

$$(b-a)\bar{f}={\int }_a^{b}f(x)dx.$$

Geometrically, $\bar{f}$ is the unique $y$-value for which the rectangle of width $b-a$ and height $\bar{f}$ has the same area as the region under the graph of $f$. Basically, the average value is the height you would flatten the graph down to if you were allowed to redistribute the area but not create or destroy any of it.

Because the integral measures signed area, so does the average value. For example the average value of $\sin$ over $[0,2\pi ]$ is $0$, not because $\sin$ is small but because the hump below the axis exactly cancels the hump above it. Over half a period the answer is much more interesting.

Example. Find the average value of $\sin$ on $[0,\pi ]$.
By the definition,

$$\begin{array}{rl}\bar{f}& =\frac{1}{\pi -0}{\int }_0^{\pi }\sin xdx\\ & =\frac{1}{\pi }[-\cos x{]}_0^{\pi }\\ & =\frac{1}{\pi }(1-(-1))\\ & =\frac{2}{\pi }.\end{array}$$

Therefore the average value of $\sin$ on $[0,\pi ]$ is $\frac{2}{\pi }\approx 0.6366$. As a sanity check, the graph sits between $0$ and $1$ and spends most of its time up near the top, so an average a bit under $\frac{2}{3}$ is exactly what you would expect; a common wrong guess is $\frac{1}{2}$, which is the average of the endpoint and peak values, not of the function.

Example. The cable illustrated at the start of this section is the curve

$$y=2\cosh (x/2),-a\le x\le a,$$

where the $x$-axis runs along the ground, the $y$-axis passes through the vertex of the curve, and $a=2{\cosh }^{-1}(3/2)$. Find, to the nearest centimetre, the average height of the cable above the ground.
Set $f(x)=2\cosh (x/2)$. The interval has length $a-(-a)=2a$, so

$$\begin{array}{rl}\bar{f}& =\frac{1}{2a}{\int }_{-a}^{a}2\cosh (x/2)dx\\ & =\frac{2}{2a}{\int }_0^{a}2\cosh (x/2)dx(\text{since }\cosh \text{ is even})\\ & =\frac{1}{a}[4\sinh (x/2){]}_0^a\\ & =\frac{4}{a}\sinh (a/2)\\ & =\frac{2}{{\cosh }^{-1}(3/2)}\sinh ({\cosh }^{-1}(3/2)).\end{array}$$

The remaining $\sinh$ of an inverse $\cosh$ is unwound with the identity ${\cosh }^{2}t-{\sinh }^{2}t=1$:

$$\sinh ({\cosh }^{-1}(3/2))=\sqrt{{\cosh }^2({\cosh }^{-1}(3/2))-1}=\sqrt{\frac{9}{4}-1}=\frac{\sqrt{5}}{2},$$

where the positive root is correct because ${\cosh }^{-1}(3/2)>0$ and $\sinh$ is positive there. Hence

$$\bar{f}=\frac{2}{{\cosh }^{-1}(3/2)}\times \frac{\sqrt{5}}{2}=\frac{\sqrt{5}}{{\cosh }^{-1}(3/2)}\approx \mathrm{2.32.}$$

Therefore the average height of the cable above the ground is about $2.32$ metres, which does indeed sit between $2$ and $3$ as predicted. Note that the fact that the $x$-axis runs along the ground is special to this example; for a general hanging cable you have to work out where the ground is before you start.

Example. The air temperature $T(t)$, measured in degrees Celsius $t$ hours after noon, is given by

$$T(t)=25+2t-\frac{t^2}{3}.$$

Find the average temperature between noon and 5 p.m.
Noon is $t=0$ and 5 p.m. is $t=5$, so

$$\begin{array}{rl}\bar{T}& =\frac{1}{5-0}{\int }_0^5\left(25+2t-\frac{t^2}{3}\right)dt\\ & =\frac{1}{5}{\left[25t+t^2-\frac{t^3}{9}\right]}_0^5\\ & =\frac{1}{5}\left(125+25-\frac{125}{9}\right)\\ & =\frac{1}{5}\cdot \frac{1350-125}{9}\\ & =\frac{1225}{45}\\ & =\frac{245}{9}.\end{array}$$

Therefore the average temperature between noon and 5 p.m. is $\frac{245}{9}=27\frac{2}{9}\approx 27.2^{\circ }$​C. Do not average the endpoint values $T(0)=25$ and $T(5)=26\frac{2}{3}$ and call it a day; that gives $25.8$, which is wrong, because the temperature does not change linearly. The endpoint average is only ever correct for a straight line.

We saw in MATH1131 that a continuous function on a closed interval attains its maximum and minimum values (the maximum-minimum theorem). The next result says such a function also attains its average value.

