Chapter 6.2, Problem 41PPS

(a)

To determine

To Find:

The inverse of each function.

Answer to Problem 41PPS

  $r=\sqrt{\frac{A}{\pi }}$

Explanation of Solution

Given:

The formula for the area of a circle is given by $A=\pi r^2$ where r is the radius of the circle.

Concept Used :

 A function  g is the inverse of a function  f if whenever

  $\begin{array}{l}y=f(x)\text{ then }x=g(y)\\ \text{That is ,}\\ \text{f and g are inverse functions if :}\\ \left[f\circ g\right](x)=x\\ and\\ \left[g\circ f\right](x)=x\end{array}$

Calculation:

  $\begin{array}{l}Given\\ A=\pi r^2\\ \text{Rewrite it by replacing A by 'y' and r by 'x':}\\ A(x)=\pi x^2\\ y=\pi x^2\\ \text{Exchange x and y:}\\ x=\pi y^2\\ \text{Solve for y :}\\ y^2=\frac{x}{\pi }\mathrm{..........}[\text{divide each side by }\pi \text{]}\\ y=\pm \sqrt{\frac{x}{\pi }}\mathrm{..........}[\text{Take square root each side]}\\ {\text{So, A}}^{-1}(x)=\pm \sqrt{\frac{x}{\pi }}\\ Or,\\ \text{If A is the area and r is the radius of the circle, then:}\\ r=\pm \sqrt{\frac{A}{\pi }}\\ \text{Since radius is always non-negative and a radical is always }\ge \text{0}\text{.}\\ \text{So, }\\ r=\sqrt{\frac{A}{\pi }}\end{array}$

(b)

To determine

To Find:

The radius of the circle whose area is 36 square centimeters.

Answer to Problem 41PPS

  $r\approx 3.39cm$

Explanation of Solution

Given:

From part (a) , we have :

  $r=\sqrt{\frac{A}{\pi }}$

is a function where we get the size radius of the circle from given area.

Calculation:

  $\begin{array}{l}Given\\ r=\sqrt{\frac{A}{\pi }}\\ and\\ A=36\\ \text{Substitute the value of A in the function:}\\ r=\sqrt{\frac{A}{\pi }}\\ \text{⇒}r=\sqrt{\frac{36}{\pi }}\\ \text{⇒}r\approx 3.39cm\\ \text{So, }the\text{ radius of the circle is 3}\text{.39 centimeters}\text{.}\end{array}$