Arc Lengths of Parametric Equations (College Board AP® Calculus BC): Revision Note

Questions on parametric arc lengths may ask you to leave your answer as a definite integral.

To get full points in a calculator FRQ you must:

• state the integral with the correct limits

• substitute the derivatives into the formula

• evalulate the integral numerically

Show that the length of the curve $x=1-\cos t$ and $y=t-\sin t$ from $t=0$ to $t=\frac{\pi }{2}$ can be written as $\sqrt{2}{\int }_0^{\frac{\pi }{2}}\sqrt{1-\cos t}dt$

Answer:

In this question, you do not need to evaluate the definite integral

Start by finding $\frac{dx}{dt}$ and $\frac{dy}{dt}$ individually

$\begin{array}{rcl}\frac{dx}{dt}& =& \sin t\\ \frac{dy}{dt}& =& 1-\cos t\end{array}$

Substitute these derivatives, and the limits $t_1=0$ and $t_2=\frac{\pi }{2}$, into the formula $L={\int }_{t_1}^{t_2}\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt$

$L={\int }_0^{\frac{\pi }{2}}\sqrt{{\left(\sin t\right)}^2+{\left(1-\cos t\right)}^2}dt$

This is not how it is given in the question, so expand and simplify under the square root

$L={\int }_0^{\frac{\pi }{2}}\sqrt{{\sin }^{2}t+1-2\cos t+{\cos }^{2}t}dt$

Here, you can use the trigonometric identity ${\sin }^{2}t+{\cos }^{2}t=1$ to simplify further

$\begin{array}{rcl}L& =& {\int }_0^{\frac{\pi }{2}}\sqrt{1+1-2\cos t}dt\\ & =& {\int }_0^{\frac{\pi }{2}}\sqrt{2-2\cos t}dt\end{array}$

This is almost the answer given, but a 2 must be factored out from inside the square root

$\begin{array}{rcl}L& =& {\int }_0^{\frac{\pi }{2}}\sqrt{2\left(1-\cos t\right)}dt\\ & =& {\int }_0^{\frac{\pi }{2}}\sqrt{2}\sqrt{1-\cos t}dt\\ & =& \sqrt{2}{\int }_0^{\frac{\pi }{2}}\sqrt{1-\cos t}dt\end{array}$

Arc length from $t=0$ to $t=\frac{\pi }{2}$ is $\sqrt{2}{\int }_0^{\frac{\pi }{2}}\sqrt{1-\cos t}dt$

Find the length of the curve $x=3t+1$ and $y=2t^{\frac{3}{2}}$ from $t=0$ to $t=8$.

Answer:

Use the formula $L={\int }_{t_1}^{t_2}\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt$

It helps to find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ individually

$\begin{array}{rcl}\frac{dx}{dt}& =& 3\\ \frac{dy}{dt}& =& 3t^{\frac{1}{2}}\end{array}$

Substitute these derivatives, and the limits $t_1=0$ and $t_2=8$, into the formula $L={\int }_{t_1}^{t_2}\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt$

$L={\int }_0^8\sqrt{{\left(3\right)}^2+{\left(3t^{\frac{1}{2}}\right)}^2}dt$

Method 1

If calculators are allowed, evaluate this definite integral on your calculator

${\int }_0^8\sqrt{{\left(3\right)}^2+{\left(3t^{\frac{1}{2}}\right)}^2}dt=52$

Method 2

If calculators are not allowed, continue by simplifying under the square root and taking out a 9

$\begin{array}{rcl}L& =& {\int }_0^8\sqrt{9+9t}dt\\ & =& {\int }_0^8\sqrt{9}\sqrt{1+t}dt\\ & =& 3{\int }_0^8\sqrt{1+t}dt\end{array}$

There are many ways to evaluate this definite integral, for example integration by substitution using $u=1+t$

Note that $\frac{du}{dt}=1$ so $du=dt$ and that $t=0\Rightarrow u=1$, $t=8\Rightarrow u=9$

$\begin{array}{rcl}L& =& 3{\int }_1^9\sqrt{u}du\\ & =& 3{\int }_1^9{u}^{\frac{1}{2}}du\\ & =& 3{\left[\frac{2}{3}{u}^{\frac{3}{2}}\right]}_1^9\\ & =& 2{\left[u^{\frac{3}{2}}\right]}_1^9\\ & =& 2\times {\left(\sqrt{9}\right)}^3-2\times {\left(\sqrt{1}\right)}^3\\ & =& 54-2\end{array}$

The length of the curve is 52 units