Chapter 11, Problem 9RE

(a)

To determine

To determine: The equation for the tangent to the curve at the pint corresponding to the given value of t .

Answer to Problem 9RE

The required equation is $y=\frac{\sqrt{3}}{2}x+\frac{1}{4}$ .

Explanation of Solution

Given information:

The equations: $x=\left(\frac{1}{2}\right)\tan t$

  $y=\left(\frac{1}{2}\right)\sec t$

The value of t = $\frac{\pi }{3}$

Formula used:

Point-slope form of a line:

  $y-y_1=m\left(x-x_1\right)$

Formulas of derivatives:

  $\frac{d}{dx}\tan x={\sec }^{2}x$

  $\frac{d}{dx}\sec x=\sec x\tan x$

Calculation:

The given equations are $x=\left(\frac{1}{2}\right)\tan t$  ..... (1)

And, $y=\left(\frac{1}{2}\right)\sec t$  ..... (2)

Substitute $\frac{\pi }{3}$ for t in equation (1) and simplify.

  $\begin{array}{c}x=\left(\frac{1}{2}\right)\tan \left(\frac{\pi }{3}\right)\\ =\frac{1}{2}\left(\sqrt{3}\right)\\ =\frac{\sqrt{3}}{2}\end{array}$

Substitute $\frac{\pi }{3}$ for t in equation (2) and simplify.

  $\begin{array}{c}y=\left(\frac{1}{2}\right)\sec \left(\frac{\pi }{3}\right)\\ =\frac{1}{2}\left(2\right)\\ =1\end{array}$

Thus, the point through which the tangent passes is $\left(\frac{\sqrt{3}}{2},1\right)$ .

Find derivative of equation (1) and equation (2) with respect to t .

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left(\frac{1}{2}\right)\tan t\\ =\frac{1}{2}{\sec }^{2}t\end{array}$

And, $\begin{array}{c}\frac{dy}{dt}=\frac{d}{dt}\left(\frac{1}{2}\right)\sec t\\ =\frac{1}{2}\sec t\tan t\end{array}$

Now,

  $\begin{array}{c}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ =\frac{\frac{1}{2}\sec t\tan t}{\frac{1}{2}{\sec }^{2}t}\\ =\frac{\tan t}{\sec t}\\ =\frac{\frac{\sin t}{\cos t}}{\frac{1}{\cos t}}\\ =\sin t\end{array}$

Substitute $\frac{\pi }{3}$ for t in the above derivative.

  $\begin{array}{c}\frac{dy}{dx}=\sin \left(\frac{\pi }{3}\right)\\ =\frac{\sqrt{3}}{2}\end{array}$

So, the slope of the tangent is $\frac{\sqrt{3}}{2}$ .

To find the equation of the tangent line, use the formula for the point-slope form of the equation of a line $y-y_1=m\left(x-x_1\right)$ .

Substitute 1 for $y_1$ , $\frac{\sqrt{3}}{2}$ for $x_1$ and $\frac{\sqrt{3}}{2}$ for m in the above formula.

  $\begin{array}{c}y-1=\frac{\sqrt{3}}{2}\left(x-\frac{\sqrt{3}}{2}\right)\\ y-1=\frac{\sqrt{3}}{2}x-\frac{3}{4}\\ y=\frac{\sqrt{3}}{2}x-\frac{3}{4}+1\\ y=\frac{\sqrt{3}}{2}x+\frac{1}{4}\end{array}$

Hence, the required equation is $y=\frac{\sqrt{3}}{2}x+\frac{1}{4}$ .