Proof. Define $F:[a,b]\to \mathbb{R}$ by

$$F(x)={\int }_a^{x}f(t)dt.$$

By the fundamental theorem of calculus, $F$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $F^{\prime }(x)=f(x)$. So $F$ satisfies the hypotheses of the ordinary mean value theorem, which supplies a $c\in (a,b)$ with

$$\frac{F(b)-F(a)}{b-a}=F^{\prime }(c).$$

But $F(a)=0$, $F(b)={\int }_a^{b}f(t)dt$ and $F^{\prime }(c)=f(c)$, so this reads

$$\frac{1}{b-a}{\int }_a^{b}f(t)dt=f(c),$$

which is the claim (in the form $\bar{f}=f(c)$). $■$

Basically, the mean value theorem for integrals says the horizontal line $y=\bar{f}$ from the rectangle picture must actually cut the graph somewhere strictly inside the interval. Continuity is doing all the work here; drop it and the theorem is false. For instance the step function equal to $0$ on $[0,1)$ and $1$ at $x=1$ has average value $0$... but a function equal to $0$ on $[0,1)$ and $2$ on $[1,2]$ has average $1$ on $[0,2]$ and never takes the value $1$ anywhere.

Example. Verify the mean value theorem for integrals for $f(x)=x^2$ on $[0,3]$ by finding the point $c$ explicitly.
The function is continuous, and its average value is

$$\bar{f}=\frac{1}{3}{\int }_0^3{x}^{2}dx=\frac{1}{3}{\left[\frac{x^3}{3}\right]}_0^3=\frac{1}{3}\times 9=3.$$

So we need $c\in (0,3)$ with $c^2=3$, i.e. $c=\sqrt{3}\approx 1.732$ (rejecting $c=-\sqrt{3}$, which is not in the interval). Therefore the theorem holds with $c=\sqrt{3}$. Notice that $c$ is not the midpoint $1.5$ of the interval; the parabola is steeper on the right, so the average value is achieved a little past halfway.

A more general version of this theorem appears in the tutorial problems for Chapter 4, and it is what is used to prove the Lagrange form of the remainder in Taylor's theorem — so this innocent-looking result is doing real work later in the course.

5.2 The arc length of a curve

Suppose $P_0(x_0,y_0)$ and $P_1(x_1,y_1)$ are two points in ${\mathbb{R}}^2$. The distance between them is

$$\mathrm{dist}(P_0,P_1)=\sqrt{(x_1-x_0{)}^2+(y_1-y_0{)}^2},$$

and if $P_0$ and $P_1$ are the endpoints of a straight line segment then we define the length of that segment to be this distance. A line segment is a very special curve; the point of this section is to measure the length of curves that bend. The idea is the obvious one — approximate the curve by a chain of little straight segments, whose lengths we can already compute, and take a limit as the segments get short.

5.2.1 An intuitive derivation of the arc length formula

Suppose $C$ is a curve in ${\mathbb{R}}^2$ which can be expressed in parametric form as

$$C=\{(x(t),y(t))\in {\mathbb{R}}^2:a\le t\le b\},$$

where $x$ and $y$ are differentiable functions of $t$. We also assume the parametrisation is chosen so that the moving point $(x(t),y(t))$ does not retrace its steps, either forwards or backwards. For $a\le s\le b$, let $\ell (s)$ denote the arc length of the partial curve

$$C_s=\{(x(t),y(t))\in {\mathbb{R}}^2:a\le t\le s\},$$

so that $\ell (a)=0$ and $\ell (b)$ is the number we want.

Take a small piece of the curve and approximate it by a secant. Fix $t$ with $a<t<b$, let $h$ be a small nonzero real number, and consider the two points $P(x(t),y(t))$ and $Q(x(t+h),y(t+h))$. The length of the arc from $P$ to $Q$ is $\ell (t+h)-\ell (t)$, and it is approximately the length of the chord $PQ$, which the distance formula gives as

$$\ell (t+h)-\ell (t)\approx \sqrt{[x(t+h)-x(t){]}^2+[y(t+h)-y(t){]}^2}.$$

Dividing both sides by $h$ (and pushing the $h$ inside the square root, where it becomes $h^2$),

$$\frac{\ell (t+h)-\ell (t)}{h}\approx \sqrt{{\left(\frac{x(t+h)-x(t)}{h}\right)}^2+{\left(\frac{y(t+h)-y(t)}{h}\right)}^2}.$$

The approximation improves as $h$ shrinks, so assuming that $\ell$ is a differentiable function, letting $h\to 0$ gives

$${\ell }^{\prime }(t)=\sqrt{[x^{\prime }(t){]}^2+[y^{\prime }(t){]}^2}.$$

By the fundamental theorem of calculus,

$$\ell (s)={\int }_a^s\sqrt{[x^{\prime }(t){]}^2+[y^{\prime }(t){]}^2}dt+K$$

for some constant $K$, and putting $s=a$ gives $0=\ell (a)=0+K$, so $K=0$. Taking $s=b$ recovers the length of the whole curve.

This is a heuristic derivation, not a proof. A rigorous treatment would first define arc length (as the supremum of the lengths of inscribed polygons) and then evaluate some genuinely awkward Riemann sums; we do not do that here. Three hypotheses were quietly used and are worth naming, because they are exactly the ones that get violated in exam questions: $x$ and $y$ must be differentiable, the parametrisation must trace the curve once and only once, and $\ell$ itself is assumed differentiable.

5.2.2 Arc length for a parametrised curve

Collecting the derivation above: if a curve $C$ is given parametrically by

$$C=\{(x(t),y(t))\in {\mathbb{R}}^2:a\le t\le b\},$$

where $x$ and $y$ are differentiable functions of $t$, then its arc length $\ell$ is

$$\boxed{\ell ={\int }_a^b\sqrt{[x^{\prime }(t){]}^2+[y^{\prime }(t){]}^2}dt.}$$

Basically, $\sqrt{[x^{\prime }(t){]}^2+[y^{\prime }(t){]}^2}dt$ is the length of an infinitesimal chord, obtained from Pythagoras applied to the horizontal displacement $x^{\prime }(t)dt$ and the vertical displacement $y^{\prime }(t)dt$; integrating adds all the little chords up.

