Chapter 3.5, Problem 29E

To determine

To find: The equation of line that are tangent and normal to the curve

  $y=\sqrt{2}\cos xat(\frac{\pi }{4},1)$

Answer to Problem 29E

The equation of tangent, $y=-x+\frac{\pi }{4}+1$

Equation of normal, $y=x-\frac{\pi }{4}+1$

Explanation of Solution

Given information: We are given equation of curve $y=\sqrt{2}\cos xand\text{point}(\frac{\pi }{4},1)$

Explanation: The equation of given curve, $y=\sqrt{2}\cos x$

Differentiate w.r.t $x$

  $\begin{array}{l}\frac{dy}{dx}=\frac{d}{dx}(\sqrt{2}\cos x)\\ =-\sqrt{2}\sin x\end{array}$

Slope of tangent (m) = $\frac{dy}{dx}$ = $-\sqrt{2}\sin x$

Slope of tangent at $(\frac{\pi }{4},1)$

  $m=-\sqrt{2}\sin \frac{\pi }{4}$

  $=-\sqrt{2}\times \frac{1}{\sqrt{2}}=-1$

Equation of tangent

  $\begin{array}{l}y-y_1=m(x-x_1)\\ y-1=-1(x-\frac{\pi }{4})\\ y-1=-x+\frac{\pi }{4}\\ y=-x+\frac{\pi }{4}+1\end{array}$

Slope of normal $(m_1)=\frac{-1}{\frac{dy}{dx}}=\frac{-1}{-\sqrt{2}\sin x}=\frac{1}{\sqrt{2}\sin x}$

Slope of normal at $(\frac{\pi }{4},1)$

  $(m_1)=\frac{1}{\sqrt{2}\sin \frac{\pi }{4}}=\frac{1}{\sqrt{2}\times \frac{1}{\sqrt{2}}}=1$

Equation of normal at $(\frac{\pi }{4},1)$ is:

  $\begin{array}{l}{y}_1-y_1=m_1(x-x_1)\\ y-1=1(x-\frac{\pi }{4})\\ y-1=x-\frac{\pi }{4}\\ y=x-\frac{\pi }{4}+1\end{array}$

Thus, equation of tangent is $y=-x+\frac{\pi }{4}+1$ and equation of normal is $y=x-\frac{\pi }{4}+1$