Chapter 2.5, Problem 32E

a.

To determine

To calculate: The tangent of curve at point P.

Answer to Problem 32E

The tangent to the curve is $y=4-\sqrt{2}\left(1+\sqrt{2}\right)\left(x-\frac{\pi }{4}\right)$ .

Explanation of Solution

Concept used:

The slope of a function is the derivative of the function.

Given information:

The function is $y=1+\sqrt{2}\csc x+\cot x$ and point P is $\left(\frac{\pi }{4},4\right)$

Calculation:

The function is $y=1+\sqrt{2}\csc x+\cot x$ .

Differentiate the equation of curve $y=1+\sqrt{2}\csc x+\cot x$ with respect to $x$ .

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(1+\sqrt{2}\csc x+\cot x\right)\\ =\frac{d}{dx}\left(1\right)+\sqrt{2}\frac{d}{dx}\left(\csc x\right)+\frac{d}{dx}\left(\cot x\right)\\ =0-\cot x\csc x-{\csc }^{2}x\\ =-\csc x\cot x-{\csc }^{2}x\end{array}$

Substitute $\frac{\pi }{2}$ for $x$ in the $m=-\csc x\cot x-{\csc }^{2}x$ .

  $\begin{array}{c}m=-\csc x\cot x-{\csc }^{2}x\\ =-\csc x\left(\cot x+\csc x\right)\\ =-\csc \left(\frac{\pi }{4}\right)\left(\cot \frac{\pi }{4}+\csc \frac{\pi }{4}\right)\\ =-\sqrt{2}\left(1+\sqrt{2}\right)\end{array}$

Substitute the values in the equation $y-y_1=m\left(x-x_1\right)$ .

  $\begin{array}{c}y-4=-\sqrt{2}\left(1+\sqrt{2}\right)\left(x-\frac{\pi }{4}\right)\\ y=4-\sqrt{2}\left(1+\sqrt{2}\right)\left(x-\frac{\pi }{4}\right)\end{array}$

Conclusion: So, tangent on the curve is $y=4-\sqrt{2}\left(1+\sqrt{2}\right)\left(x-\frac{\pi }{4}\right)$ .

b.

To determine

To calculate: The horizontal tangent of the curve at Q.

Answer to Problem 32E

The horizontal tangent to the curve is $y=undefined$ .

Explanation of Solution

Given information:

The function is $y=1+\sqrt{2}\csc x+\cot x$ and point P is $\left(\frac{\pi }{4},4\right)$

Concept used:

The function is said to have horizontal tangent if $\frac{dy}{dx}=0$ .

Calculation:

The function is $f\left(x\right)={\pi }^2$ .

The function is constant with respect to $x$

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(1+\sqrt{2}\csc x+\cot x\right)\\ =\frac{d}{dx}\left(1\right)+\sqrt{2}\frac{d}{dx}\left(\csc x\right)+\frac{d}{dx}\left(\cot x\right)\\ =0-\cot x\csc x-{\csc }^{2}x\\ =-\csc x\cot x-{\csc }^{2}x\end{array}$

Substitute $0$ for $x$ in the derivative.

  $\begin{array}{c}-{\csc }^{2}x=\cot x\csc x\\ \frac{\csc x}{\cot x}=-1\\ \sec x=-1\\ x={\sec }^{-1}\left(-1\right)\\ x=\pi \end{array}$

Substitute $\pi$ for $x$ in $y=1+\sqrt{2}\csc x+\cot x$ .

  $\begin{array}{c}y=1+\sqrt{2}\csc \left(\pi \right)+\cot \left(\pi \right)\\ y=undefined\end{array}$