Spaceballs (Parts 4 and 5)

Refer to Spaceballs (Part 1) to derive the graph that both balls follow on the xy-plane here .

$y=\frac{-x^{2}}{40}$

Notice that since the balls follow the same graph, they will travel the same distance. To find how much distance the balls cover in the air, we first need to find when they will land again. We do this by defining the plank's position as $y=-x$ on our xy-plane. Now, we need to see when each ball's path will hit the plank.

$-x=\frac{-x^{2}}{40}$ $x=40m$ $y=-40m$

To use the formula to find the arc length of a segment of a graph, we first need to find the derivative of the graphed function.

$f(x)=\frac{-x^{2}}{40}$ $f'(x)=\frac{-x}{20}$

We will plug this derivative (along with our x-boundaries of $0$ and $40$ into the formula for arc length.

$S=\int_{a}^{b} \sqrt{1+[f'(x)]^{2}}\,\mathrm{d}x$ $S=\int_0^{40} \sqrt{1+[\frac{-x}{20}]^{2}}\,\mathrm{d}x$ $S=\int_0^{40} \sqrt{1+\frac{x^{2}}{400}}\,\mathrm{d}x$ $S=\int_0^{40} \sqrt{\frac{400+x^{2}}{400}}\,\mathrm{d}x$ $S=\frac{1}{20}\int_0^{40} \sqrt{400+x^{2}}\,\mathrm{d}x$

Perform a trigonometric substitution with $x=20\tan u$ .

$x=20\tan u$ $\mathrm{d}x=20\sec^{2} u\mathrm{d}u$ $u=\tan^{-1}\frac{x}{20}$ $S=\frac{20}{20}\int_0^{\tan^{-1}(2)} \sqrt{400+400\tan^{2}u}\,\sec^{2} u\mathrm{d}u$ $S=\int_0^{\tan^{-1}(2)} \sqrt{400(1+\tan^{2}u)}\sec^{2} u\,\mathrm{d}u$ $S=20\int_0^{\tan^{-1}(2)} \sqrt{\sec^{2} u}\sec^{2} u\,\mathrm{d}u$ $S=20\int_0^{\tan^{-1}(2)} \sec^{3} u\,\mathrm{d}u$

Use Integration by Parts to integrate $\int\sec^{3} u\,\mathrm{d}u$ .

$m=\sec u$ $\mathrm{d}m=\sec u\tan u\,\mathrm{d}u$ $\mathrm{d}n=\sec^{2} u\,\mathrm{d}u$ $n=\tan u$

$\int\sec^{3} u\,\mathrm{d}u=\sec u\tan u-\int\sec u\tan^{2} u\,\mathrm{d}u$ $\int\sec^{3} u\,\mathrm{d}u=\sec u\tan u-\int\sec u(\sec^{2} u-1)\,\mathrm{d}u$ $\int\sec^{3} u\,\mathrm{d}u=\sec u\tan u-\int\sec^{3} u\,\mathrm{d}u+\int\sec u\,\mathrm{d}u$ $2\int\sec^{3} u\,\mathrm{d}u=\sec u\tan u+\ln|\sec u+ \tan u| + C_{0}$ $\int\sec^{3} u\,\mathrm{d}u=\frac{1}{2}(\sec u\tan u+\ln|\sec u+ \tan u|)+C$

Plug that whole thing back in.

$S=20[\frac{1}{2}(\sec u\tan u+\ln|\sec u+\tan u|)]|_{0}^{\tan^{-1}(2)}$ $S=10[(\sqrt{5})(2)+\ln|(\sqrt{5})+(2)|]$ $S=59.1577m$

Since both balls follow the same path, each ball travels a distance of $\boxed{59.1577m}$ .