Chapter 6.4, Problem 15E

To determine

Find the exact length of the curves:

Answer to Problem 15E

The length of the curve is

Explanation of Solution

Given information: Given curves

Calculation:

Graph of the curves

  

If a smooth curve with parametric equations

Is traversed exactly once as increases from then its

Length is $L={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}}dt$

  $\begin{array}{l}x=e^t\cos t\\ y=e^t\sin et,0\le \theta \le \pi \\ {\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2={\left[e^t-\left(\cos t+\sin t\right)\right]}^2+{\left[e^t\left(\sin t+\cos t\right)\right]}^2\\ ={\left(e^t\right)}^2\left({\cos }^{2}t-2\cos t\sin t+{\sin }^{2}t\right)+{\left(e^t\right)}^2\left({\sin }^{2}t+2\sin t\cos t+{\cos }^{2}t\right)\\ =e^{2t}\left(2{\cos }^{2}t+2{\sin }^{2}t\right)\\ =2e^{2t}\end{array}$

So,

  $\begin{array}{l}L={\displaystyle \underset{0}{\overset{\pi }{\int }}\sqrt{2e^{2t}}dt}\\ ={\displaystyle \underset{0}{\overset{\pi }{\int }}\sqrt{2}{e}^{t}dt}\\ =\sqrt{2}{\left[e^t\right]}_0^{\pi }\\ =\sqrt{2}\left(e^{\pi }-1\right)\end{array}$

Thus, the length of the curve is