Chapter 3.5, Problem 32E

To determine

To find: The equation of tangent at P $(\frac{\pi }{4},4)$ and horizontal tangent at point Q.

Answer to Problem 32E

The equation of tangent at P

  $y=-4x+\pi +4$

Equation of tangent at Q

  $y-2=0$

Explanation of Solution

Given information: We are given equation of curve $y=1+\sqrt{2}\text{cosec}x+\cot x$ and point $p=(\frac{\pi }{4},4)$

The equation of given curve is

  $\begin{array}{l}y=1+\sqrt{2}\text{cosec}x+\cot x\\ \frac{dy}{dx}=-\sqrt{2}\text{cosec}x\cot x-{\text{cosec}}^{2}x\\ \frac{dy}{dx}atP(\frac{\pi }{4},4)\\ \frac{dy}{dx}=-\sqrt{2}\text{cosec}\frac{\pi }{4}\cot \frac{\pi }{4}-{\text{cosec}}^2\frac{\pi }{4}\\ $ =-\sqrt{2}\times \sqrt{2}\times 1-{(\sqrt{2})}^2$=-2-2\\ =-4\end{array}$

Slope of tangent at point P

  $m=-4$

Equation of tangent at point P

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

Horizontal tangent is parallel to $x$ -axis

Slope at point Q = 0

  $-\sqrt{2}\text{cosec}x\cot x-{\text{cosec}}^{2}x=0$

  $\text{-cosec}x(\sqrt{2}\cot x+\text{cosec}x)=0$

  $\text{cosec}x=0\sqrt{2}\cot x+\text{cosec}x=0$

  $notpossible\sqrt{2}\frac{\cos x}{\sin x}=\frac{-1}{\sin x}$

  $\sqrt{2}\sin x\cos x=-\sin x$

  $\sqrt{2}\sin x\cos x+\sin x=0$

  $\sin x(\sqrt{2}\cos x+1)=0$

  $\sin x=0\sqrt{2}\cos x+1=0$

  $x=0\cos x=\frac{1}{\sqrt{2}}$

  $notpossiblex=\frac{3\pi }{4}$

  $Whenx=\frac{3\pi }{4}$

  $y=1+\sqrt{2}\text{cosec}\frac{3\pi }{4}+\cot \frac{3\pi }{4}$

  $=1+\sqrt{2}\times \sqrt{2}+(-1)$

  $=1+2-1=2$

  $Q(\frac{3\pi }{4},2)$

Equation of horizontal tangent

  $\begin{array}{l}y-y_1=0(x-x_1)\\ y-2=0(x-\frac{3\pi }{4})\\ y-2=0\end{array}$