Example. Find the arc length of one arch of the cycloid

$$x(t)=r(t-\sin t),y(t)=r(1-\cos t),0\le t\le 2\pi .$$

(This is the curve traced by a point on the rim of a wheel of radius $r$ rolling along the $x$-axis, and it is closely related to the 'curve of fastest descent' from MATH1131.) Differentiating,

$$x^{\prime }(t)=r(1-\cos t),y^{\prime }(t)=r\sin t,$$

so

$$\begin{array}{rl}[x^{\prime }(t){]}^2+[y^{\prime }(t){]}^2& =r^2(1-2\cos t+{\cos }^{2}t)+r^2{\sin }^{2}t\\ & =r^2(1-2\cos t+1)(\text{since }{\cos }^{2}t+{\sin }^{2}t=1)\\ & =2r^2(1-\cos t).\end{array}$$

Before substituting this into the formula it is best to express $1-\cos t$ as a square, since we are about to take a square root. The double angle identity $1-\cos 2\theta =2{\sin }^2\theta$ with $\theta =t/2$ gives

$$[x^{\prime }(t){]}^2+[y^{\prime }(t){]}^2=2r^2\cdot 2{\sin }^2(t/2)=4r^2{\sin }^2(t/2).$$

Now $\sqrt{A^2}=|A|$, not $A$ — but $\sin (t/2)$ is positive for $0<t<2\pi$, so here the naive square root causes no problems. Hence

$$\begin{array}{rl}\ell & ={\int }_0^{2\pi }\sqrt{4r^2{\sin }^2(t/2)}dt\\ & =2r{\int }_0^{2\pi }\sin (t/2)dt\\ & =2r[-2\cos (t/2){]}_0^{2\pi }\\ & =2r(2+2)\\ & =8r.\end{array}$$

Therefore one arch of the cycloid has length $8r$. Notice how clean that is: the arch spans a horizontal distance of $2\pi r\approx 6.28r$ and the answer is exactly $8r$, with no $\pi$ in sight at all.

Example. Use the parametrisation $x(t)=r\cos t$, $y(t)=r\sin t$ with $0\le t\le 2\pi$ to find the circumference of a circle of radius $r$. What goes wrong with the parametrisation $x(t)=r\cos 2t$, $y(t)=r\sin 2t$ on the same interval?
For the first parametrisation, $x^{\prime }(t)=-r\sin t$ and $y^{\prime }(t)=r\cos t$, so

$$\begin{array}{rl}\ell & ={\int }_0^{2\pi }\sqrt{r^2{\sin }^{2}t+r^2{\cos }^{2}t}dt\\ & ={\int }_0^{2\pi }rdt\\ & =2\pi r,\end{array}$$

which is the circumference we expect. For the second, $x^{\prime }(t)=-2r\sin 2t$ and $y^{\prime }(t)=2r\cos 2t$, and the same computation gives

$${\int }_0^{2\pi }\sqrt{4r^2{\sin }^{2}2t+4r^2{\cos }^{2}2t}dt={\int }_0^{2\pi }2rdt=4\pi r,$$

which is twice the circumference. The reason is that as $t$ runs from $0$ to $2\pi$, the point $(r\cos 2t,r\sin 2t)$ goes around the circle twice; the integral faithfully reports the total distance travelled, which is not the length of the curve. To use this parametrisation correctly you would integrate from $0$ to $\pi$.

For a closed curve (or any curve at all), check that your parametrisation traverses it exactly once before integrating; the formula cannot tell the difference between a long curve and a short curve walked twice.

5.2.3 Arc length for the graph of a function

Most curves you meet are given as $y=f(x)$ rather than parametrically, but that is just a special case: parametrise the graph of $f$ on $[a,b]$ by

$$x(t)=t,y(t)=f(t),a\le t\le b.$$

Then $x^{\prime }(t)=1$ and $y^{\prime }(t)=f^{\prime }(t)$, so the parametric formula immediately gives $\ell ={\int }_a^b\sqrt{1+[f^{\prime }(t){]}^2}dt$. Renaming the variable of integration $x$, as one usually does, the arc length $\ell$ of the graph of $f$ on $[a,b]$ is

$$\boxed{\ell ={\int }_a^b\sqrt{1+[f^{\prime }(x){]}^2}dx.}$$

Notice immediately that $\ell \ge b-a$, since the integrand is at least $1$: the curve is never shorter than the interval it spans, with equality only for a horizontal line. That is the cheapest sanity check available on any arc length answer, and it costs nothing to apply.

