Chapter 4.1, Problem 41E

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 41E

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

Explanation of Solution

Given information: $x=2\cos t,y=2\sin t,t=\pi /4.$

Calculation:

The equation of the line tangent is,

  $\begin{array}{l}y-y_1=m(x-x_1)\mathrm{......}(1).\\ x_1=2\cos t=2\cos (\pi /4)=2\times (\sqrt{2}/2)=\sqrt{2}.\\ y_1=2\sin t=2\sin (\pi /4)=2\times (\sqrt{2}/2)=\sqrt{2}.\\ m=dy/dx=(dy/dt)/(dx/dt)=(d(2\sin t)/dt)/(d(2\cos t)/dt).\\ m=2\cos t/2(-\sin t)=-\cot t=-1\times 1=-1.\\ \text{putting}\text{the}\text{value}\text{of}{\text{x}}_{\text{1}}{\text{,y}}_{\text{1}}\text{and}\text{m}\text{in}\text{equation}\text{(1)}\text{.}\\ y-\sqrt{2}=(-1)(x-\sqrt{2}).\\ y-\sqrt{2}=-x+\sqrt{2}.\\ y=-x+2\sqrt{2}.\end{array}$

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