Chapter 4.1, Problem 47E

To determine

To find:the equation of tangent line to the curve at a point.

Answer to Problem 47E

Required equation of the tangent to the given curve at $t=\frac{\pi }{3}$ is $y=\sqrt{3}x+2-\frac{\pi }{\sqrt{3}}$ .

Explanation of Solution

Given information:

1. The given equation of the curve is as follows

  $x=t-\sin t,y=1-\cos t$ .

2. Need to find the equation of tangent at point $t=\frac{\pi }{3}$ .

Formula used:

1. The slope of the tangent of the curve is $m=\frac{dy}{dx}$ .
2. equation of the tangent line is as follows

  $\left(y-y_0\right)=m\left(x-x_0\right)$ .

3. $\frac{dy}{dx}=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}$

Calculation:

Give equation of curve is

  $x=t-\sin t,y=1-\cos t$

As $x=t-\sin t$ , $\frac{dx}{dt}=\frac{d}{dt}\left(t-\sin t\right)$

  $=\frac{d}{dt}t-\frac{d}{dt}\sin t$

  $\frac{dx}{dt}=1-\cos t$

and $y=1-\cos t$ , $\frac{dy}{dt}=\frac{d}{dt}\left(1-\cos t\right)$

  $=\frac{d}{dt}1-\frac{d}{dt}\cos t$

  $\frac{dy}{dt}=\sin t$

Calculate slope of the tangent to the curve that is $m$ .

  $m=\frac{dy}{dx}$

  $=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}$

  $=\frac{\sin t}{1-\cos t}$

  $m=\frac{\sin t}{1-\cos t}$

Calculate value of $x_0,y_0\text{and}m$ at $t=\frac{\pi }{3}$ .

  $x_0=t-\sin t$

  $=\frac{\pi }{3}-\sin \left(\frac{\pi }{3}\right)$

  $=\frac{\pi }{3}-\frac{\sqrt{3}}{2}$

  $y_0=1-\cos t$

  $=1-\cos \left(\frac{\pi }{3}\right)$

  $=1-\frac{1}{2}$

  $=\frac{1}{2}$

  $m=\frac{\sin t}{1-\cos t}$

  $=\frac{\sin \left(\frac{\pi }{3}\right)}{1-\cos \left(\frac{\pi }{3}\right)}$

  $=\frac{\frac{\sqrt{3}}{2}}{1-\frac{1}{2}}$

  $m=\sqrt{3}$

Substitute calculated value of $x_0,y_0\text{and}m$ in equation of tangent line.

  $\left(y-y_0\right)=m\left(x-x_0\right)$

  $\left(y-\frac{1}{2}\right)=\sqrt{3}\left(x-\left(\frac{\pi }{3}-\frac{\sqrt{3}}{2}\right)\right)$

  $y=\sqrt{3}x-\frac{\sqrt{3}\pi }{3}+\frac{3}{2}+\frac{1}{2}$

  $y=\sqrt{3}x+2-\frac{\pi }{\sqrt{3}}$

Hence required equation of the tangent to the given curve at $t=\frac{\pi }{3}$ is $\underset{\_}{y=\sqrt{3}x+2-\frac{\pi }{\sqrt{3}}}$ .