Example. Calculate the length of the arc $y=x^{3/2}$ for $0\le x\le 1$.
Here $f^{\prime }(x)=\frac{3}{2}{x}^{1/2}$, so $[f^{\prime }(x){]}^2=\frac{9x}{4}$ and

$$\begin{array}{rl}\ell & ={\int }_0^1\sqrt{1+\frac{9x}{4}}dx\\ & =\frac{4}{9}\cdot \frac{2}{3}{\left[{\left(1+\frac{9x}{4}\right)}^{3/2}\right]}_0^1\\ & =\frac{8}{27}\left({\left(\frac{13}{4}\right)}^{3/2}-1\right)\\ & =\frac{8}{27}\left(\frac{13\sqrt{13}}{8}-1\right)\\ & =\frac{13\sqrt{13}-8}{27}.\end{array}$$

Therefore the arc length is $\frac{13\sqrt{13}-8}{27}\approx 1.4397$. The chord from $(0,0)$ to $(1,1)$ has length $\sqrt{2}\approx 1.4142$, and the curve should be slightly longer than the chord; it is, by about $2\%$.

Example. Find the arc length of $y=\frac{1}{3}(x^2+2{)}^{3/2}$ for $0\le x\le 3$.
Differentiating with the chain rule, $f^{\prime }(x)=\frac{1}{3}\cdot \frac{3}{2}(x^2+2{)}^{1/2}\cdot 2x=x\sqrt{x^2+2}$, so

$$\begin{array}{rl}1+[f^{\prime }(x){]}^2& =1+x^2(x^2+2)\\ & =x^4+2x^2+1\\ & =(x^2+1{)}^2.\end{array}$$

The square root is now trivial (and $x^2+1>0$, so no absolute values are needed), giving

$$\begin{array}{rl}\ell & ={\int }_0^3(x^2+1)dx\\ & ={\left[\frac{x^3}{3}+x\right]}_0^3\\ & =9+3\\ & =12.\end{array}$$

Therefore the arc length is exactly $12$ units.

That example is worth pausing on. Arc-length integrands in this course are always engineered so that $1+[f^{\prime }(x){]}^2$ collapses to a perfect square, or so that a Pythagorean or hyperbolic identity flattens it; if yours does not collapse, you have almost certainly made an algebra slip — go back and check $f^{\prime }$ before you start hunting for exotic substitutions. The reason is that $\int \sqrt{1+[f^{\prime }(x){]}^2}dx$ is hopeless for almost every $f$ you could write down: even $y=x^2$ leads to $\int \sqrt{1+4x^2}dx$, which needs a hyperbolic substitution from the substitution table, and $y=\sin x$ leads to an elliptic integral that has no elementary antiderivative at all.

Example. Find the arc length of $y=\frac{x^2}{4}-\frac{\ln x}{2}$ for $1\le x\le 2$.
Differentiating, $f^{\prime }(x)=\frac{x}{2}-\frac{1}{2x}$, so

$$\begin{array}{rl}1+[f^{\prime }(x){]}^2& =1+\frac{x^2}{4}-\frac{1}{2}+\frac{1}{4x^2}\\ & =\frac{x^2}{4}+\frac{1}{2}+\frac{1}{4x^2}\\ & ={\left(\frac{x}{2}+\frac{1}{2x}\right)}^2,\end{array}$$

where the middle term flipped sign from $-\frac{1}{2}$ to $+\frac{1}{2}$ precisely because of the $+1$. Since $x>0$ the bracket is positive, so

$$\begin{array}{rl}\ell & ={\int }_1^2\left(\frac{x}{2}+\frac{1}{2x}\right)dx\\ & ={\left[\frac{x^2}{4}+\frac{\ln x}{2}\right]}_1^2\\ & =\left(1+\frac{\ln 2}{2}\right)-\frac{1}{4}\\ & =\frac{3}{4}+\frac{\ln 2}{2}.\end{array}$$

Therefore the arc length is $\frac{3}{4}+\frac{\ln 2}{2}\approx 1.0966$, comfortably more than the span $b-a=1$. This is the standard shape of a hand-made arc length question: $f^{\prime }$ is of the form $u-\frac{1}{4u}$ or $\frac{u}{2}-\frac{1}{2u}$, so that the cross term in $[f^{\prime }{]}^2$ is exactly $-1$ and the $+1$ repairs the square.

Example. Find the arc length of the catenary

$$y=\frac{1}{a}\cosh (ax),x\in [-b,b].$$

A catenary is the shape of a hanging cable, which is why this function keeps reappearing. Here $f^{\prime }(x)=\sinh (ax)$, and the hyperbolic Pythagorean identity ${\cosh }^{2}t-{\sinh }^{2}t=1$ does the collapsing for us:

$$\begin{array}{rl}\ell & ={\int }_{-b}^b\sqrt{1+{\sinh }^2(ax)}dx\\ & ={\int }_{-b}^b\sqrt{{\cosh }^2(ax)}dx\\ & ={\int }_{-b}^b\cosh (ax)dx(\text{since }\cosh \text{ is always positive})\\ & =2{\int }_0^b\cosh (ax)dx(\text{since }\cosh \text{ is even})\\ & =2{\left[\frac{1}{a}\sinh (ax)\right]}_0^b\\ & =\frac{2}{a}\sinh (ab)\\ & =\frac{1}{a}\left(e^{ab}-e^{-ab}\right).\end{array}$$

Therefore the catenary has arc length $\frac{1}{a}\left(e^{ab}-e^{-ab}\right)$ units. Notice how this is the only function whose arc length integrand simplifies without any factoring at all — the identity $1+{\sinh }^2={\cosh }^2$ is built into it, which is one reason the hyperbolic functions were invented in the first place.

