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.

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.

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 $$\int_0^{2\pi} 2\sin(t/2)\;dt = 8.$$

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)

Ex 12.5.11 Find the length of the curve from the previous exercise, shown in figure 12.5.1. (answer)

Ex 12.5.12 Find the length of the spiral of Archimedes (figure 12.3.4) for $0\le\theta\le2\pi$. (answer)

Figure 12.5.1. A region formed by the end of a string.