
# 10.5: Calculus with Parametric Equations

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

Example $$\PageIndex{1}$$

Find the slope of the cycloid $$x=t-\sin t$$, $$y=1-\cos t$$.

Solution

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 10.4.1. 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 $$\PageIndex{2}$$

Find the area under one arch of the cycloid $$x=t-\sin t$$, $$y=1-\cos t$$.

Solution

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 9.9 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 $$\PageIndex{3}$$:

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

$$\int_0^{2\pi} \sqrt{(1-\cos t)^2+\sin^2 t}\;dt= \int_0^{2\pi} \sqrt{2-2\cos t}\;dt.$$

Solution

Now we use the formula $$\sin^2(t/2)=(1-\cos(t))/2$$ or $$4\sin^2(t/2)=2-2\cos t$$ 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

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

### Contributors

• Integrated by Justin Marshall.