Chapter 12, Problem 12CR

To determine

To define: The $r$ when $s=12$ and $t=4$ , $r=\frac{1}{2}$ when $s=6$ and $t=2$ ,where $r$ varies directly as the cube of $t$ and inversely as the square of $s$ .

The options are,

  $\text{a}\text{.}\frac{1}{4}$

  $\text{b}.0$

  $\text{c}\text{.}\frac{1}{2}$

  $\text{d}.1$

Answer to Problem 12CR

The option $\text{d}$ is correct.

Explanation of Solution

Given information:

The variable $r$ varies directly as the cube of $t$ and inversely as the square of $s$ and $r=\frac{1}{2}$ when $s=6$ and $t=2$ .

If a variable varies directly as one variable and inversely as another then resultant is known as combined variation.

The relationship expressed in the form,

  $zy=kx$ , where $k$ is a nonzero constant.

The variable $r$ varies directly as the cube of $t$ and inversely as the square of $s$ expressed as,

  $\begin{array}{l}r=\frac{k{t}^3}{s^2}\\ s^{2}r=k{t}^3\end{array}$

The variable $r=\frac{1}{2}$ when $s=6$ and $t=2$ then put the values in equation and solve for the value of $k$ ,

  $\begin{array}{l}{\left(6\right)}^2\frac{1}{2}=k{\left(2\right)}^3\\ 18=k{\left(2\right)}^3\\ k=\frac{18}{8}\\ k=\frac{9}{4}\end{array}$

Furthermore,

Solve for $r$ when $s=12$ and $t=4$ , $k=\frac{9}{4}$ put the all values in equation $s^{2}r=k{t}^3$ as,

  $\begin{array}{c}{\left(12\right)}^{2}r=\left(\frac{9}{4}\right){\left(4\right)}^3\\ 144r=\frac{9\cdot 64}{4}\\ r=\frac{576}{576}\\ r=1\end{array}$

The variable $r=1$ when $s=12$ and $t=4$ , $r=\frac{1}{2}$ .

Thus , option $\text{d}$ is correct.