Example. Hence find the length of the suspended cable of Section 5.1, namely $y=2\cosh (x/2)$ on $[-a,a]$ where $a=2{\cosh }^{-1}(3/2)$.
Comparing with the general catenary, $\frac{1}{c}\cosh (cx)=2\cosh (x/2)$ forces $c=\frac{1}{2}$, and the interval endpoint is $b=a$. So the formula gives

$$\begin{array}{rl}\ell & =\frac{2}{c}\sinh (ca)\\ & =4\sinh \left(\frac{a}{2}\right)\\ & =4\sinh ({\cosh }^{-1}(3/2))\\ & =4\times \frac{\sqrt{5}}{2}\\ & =2\sqrt{5},\end{array}$$

reusing the value $\sinh ({\cosh }^{-1}(3/2))=\frac{\sqrt{5}}{2}$ computed earlier. Therefore the cable is $2\sqrt{5}\approx 4.47$ metres long. As a sanity check, the poles are $2a=4{\cosh }^{-1}(3/2)\approx 3.85$ metres apart, so the cable is longer than the gap it spans — as it must be, since it sags.

5.2.4 Arc length for a polar curve

Suppose a curve is described in polar coordinates by $r=f(\theta )$ for ${\theta }_0\le \theta \le {\theta }_1$. Since

$$x=r\cos \theta =f(\theta )\cos \theta \text{and}y=r\sin \theta =f(\theta )\sin \theta ,$$

this is a parametrisation of the curve, with $\theta$ as the parameter, so there is nothing new to derive — just some algebra to do. By the product rule,

$$x^{\prime }(\theta )=f^{\prime }(\theta )\cos \theta -f(\theta )\sin \theta ,y^{\prime }(\theta )=f^{\prime }(\theta )\sin \theta +f(\theta )\cos \theta ,$$

and hence

$$\begin{array}{rl}[x^{\prime }(\theta ){]}^2+[y^{\prime }(\theta ){]}^2& =[f^{\prime }(\theta ){]}^2{\cos }^2\theta -2f(\theta )f^{\prime }(\theta )\sin \theta \cos \theta +[f(\theta ){]}^2{\sin }^2\theta \\ & +[f^{\prime }(\theta ){]}^2{\sin }^2\theta +2f(\theta )f^{\prime }(\theta )\sin \theta \cos \theta +[f(\theta ){]}^2{\cos }^2\theta \\ & =[f(\theta ){]}^2+[f^{\prime }(\theta ){]}^2,\end{array}$$

where the cross terms cancelled and ${\sin }^2\theta +{\cos }^2\theta =1$ mopped up the rest. Writing $f(\theta )$ as $r$ and $f^{\prime }(\theta )$ as $\frac{dr}{d\theta }$, the arc length $\ell$ of a polar curve is

$$\boxed{\ell ={\int }_{{\theta }_0}^{{\theta }_1}\sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}d\theta .}$$

Basically, moving through an angle $d\theta$ carries you $rd\theta$ around the circle and $\frac{dr}{d\theta }d\theta$ outwards, and these two displacements are perpendicular, so Pythagoras gives the little bit of arc. The $r^2$ term is not optional and it is not ${\left(\frac{dr}{d\theta }\right)}^2$ in disguise; forgetting it is the classic polar arc length mistake (it would say that a circle $r=1$, on which $\frac{dr}{d\theta }=0$, has length zero).

Example. Find the length of the cardioid $r=1+\cos \theta$.
The full cardioid is traced once as $\theta$ runs from $0$ to $2\pi$. Here $\frac{dr}{d\theta }=-\sin \theta$, so

$$\begin{array}{rl}{r}^2+{\left(\frac{dr}{d\theta }\right)}^2& =(1+\cos \theta {)}^2+{\sin }^2\theta \\ & =1+2\cos \theta +{\cos }^2\theta +{\sin }^2\theta \\ & =2+2\cos \theta \\ & =4{\cos }^2(\theta /2),\end{array}$$

using the double angle identity $1+\cos 2\alpha =2{\cos }^2\alpha$ with $\alpha =\theta /2$ in the last line. Now be careful with the square root: $\sqrt{4{\cos }^2(\theta /2)}=2|\cos (\theta /2)|$, and $\cos (\theta /2)$ is positive on $(0,\pi )$ but negative on $(\pi ,2\pi )$. The cardioid is symmetric about the $x$-axis, so the cleanest fix is to integrate over the top half and double:

$$\begin{array}{rl}\ell & =2{\int }_0^{\pi }2\cos (\theta /2)d\theta \\ & =4[2\sin (\theta /2){]}_0^{\pi }\\ & =8(1-0)\\ & =8.\end{array}$$

Therefore the cardioid has length $8$ units. If you drop the absolute value and integrate $2\cos (\theta /2)$ straight from $0$ to $2\pi$ you get $0$, which is obviously nonsense; whenever a half-angle appears inside a square root, check the sign on the second half of the range.

Example. Find the length of the spiral $r=e^{\theta }$ for $0\le \theta \le 2\pi$.
Here $\frac{dr}{d\theta }=e^{\theta }=r$, so the integrand is $\sqrt{e^{2\theta }+e^{2\theta }}=\sqrt{2}{e}^{\theta }$ and

$$\begin{array}{rl}\ell & =\sqrt{2}{\int }_0^{2\pi }{e}^{\theta }d\theta \\ & =\sqrt{2}[e^{\theta }{]}_0^{2\pi }\\ & =\sqrt{2}\left(e^{2\pi }-1\right).\end{array}$$

Therefore the length is $\sqrt{2}\left(e^{2\pi }-1\right)\approx 755.9$ units — the exponential spiral grows fast enough that a single turn is already enormous.

