Short and Sweet Calculus

## 7.5 Arc length

In this section, we want to find the length of the curve $$y=f(x)$$ from $$x=a$$ to $$x=b$$. A piece of a curve that lies between two specific points is called an arc. Like the previous sections of this chapter, we construct a formula for the length of an arc by considering infinitesimals.

Let $$s$$ be the length of the arc from the fixed point $$(a,f(a))$$ to a variable point $$(x,f(x))$$ as shown in Figure 7.15. Assume $$s$$ increases by an infinitesimally small amount $$ds$$, and $$x$$ and $$y$$ by $$dx$$ and $$dy$$, respectively. Because $$ds$$ is so small, this part of the curve is virtually straight and by the Pythagorean theorem we can write $\boxed{ds=\sqrt{(dx)^{2}+(dy)^{2}}}$ or $ds=\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}dx.$ But $$dy/dx=f'(x).$$ Therefore, $ds=\sqrt{1+[f'(x)]^{2}}dx.$ As $$ds$$ sweeps along the curve from $$(a,f(a))$$ to $$(b,f(b))$$, we add up all the infinitesimal lengths: \begin{aligned} \text{length of arc} & =\int_{a}^{b}ds=\int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}dx\\ & =\int_{a}^{b}\sqrt{1+[f'(x)]^{2}}dx.\end{aligned}

If we want to compute the length of the curve $$x=h(y)$$ between $$(h(c),c)$$ and $$(h(d),d)$$, then we can still use $ds=\sqrt{(dx)^{2}+(dy)^{2}}.$ Therefore, $ds=\sqrt{\left(\frac{dx}{dy}\right)^{2}+1}\ dy=\sqrt{[h'(y)]^{2}+1}\ dy$ and $\text{length of arc}=\int_{c}^{d}ds=\int_{c}^{d}\sqrt{(dx)^{2}+(dy)^{2}}=\int_{c}^{d}\sqrt{\left(\frac{dx}{dy}\right)^{2}+1}\ dy=\int_{c}^{d}\sqrt{[h'(y)]^{2}+1}\ dy.$

Example 7.7. Find the length of the curve $$y^{2}=x^{3}$$ between the points $$(0,0)$$ and $$(4,8)$$.

Solution

Solving $$y^{2}=x^{3}$$ for $$y$$, we obtain $y=x^{\frac{3}{2}}\ (\text{for }y>0)\Rightarrow\frac{dy}{dx}=\frac{3}{2}x^{1/2}.$ The arc length formula yields $L=\int_{0}^{4}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}dx=\int_{0}^{4}\sqrt{1+\frac{9}{4}x}\ dx.$ Let $$u=1+\frac{9}{4}x$$. Then $$du=\frac{9}{4}dx$$ or $$dx=\frac{4}{9}du$$ and $x=0\quad\Leftrightarrow\quad u=1$ $x=4\quad\Leftrightarrow\quad u=10$ Therefore, $L=\int_{1}^{10}\frac{4}{9}\sqrt{u}\ du=\frac{4}{9}\left(\frac{2}{3}\right)u^{\frac{3}{2}}\Bigg|_{u=1}^{u=10}=\frac{8}{27}(\sqrt{1000}-1)\approx9.07342.$