Chapter 4.1, Problem 48E

To determine

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

Answer to Problem 48E

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

Explanation of Solution

Given information:

1. The given equation of the curve is as follows

  $x=\cos t,y=1+\sin t$ .

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

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

  $x=\cos t,y=1+\sin t$

As $x=\cos t$ , $\frac{dx}{dt}=\frac{d}{dt}\cos t$

  $=\frac{d}{dt}\cos t$

  $\frac{dx}{dt}=-\sin t$

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

  $=\frac{d}{dt}1+\frac{d}{dt}\sin t$

  $\frac{dy}{dt}=\cos t$

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

  $m=\frac{dy}{dx}$

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

  $=\frac{\cos t}{-\sin t}$

  $m=-\cot t$

Calculate value of $x_0,y_0\text{and}m$ at $t=\frac{\pi }{2}$ .

  $x_0=\cos \frac{\pi }{2}$

  $=0$

  $y_0=1+\sin t$

  $=1+\sin \left(\frac{\pi }{2}\right)$

  $=1+1$

  $=2$

  $m=-\cot t$

  $=-\cot \left(\frac{\pi }{2}\right)$

  $m=0$

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

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

  $\left(y-2\right)=0\left(x-0\right)$

  $y=2$

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