Example. The spiral $r=e^{-\theta /10}$, where $\theta \ge 0$, winds infinitely many times around the origin. Is its total arc length finite? Explain.
Compute the arc length of the segment $0\le \theta \le {\theta }_1$ first, and take a limit afterwards. Here $\frac{dr}{d\theta }=-\frac{1}{10}{e}^{-\theta /10}$, so

$$\begin{array}{rl}\ell ({\theta }_1)& ={\int }_0^{{\theta }_1}\sqrt{{\left(e^{-\theta /10}\right)}^2+{\left(-\frac{1}{10}{e}^{-\theta /10}\right)}^2}d\theta \\ & ={\int }_0^{{\theta }_1}{e}^{-\theta /10}\sqrt{1+\frac{1}{100}}d\theta \\ & =\frac{\sqrt{101}}{10}{\int }_0^{{\theta }_1}{e}^{-\theta /10}d\theta \\ & =\frac{\sqrt{101}}{10}[-10e^{-\theta /10}{]}_0^{{\theta }_1}\\ & =\sqrt{101}\left(1-e^{-{\theta }_1/10}\right).\end{array}$$

As ${\theta }_1\to \infty$ we have $e^{-{\theta }_1/10}\to 0$, so $\ell ({\theta }_1)\to \sqrt{101}$. Therefore the total arc length is finite and equals $\sqrt{101}\approx 10.05$ units. This is a curve of infinite extent (in the sense that it never stops winding) with finite length; the windings shrink geometrically, and a geometric series converges. Keep this example in mind — the very last example of the chapter is its evil twin.

5.3 The speed of a moving particle

In Chapter 4 of the MATH1131 calculus notes we discussed the speed of a particle moving along a straight line. Now we can handle a particle moving along any curve in the plane, because we can measure the length of that curve.

Suppose a particle $P$ moves in the plane with position $(x(t),y(t))$ at time $t$. The distance $s(t)$ travelled from time zero to time $t$ is exactly the arc length of the path traversed in that time,

$$s(t)={\int }_0^t\sqrt{[x^{\prime }(u){]}^2+[y^{\prime }(u){]}^2}du.$$

By definition the speed of $P$ is the rate of change of distance travelled with respect to time, so by the fundamental theorem of calculus the speed $v(t)$ is

$$\boxed{v(t)=s^{\prime }(t)=\sqrt{[x^{\prime }(t){]}^2+[y^{\prime }(t){]}^2}.}$$

If we bundle the position into a vector $\mathbf{r}(t)=(x(t),y(t))$, then ${\mathbf{r}}^{\prime }(t)=(x^{\prime }(t),y^{\prime }(t))$ is the velocity vector and the formula just says $v(t)=|{\mathbf{r}}^{\prime }(t)|$: speed is the length of the velocity. Basically, arc length and speed are the same computation read in two directions — the arc length integrand is the speed, and distance travelled is the integral of speed:

$$\frac{ds}{dt}=v(t)\text{and}s=\int v(t)dt.$$

Example. A stone is thrown horizontally from the deck of the Sydney Harbour Bridge at $20$ metres per second. Its position $t$ seconds after the throw is

$$x(t)=20t,y(t)=50-5t^2,0\le t\le \sqrt{10},$$

where $y(t)$ is the height above the water. Find the speed of the stone an instant before it hits the water.
Differentiating, $x^{\prime }(t)=20$ and $y^{\prime }(t)=-10t$, so the speed at time $t$ is

$$v(t)=\sqrt{20^2+100t^2}=\sqrt{400+100t^2}.$$

The stone hits the water when $y(t)=0$, that is when $5t^2=50$, i.e. $t=\sqrt{10}$. Hence

$$\underset{t\to (\sqrt{10}{)}^{-}}{\lim }v(t)=\sqrt{400+100\times 10}=\sqrt{1400}=10\sqrt{14}\approx \mathrm{37.42.}$$

Therefore the stone is travelling at about $37.42$ metres per second just before impact. Notice that the horizontal component of the velocity is a constant $20$ throughout — gravity only ever changes the vertical component — so the speed never drops below $20$.

Example. A projectile is fired from an elevated cannon. Its horizontal distance $x$ (in metres) from the cannon and height $y$ (in metres) above the ground, exactly $t$ seconds after firing, are

$$x(t)=40t,y(t)=-5t^2+40t+45,0\le t\le t_1,$$

where $t_1$ is the time of impact.
(a) Find $t_1$.
(b) Find the speed of the projectile immediately prior to impact.
(c) What was the average height of the projectile above the ground during this period?

(a) Impact happens when $y=0$:

$$\begin{array}{rl}-5t^2+40t+45& =0\\ t^2-8t-9& =0\\ (t-9)(t+1)& =0,\end{array}$$

so $t=9$ or $t=-1$; discarding the negative time, $t_1=9$ seconds.

