Chapter 3.1, Problem 48E

To determine

To find : the equation of line tangent to the curve at the point $t=\pi /2$ .

Answer to Problem 48E

The equation of line tangent to the curve at point $t=\pi /2$ is $y=2$ .

Explanation of Solution

Given information :

  $x=\cos t,y=1+\sin t$ .

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 $t$ .

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

So,

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

Slope is given by $m={\left.\frac{dy}{dx}\right|}_{t=\pi /2}$ .

  $\begin{array}{c}m={\left.\frac{dy}{dx}\right|}_{t=\pi /2}\\ =-\cot \left(\frac{\pi }{2}\right)\\ =0\end{array}$

The value of $x$ and $y$ at $t=\pi /2$ is given by $x_1=\cos \frac{\pi }{2},y_1=1+\sin \frac{\pi }{2}$ .

  $x_1=0$

And

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

Thus, the equation of line tangent to the curve at point $t=\pi /2$ is $y-2=0\left(x-0\right)$ .