Chapter 11.1, Problem 50E

To determine

To determine: The correct option for the total length of the curve.

Answer to Problem 50E

The correct option is (D).

Explanation of Solution

Given information:

The parametric equations: $x=6\sin t-3\sin \left(7t\right)$ , $y=6\cos t-3\cos \left(7t\right)$

The graph:

  

The options:

(A) ${\displaystyle \underset{0}{\overset{2\pi }{\int }}\sqrt{{\left(6\sin t-3\sin \left(7t\right)\right)}^2+{\left(6\cos t-3\cos \left(7t\right)\right)}^2}dt}$

(B) ${\displaystyle \underset{0}{\overset{2\pi }{\int }}\sqrt{{\left(6\cos t-3\cos \left(7t\right)\right)}^2+{\left(6\sin t-3\sin \left(7t\right)\right)}^2}dt}$

(C) ${\displaystyle \underset{0}{\overset{2\pi }{\int }}\sqrt{{\left(6\cos t-21\cos \left(7t\right)\right)}^2-{\left(6\sin t-21\sin \left(7t\right)\right)}^2}dt}$

(D) ${\displaystyle \underset{0}{\overset{2\pi }{\int }}\sqrt{{\left(6\cos t-21\cos \left(7t\right)\right)}^2+{\left(-6\sin t+21\sin \left(7t\right)\right)}^2}dt}$

(E) ${\displaystyle \underset{0}{\overset{2\pi }{\int }}\sqrt{{\left(6\cos t-3\cos \left(7t\right)\right)}^2+{\left(6\sin t+3\sin \left(3t\right)\right)}^2}dt}$

Calculation:

The parametric equations are $x=6\sin t-3\sin \left(7t\right)$ and $y=6\cos t-3\cos \left(7t\right)$ .

Differentiate the parametric equations with respect to t .

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left(6\sin t-3\sin 7t\right)\\ =6\cos t-21\cos 7t\end{array}$

And, $\begin{array}{c}\frac{dy}{dt}=\frac{d}{dt}\left(6\cos t-3\cos 7t\right)\\ \frac{dy}{dt}=-6\sin t+21\sin 7t\end{array}$

Consider that L be the length of the parametric curve traversed when t increases $t_1$ to $t_2$ .

So, $L={\displaystyle \underset{t_2}{\overset{t_2}{\int }}\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}}dt$

Write the length of the path traveled by the projectile as:

  $L={\displaystyle \underset{0}{\overset{2\pi }{\int }}\sqrt{\left(6\cos t-21\cos 7t\right)+\left(-6\sin t+21\sin 7t\right)}dt}$

Hence, the correct option is (D).