We have already seen how to compute slopes of curves given by parametric equations—it is how we computed slopes in polar coordinates.

Example 12.5.1 Find the slope of the cycloid $x=t-\sin t$, $y=1-\cos t$. We compute $x'=1-\cos t$, $y'=\sin t$, so $${dy\over dx} ={\sin t\over 1-\cos t}.$$ Note that when $t$ is an odd multiple of $\pi$, like $\pi$ or $3\pi$, this is $(0/2)=0$, so there is a horizontal tangent line, in agreement with figure 12.4.1. At even multiples of $\pi$, the fraction is $0/0$, which is undefined. The figure shows that there is no tangent line at such points. $\square$

Areas can be a bit trickier with parametric equations, depending on the curve and the area desired. We can potentially compute areas between the curve and the $x$-axis quite easily.

Example 12.5.2 Find the area under one arch of the cycloid $x=t-\sin t$, $y=1-\cos t$. We would like to compute $$\int_0^{2\pi} y\;dx,$$ but we do not know $y$ in terms of $x$. However, the parametric equations allow us to make a substitution: use $y=1-\cos t$ to replace $y$, and compute $dx=(1-\cos t)\;dt$. Then the integral becomes $$\int_0^{2\pi} (1-\cos t)(1-\cos t)\;dt=3\pi.$$ Note that we need to convert the original $x$ limits to $t$ limits using $x=t-\sin t$. When $x=0$, $t=\sin t$, which happens only when $t=0$. Likewise, when $x=2\pi$, $t-2\pi=\sin t$ and $t=2\pi$. Alternately, because we understand how the cycloid is produced, we can see directly that one arch is generated by $0\le t\le 2\pi$. In general, of course, the $t$ limits will be different than the $x$ limits. $\square$

This technique will allow us to compute some quite interesting areas, as illustrated by the exercises.

As a final example, we see how to compute the length of a curve given by parametric equations. Section 11.4 investigates arc length for functions given as $y$ in terms of $x$, and develops the formula for length: $$\int_a^b \sqrt{1+\left({dy\over dx}\right)^2}\;dx.$$ Using some properties of derivatives, including the chain rule, we can convert this to use parametric equations $x=f(t)$, $y=g(t)$: $$\eqalign{ \int_a^b \sqrt{1+\left({dy\over dx}\right)^2}\;dx&= \int_a^b \sqrt{\left({dx\over dt}\right)^2 +\left({dx\over dt}\right)^2\left({dy\over dx}\right)^2}\;{dt\over dx}\;dx\cr &=\int_u^v \sqrt{\left({dx\over dt}\right)^2+ \left({dy\over dt}\right)^2}\;dt\cr &=\int_u^v \sqrt{(f'(t))^2+(g'(t))^2}\;dt.\cr} $$ Here $u$ and $v$ are the $t$ limits corresponding to the $x$ limits $a$ and $b$.

Example 12.5.3 Find the length of one arch of the cycloid. From $x=t-\sin t$, $y=1-\cos t$, we get the derivatives $f'=1-\cos t$ and $g'=\sin t$, so the length is $$ \int_0^{2\pi} \sqrt{(1-\cos t)^2+\sin^2 t}\;dt= \int_0^{2\pi} \sqrt{2-2\cos t}\;dt. $$ Now we use the formula $\ds \sin^2(t/2)=(1-\cos(t))/2$ or $\ds 4\sin^2(t/2)=2-2\cos t$ to get $$\int_0^{2\pi} \sqrt{4\sin^2(t/2)}\;dt.$$ Since $0\le t\le2\pi$, $\sin(t/2)\ge 0$, so we can rewrite this as $$\int_0^{2\pi} 2\sin(t/2)\;dt = 8.$$ $\square$

## Exercises 12.5

**Ex 12.5.1**
Consider the curve of
exercise 6 in
section 12.4. Find all values of
$t$ for which the curve has a horizontal tangent line.
(answer)

**Ex 12.5.2**
Consider the curve of
exercise 6 in
section 12.4. Find the area under
one arch of the curve.
(answer)

**Ex 12.5.3**
Consider the curve of
exercise 6 in
section 12.4. Set up an integral
for the length of one arch of the curve.
(answer)

**Ex 12.5.4**
Consider the hypercycloid of
exercise 7 in
section 12.4. Find all points at
which the curve has a horizontal tangent line.
(answer)

**Ex 12.5.5**
Consider the hypercycloid of
exercise 7 in
section 12.4. Find the area between the
large circle and
one arch of the curve.
(answer)

**Ex 12.5.6**
Consider the hypercycloid of
exercise 7 in section 12.4. Find the length of one arch of the curve.
(answer)

**Ex 12.5.7**
Consider the hypocycloid of
exercise 8 in
section 12.4. Find the area inside the curve.
(answer)

**Ex 12.5.8**
Consider the hypocycloid of
exercise 8 in section 12.4. Find the length of one arch of the curve.
(answer)

**Ex 12.5.9**
Recall the involute of a circle from
exercise 9 in
section 12.4. Find the point in the first
quadrant in figure 12.4.4
at which the tangent line is vertical.
(answer)

**Ex 12.5.10**
Recall the involute of a circle from
exercise 9 in
section 12.4. Instead of an infinite
string, suppose we have a string of length $\pi$ attached to the unit
circle at $(-1,0)$, and initially laid around the top of the circle
with its end at $(1,0)$. If we grasp the end of the string and begin
to unwind it, we get a piece of the involute, until the string is
vertical. If we then keep the string taut and continue to rotate it
counter-clockwise, the end traces out a semi-circle with center at
$(-1,0)$, until the string is vertical again. Continuing, the end of
the string traces out the mirror image of the initial portion of the
curve; see figure 12.5.1. Find the area
of the region inside this curve and outside the unit circle.
(answer)