Chapter 12.8, Problem 9WE

To determine

To calculate: How does $h$ changes when $r$ is doubled and $d$ is halved.

In the formula $h^2=\frac{3r^2}{\pi d}$ .

Answer to Problem 9WE

The value of $h$ is $\sqrt{8}$ times as grater as the original value of $h$ , Then to more simplified $\sqrt{8}$ gives $2\sqrt{2}\approx 2.8$ .

Explanation of Solution

Given information:

How does $h$ changes when $r$ is doubled and $d$ is halved.

In the formula $h^2=\frac{3r^2}{\pi d}$ .

Calculation:

Consider does $h$ changes when $r$ is doubled and $d$ is halved.

In the formula $h^2=\frac{3r^2}{\pi d}$ .

As $y$ varies directly as $x$ , then the variation is direct and its equation is in the form $y=kx$ .

The variation as:

  $h^2=\frac{3r^2}{\pi d}$

Now find $h$ by take the square root on both sides:

  $\begin{array}{c}{h}^2=\frac{3r^2}{\pi d}\\ \sqrt{h^2}=\pm \sqrt{\frac{3r^2}{\pi d}}\\ h=\pm \sqrt{\frac{3r^2}{\pi d}}\end{array}$

Apply $r=2r,d=0.5d$ in the above result:

  $\begin{array}{c}h=\pm \sqrt{\frac{3r^2}{\pi d}}\\ h=\pm \sqrt{\frac{3{\left(2r\right)}^2}{\pi 0.5d}}\\ h=\pm \sqrt{\frac{3\left(4r^2\right)}{0.5\pi d}}\end{array}$

Simplified further to get the value of $h$:

  $\begin{array}{l}h=\pm \sqrt{\frac{12r^2}{0.5\pi d}}\\ h=\pm \sqrt{\frac{24r^2}{\pi d}}\\ h=\pm \sqrt{\frac{3r^2}{\pi d}\cdot 8}\\ h=\pm \sqrt{\frac{3r^2}{\pi d}}\cdot \sqrt{8}\end{array}$

The value of $h$ is:

  $h=\pm \sqrt{\frac{3r^2}{\pi d}}\cdot \sqrt{8}$

Then, the value of $h$ is $\sqrt{8}$ times as grater as the original value of $h$ .

Then to more simplified $\sqrt{8}$ gives $2\sqrt{2}\approx 2.8$ .

Thus, the value of $h$ is $2\sqrt{2}\approx 2.8$ .