(b) Here $x^{\prime }(t)=40$ and $y^{\prime }(t)=-10t+40$, so at $t=9$ we have $y^{\prime }(9)=-50$ and

$$v(9)=\sqrt{40^2+(-50{)}^2}=\sqrt{1600+2500}=\sqrt{4100}=10\sqrt{41}\approx \mathrm{64.03.}$$

Therefore the projectile is travelling at about $64.03$ metres per second immediately prior to impact.

(c) This is a job for the average value formula applied to $y$ on $[0,9]$:

$$\begin{array}{rl}\bar{y}& =\frac{1}{9}{\int }_0^9\left(-5t^2+40t+45\right)dt\\ & =\frac{1}{9}{\left[-\frac{5t^3}{3}+20t^2+45t\right]}_0^9\\ & =\frac{1}{9}\left(-1215+1620+405\right)\\ & =\frac{810}{9}\\ & =90.\end{array}$$

Therefore the average height of the projectile was $90$ metres. Note carefully that this is the average height with respect to time, which is what the phrasing asks for; averaging with respect to horizontal distance $x$ would be a different integral (though here it happens to agree, since $x$ is a linear function of $t$).
For completeness, the distance actually flown is ${\int }_0^9\sqrt{1600+(40-10t{)}^2}dt\approx 427.53$ metres, which needs the substitution $u=40-10t$ followed by a hyperbolic substitution from the substitution table; it is marked as extension material precisely because the algebra is grim, and it is a good illustration that a physically natural arc length usually does not simplify.

Since speed is the derivative of distance travelled, the two quantities determine each other, and it is natural to ask for a parametrisation in which the speed is always $1$. Such a parametrisation is said to be by arc length: if $v(t)=1$ for all $t$ then $s(t)={\int }_0^{t}1du=t$, so the parameter is the distance travelled along the curve.

Example. A particle moves on the circle $x(t)=3\cos 2t$, $y(t)=3\sin 2t$. Find its speed and the distance travelled by time $t$, and hence reparametrise the circle by arc length.
Differentiating, $x^{\prime }(t)=-6\sin 2t$ and $y^{\prime }(t)=6\cos 2t$, so

$$v(t)=\sqrt{36{\sin }^{2}2t+36{\cos }^{2}2t}=6,$$

a constant. Hence $s(t)={\int }_0^{t}6du=6t$, i.e. $t=\frac{s}{6}$. Substituting this back into the position,

$$x=3\cos \left(\frac{s}{3}\right),y=3\sin \left(\frac{s}{3}\right),$$

and it is easily checked that this new parametrisation has speed $\sqrt{{\sin }^2(s/3)+{\cos }^2(s/3)}=1$. Therefore the arc length parametrisation of the circle of radius $3$ is $\left(3\cos \frac{s}{3},3\sin \frac{s}{3}\right)$, and running $s$ from $0$ to $6\pi$ traverses it once — recovering the circumference $2\pi \times 3=6\pi$ for free. In general the arc-length parametrisation of a circle of radius $R$ is $\left(R\cos \frac{s}{R},R\sin \frac{s}{R}\right)$; the $\frac{1}{R}$ inside is exactly what makes the speed $1$ rather than $R$.

Example. The position of a particle $P$ at time $t\ge 0$ is

$$x(t)=\cos \left(\frac{\pi }{2}(\cos \pi t-1)\right),y(t)=-\sin \left(\frac{\pi }{2}(\cos \pi t-1)\right).$$

(a) Find a formula for the speed $v(t)$.
(b) What curve does the trajectory trace out, and what is its length?
(c) How far does the particle travel during the time interval $[0,3]$?

(a) Write $\varphi (t)=\frac{\pi }{2}(\cos \pi t-1)$, so that $x=\cos \varphi$ and $y=-\sin \varphi$. By the chain rule,

$$x^{\prime }(t)=-\sin \varphi \cdot {\varphi }^{\prime }(t),y^{\prime }(t)=-\cos \varphi \cdot {\varphi }^{\prime }(t),$$

and hence

$$\begin{array}{rl}v(t)& =\sqrt{{\varphi }^{\prime }(t{)}^2{\sin }^2\varphi +{\varphi }^{\prime }(t{)}^2{\cos }^2\varphi }\\ & =|{\varphi }^{\prime }(t)|\\ & =\left|\frac{\pi }{2}\times (-\pi \sin \pi t)\right|\\ & =\frac{{\pi }^2}{2}|\sin \pi t|.\end{array}$$

Therefore $v(t)=\frac{{\pi }^2}{2}|\sin \pi t|$; in particular the particle is momentarily at rest whenever $t$ is an integer, and fastest when $t$ is a half-integer.

(b) Since $\cos \pi t-1$ takes every value in $[-2,0]$, the angle $\varphi$ ranges over $[-\pi ,0]$, and $(\cos \varphi ,-\sin \varphi )$ then sweeps out the upper half of the unit circle. So the trajectory is the semicircle of centre $(0,0)$ and radius $1$ lying in the upper half-plane, whose length is half the circumference, namely $\pi$.

