Chapter 4.3, Problem 14E

To determine

To calculate : The points of inflection of the function $y=x\sqrt{9-x^2}$ .

Answer to Problem 14E

The points of inflection of the function $y=x\sqrt{9-x^2}$ is $\left(0,0\right)$ .

Explanation of Solution

Given information : The function $y=x\sqrt{9-x^2}$ .

Formula used : The point of inflection of the graph is the point where the function has a tangent line an where the concavity of the graph changes.

Use the product, power, and quotient rule to differentiate.

Calculation :

Consider the function $y=x\sqrt{9-x^2}$ .

Calculate the first derivative of the function using product and power rule.

  $\begin{array}{c}{y}^{\prime }=x\left(\frac{-2x}{2\sqrt{9-x^2}}\right)+\sqrt{9-x^2}\\ =\frac{-x^2}{\sqrt{9-x^2}}+\left(9-x^2\right)\\ =\frac{-2x^2+9}{\sqrt{9-x^2}}\end{array}$

Calculate the second derivative of the function using quotient and power rule.

  $\begin{array}{c}{y}^{″}=\frac{\left(\sqrt{9-x^2}\right)\frac{d}{dx}\left(-2x^2+9\right)-\left(-2x^2+9\right)\frac{d}{dx}\left(\sqrt{9-x^2}\right)}{{\left(\sqrt{9-x^2}\right)}^2}\\ =\frac{\left(\sqrt{9-x^2}\right)\left(-4x\right)-\left(-2x^2+9\right)\left(\frac{-2x}{2\sqrt{9-x^2}}\right)}{\left(9-x^2\right)}\\ =\frac{\left(9-x^2\right)\left(-4x\right)+x\left(-2x^2+9\right)}{{\left(9-x^2\right)}^{\frac{3}{2}}}\\ =\frac{-36x+4x^3-2x^3+9x}{{\left(9-x^2\right)}^{\frac{3}{2}}}\end{array}$

Simplify further.

  $\begin{array}{c}{y}^{″}=\frac{-27x+2x^3}{{\left(9-x^2\right)}^{\frac{3}{2}}}\\ =\frac{x\left(2x^2-27\right)}{{\left(9-x^2\right)}^{\frac{3}{2}}}\end{array}$

Set $y^{″}=0$ and solve for x.

  $\begin{array}{c}\frac{x\left(2x^2-27\right)}{{\left(9-x^2\right)}^{\frac{3}{2}}}=0\\ x\left(2x^2-27\right)=0\\ x=0\end{array}$

And,

  $\begin{array}{c}2x^2-27=0\\ 2x^2=27\\ x^2=\frac{27}{2}\\ x=\pm 3.674\end{array}$

Draw the graph of the function $y=x\sqrt{9-x^2}$ .

  

It can be seen from the graph that the concavity of the graph changes at the point $\left(0,0\right)$ .

Thus, the points of inflection of the function $y=x\sqrt{9-x^2}$ is $\left(0,0\right)$ .