Chapter 5.5, Problem 8QR

To determine

To check: whether the given curve is concave up or concave down on the given interval.

Answer to Problem 8QR

Concave up.

Explanation of Solution

Given:

  $y=\csc x$ on $\left[0,\pi \right]$ .

Concept used:

A curve $y=f\left(x\right)$ is said to be concave down on the interval $\left[a,b\right]$ if $f^{″}\left(x\right)<0$ on $\left[a,b\right]$ and concave up if $f^{″}\left(x\right)>0$ on $\left[a,b\right]$ .

Calculation:

Consider the given equation,

  $y=\csc x$

Differentiating with respect to $x$ , it follows that

  $\frac{dy}{dx}=-\csc x\cot x$

Again differentiating with respect to $x$ , it follows that

  $\begin{array}{c}\frac{d^{2}y}{d{x}^2}=-\left[\frac{d\left(\csc x\right)}{dx}\cdot \cot x+\frac{d\left(\cot x\right)}{dx}\cdot \csc x\right]\\ =-\left[\left(-\csc x\cot x\right)\cot x+\left(-{\csc }^{2}x\right)\csc x\right]\\ =\csc x{\cot }^{2}x+{\csc }^{3}x\\ =\csc x\left({\cot }^{2}x+{\csc }^{2}x\right)\end{array}$

Now,

  $\begin{array}{c}\csc x>0\text{on}\left[0,\pi \right]\\ \csc x\left({\cot }^{2}x+{\csc }^{2}x\right)>0\text{on}\left[0,\pi \right]\end{array}$

So, $\frac{d^{2}y}{d{x}^2}>0$ on $\left[0,\pi \right]$ .

Hence the curve is concave up on $\left[0,\pi \right]$ .