Chapter 3.1, Problem 47E

To determine

To find : the equation of line tangent to the curve at the point $t=\pi /3$ .

Answer to Problem 47E

The equation of line tangent to the curve at point $t=\pi /3$ is $y-\frac{1}{2}=\sqrt{3}\left(x-\frac{\pi }{3}+\frac{\sqrt{3}}{2}\right)$ .

Explanation of Solution

Given information :

  $x=t-\sin t,y=1-\cos t$ .

Formula needed :

The equation of line tangent to the curve at point $t=a$ is $y-y_1=m\left(x-x_1\right)$ where $x_1,y_1$ are the value of $x$ and $y$ at $t=a$ respectively. Slope of line is $m={\left.\frac{dy}{dx}\right|}_{t=a}$ .

  $\frac{dy}{dx}$ partially expressed as $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ .

Partially differentiate $x$ and $y$ with respect to $t$ .

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left(t-\sin t\right)\\ =1-\cos t\\ \frac{dy}{dt}=\frac{d}{dt}\left(1-\cos t\right)\\ =\sin t\end{array}$

So,

  $\begin{array}{c}\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\\ =\frac{\sin t}{1-\cos t}\end{array}$

Slope is given by $m={\left.\frac{dy}{dx}\right|}_{t=\pi /3}$ .

  $\begin{array}{c}m={\left.\frac{dy}{dx}\right|}_{t=\pi /3}\\ =\frac{\sin \left(\frac{\pi }{3}\right)}{1-\cos \left(\frac{\pi }{3}\right)}\\ =\frac{\sqrt{3}/2}{1-1/2}\\ =\sqrt{3}\end{array}$

The value of $x$ and $y$ at $t=\pi /3$ is given by $x_1=\frac{\pi }{3}-\sin \left(\frac{\pi }{3}\right),y_1=1-\cos \left(\frac{\pi }{3}\right)$ .

  $x_1=\frac{\pi }{3}-\frac{\sqrt{3}}{2}$

And

  $\begin{array}{c}{y}_1=1-\frac{1}{2}\\ =\frac{1}{2}\end{array}$

Thus, the equation of line tangent to the curve at point $t=\pi /3$ is $y-\frac{1}{2}=\sqrt{3}\left(x-\left(\frac{\pi }{3}-\frac{\sqrt{3}}{2}\right)\right)$ .