Chapter 4, Problem 51RE

To determine

Find an equation for the line tangent to the curve at the point defined by the given value of t.

Answer to Problem 51RE

The equation of tangent line is $y=\frac{10}{3}x-\frac{15}{\sqrt{3}}$ .

Explanation of Solution

Given information: $x=3\sec t,y=5\tan t,t=\frac{\pi }{6}.$

Calculation:

Formula:

The equation of the tangent line is,

  $\begin{array}{l}y-y_1=m(x-x_1).\\ x=3\sec t,y=5\tan t,t=\frac{\pi }{6}so,\\ x_1=3\sec (\frac{\pi }{6})=3\times \frac{2}{\sqrt{3}}=\frac{6}{\sqrt{3}}.\\ y_1=5\tan (\frac{\pi }{6})=5\times \frac{1}{\sqrt{3}}=\frac{5}{\sqrt{3}}.\\ x=3\sec t\Rightarrow \frac{dx}{dt}=3\sec t.\tan t.\\ y=5\tan t\Rightarrow \frac{dy}{dt}=5{\sec }^{2}t.\\ \frac{dy}{dx}=\frac{5{\sec }^{2}t}{3\sec t.\tan t}=\frac{5}{3}\cos ect=\frac{5}{3}\cos ec(\frac{\pi }{6})=\frac{5}{3}\times 2=\frac{10}{3}.\\ \frac{dy}{dx}=m=\frac{10}{3}.\\ \text{So}\text{,}\text{the}\text{equation}\text{of}\text{tanget}\text{line}\text{is,}\\ (y-\frac{5}{\sqrt{3}})=\frac{10}{3}(x-\frac{6}{\sqrt{3}}).\\ y=\frac{10}{3}x-\frac{60}{3\sqrt{3}}+\frac{5}{\sqrt{3}}=\frac{10}{3}x-\frac{15}{\sqrt{3}}.\\ y=\frac{10}{3}x-\frac{15}{\sqrt{3}}.\end{array}$