Chapter 6, Problem 9P

To determine

To find : The length of the arc of the curve from the origin to the nearest point where there is a vertical tangent line.

Answer to Problem 9P

The required length is $\ln \left(\frac{\pi }{2}\right)$

Explanation of Solution

Given information :

The parametric equations are:

  $x={\displaystyle \underset{1}{\overset{t}{\int }}\frac{\cos u}{u}du}$ , $y={\displaystyle \underset{1}{\overset{t}{\int }}\frac{\sin u}{u}du}$

Formula used :

  $\begin{array}{l}x={\displaystyle \underset{1}{\overset{t}{\int }}f\left(u\right)}du\\ \Rightarrow \frac{dx}{du}=f\left(t\right)\end{array}$

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

Calculation :

Given that , $x={\displaystyle \underset{1}{\overset{t}{\int }}\frac{\cos u}{u}du}$ , $y={\displaystyle \underset{1}{\overset{t}{\int }}\frac{\sin u}{u}du}$

Now, $\frac{dx}{dt}=\frac{\cos t}{t}$ and $\frac{dy}{dt}=\frac{\sin t}{t}$

Vertical tangent occurs at $\frac{dx}{dt}=0$

So, $\frac{\cos t}{t}=0\Rightarrow \cos t=0\Rightarrow t=\frac{\pi }{2}$

Thus parameter corresponding to $\left(x,y\right)=\left(0,0\right)$ is $t=1$ .

So, the nearest vertical tangent occurs when, $t=\frac{\pi }{2}$

So, the length of the tangent between $t=1\text{ and }t=\frac{\pi }{2}$ is

  $\begin{array}{c}L={\displaystyle \underset{1}{\overset{\frac{\pi }{2}}{\int }}\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt}\\ ={\displaystyle \underset{1}{\overset{\frac{\pi }{2}}{\int }}\sqrt{\frac{{\cos }^{2}t}{t^2}+\frac{{\sin }^{2}t}{t^2}}}dt\\ ={\displaystyle \underset{1}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{t}}dt\\ ={{\displaystyle \left[\ln t\right]}}_1^{\frac{\pi }{2}}\\ =\ln \left(\frac{\pi }{2}\right)-0=\ln \left(\frac{\pi }{2}\right)\end{array}$