(c) Distance travelled is the integral of speed, and $|\sin \pi t|$ has period $1$ with ${\int }_0^1|\sin \pi t|dt={\left[-\frac{\cos \pi t}{\pi }\right]}_0^1=\frac{2}{\pi }$. Hence

$${\int }_0^{3}v(t)dt=\frac{{\pi }^2}{2}\times 3\times \frac{2}{\pi }=3\pi .$$

Therefore the particle travels $3\pi$ units in the first three seconds. Compare (b) and (c): the curve is only $\pi$ units long, but the particle covers $3\pi$ units, because it sweeps back and forth along the same semicircle three times. This is the same trap as the doubly-traversed circle of Section 5.2.2, seen from the other side: arc length is a property of the curve, distance travelled is a property of the motion, and only a parametrisation which does not retrace makes them equal.

5.4 Surface area

Finding the surface area of a general surface in ${\mathbb{R}}^3$ is genuinely hard, and the machinery needed for it (the partial derivatives of Functions of several variables, together with double integrals) is second year material. In this section we restrict to surfaces obtained by rotating a plane curve about one of the axes, which is exactly the case where a one-variable integral suffices.

Everything rests on one piece of geometry. The frustum of a right circular cone is what you get by slicing the top off a cone parallel to its base.

This is proved by cutting the cone open and flattening it into a sector of a circle; it is an exercise in the tutorial problems, and it needs nothing beyond the formula for the area of a sector. Basically, $\pi (r+R)s=2\pi \left(\frac{r+R}{2}\right)s$, i.e. (circumference at the average radius) $\times$ (slant height) — the frustum is a rectangle of width $2\pi \bar{r}$ and height $s$, rolled up.

5.4.1 An heuristic derivation for the surface area of a surface of revolution

Suppose a curve $C$ has parametrisation

$$C=\{(x(t),y(t))\in {\mathbb{R}}^2:a\le t\le b\}.$$

We assume that the curve lies in the upper half-plane (more precisely $y(t)\ge 0$, meeting the $x$-axis at only finitely many points if at all), and that $C$ is simple: if $(x(t_0),y(t_0))=(x(t_1),y(t_1))$ then $t_0=t_1$, i.e. the curve does not cross itself. Rotating $C$ about the $x$-axis produces a surface of revolution, and we want its area.

The derivation runs exactly parallel to the arc length one. Let $A(s)$ denote the area of the surface formed by rotating the partial curve $\{(x(t),y(t)):a\le t\le s\}$, and assume $A$ is differentiable. Fix $t\in (a,b)$, let $h$ be small and nonzero, and consider $P(x(t),y(t))$ and $Q(x(t+h),y(t+h))$. Then $A(t+h)-A(t)$ is the area swept out by the arc from $P$ to $Q$, which for small $h$ is approximately the area swept by the chord $PQ$ — and a rotated chord is precisely a frustum, of radii $y(t)$ and $y(t+h)$ and slant height equal to the length of $PQ$. So the frustum formula gives

$$A(t+h)-A(t)\approx \pi (y(t+h)+y(t))\sqrt{[x(t+h)-x(t){]}^2+[y(t+h)-y(t){]}^2}.$$

Dividing by $h$,

$$\frac{A(t+h)-A(t)}{h}\approx \pi (y(t+h)+y(t))\sqrt{{\left(\frac{x(t+h)-x(t)}{h}\right)}^2+{\left(\frac{y(t+h)-y(t)}{h}\right)}^2}.$$

Since $y$ is differentiable at $t$ it is continuous there, so $y(t+h)\to y(t)$ as $h\to 0$, and the limit is

$$A^{\prime }(t)=\pi (y(t)+y(t))\sqrt{[x^{\prime }(t){]}^2+[y^{\prime }(t){]}^2}=2\pi y(t)\sqrt{[x^{\prime }(t){]}^2+[y^{\prime }(t){]}^2}.$$

Applying the fundamental theorem of calculus and using $A(a)=0$ to kill the constant of integration exactly as before,

$$A(s)={\int }_a^{s}2\pi y(t)\sqrt{[x^{\prime }(t){]}^2+[y^{\prime }(t){]}^2}dt,$$

and putting $s=b$ gives the area of the whole surface.

5.4.2 Surface area formulae and examples

Assume throughout that $C$ lies in the upper half-plane and is simple, and that all the derivatives written down exist. Rotating about the $x$-axis, the area $A$ of the surface of revolution is

$$\boxed{A={\int }_a^{b}2\pi y(t)\sqrt{[x^{\prime }(t){]}^2+[y^{\prime }(t){]}^2}dt}$$

if $C$ is given parametrically by $(x(t),y(t))$ with $a\le t\le b$;

$$\boxed{A={\int }_a^{b}2\pi f(x)\sqrt{1+[f^{\prime }(x){]}^2}dx}$$

if $C$ is the graph $y=f(x)$ on $[a,b]$; and

$$\boxed{A={\int }_{{\theta }_0}^{{\theta }_1}2\pi r\sin \theta \sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}d\theta }$$

if $C$ is the polar curve $r=f(\theta )$ with ${\theta }_0\le \theta \le {\theta }_1$. Only the first of these needs deriving; the other two follow from it via the parametrisations $x(t)=t$, $y(t)=f(t)$ and $x(\theta )=f(\theta )\cos \theta$, $y(\theta )=f(\theta )\sin \theta$ used in Sections 5.2.3 and 5.2.4. Rotating about the $y$-axis instead, the parametric formula becomes

$$\boxed{A={\int }_a^{b}2\pi x(t)\sqrt{[x^{\prime }(t){]}^2+[y^{\prime }(t){]}^2}dt,}$$