Chapter 4.1, Problem 42E

To determine

To find: the equation of line tangent to the curve at the point defined by the given value of t.

Answer to Problem 42E

The equation of line tangent $y=\sqrt{3}x+2.$

Explanation of Solution

Given information: $x=\sin 2\pi t,y=\cos 2\pi t,t=-1/6.$

Calculation:

The equation of the line tangent is,

  $\begin{array}{l}y-y_1=m(x-x_1)\mathrm{......}(1).\\ x_1=\sin (2\pi t)=\sin (2\pi \times (-1/6)=\sin (-\pi /3)=-\sqrt{3}/2.\\ y_1=\cos (2\pi t)=\cos (2\pi \times (-1/6)=\cos (-\pi /3)=1/2.\\ m=dy/dx=(dy/dt)/(dx/dt)=(d(\cos (2\pi t))/dt)/(d(\sin (2\pi t))/dt).\\ m=-2\pi \sin (2\pi t)/2\pi \cos (2\pi t)=-\tan (2\pi t)=-(-\sqrt{3})=\sqrt{3}.\\ \text{putting}\text{the}\text{value}\text{of}{\text{x}}_{\text{1}}{\text{,y}}_{\text{1}}\text{and}\text{m}\text{in}\text{equation}\text{(1)}.\\ y-1/2=\sqrt{3}(x+\sqrt{3}/2).\\ y-1/2=\sqrt{3}x+3/2.\\ y=\sqrt{3}x+3/2+1/2.\\ y=\sqrt{3}x+2.\end{array}$

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