# Suppose r varies directly as the square of m, and inversely as s. If r = 15 when m = 15 and s =...

## Question:

Suppose r varies directly as the square of m, and inversely as s. If {eq}r = 15 \text{ when } m = 15 \text{ and } s = 9, {/eq} find r when {eq}m = 60 \text{ and } s = 9. {/eq}

## Proportional Equation:

At first, we will write a proportionality equation following the condition of the problem. By using the initial value of the variables, we can find the value of the constant of proportionality and then we will get the exact equation. Once we get this equation, we can find the required value of any variable knowing the remaining.

Following the condition of the problem,

{eq}\begin{align} r\propto \frac{m^2}{s}\\\ r &= k \frac{m^2}{s} & \left[k \text{ is constant of proportionality} \right]\\ 15 &= k \frac{15^2}{9} & \left[\text{ As we know } r = 15 \text{ when } m = 15 \text{ and } s = 9 \right]\\ 1 &= k \frac{15}{9} \\ k &= \frac {9} {15} \end{align} {/eq}

Now, the equation becomes

{eq}\begin{align} r &= \frac {9} {15} \frac{m^2}{s} & \left[ \text{ Plug in the value of constant of proportionality}\, k = \frac {9} {15} \right]\\ &= \frac {9} {15} \frac{60^2}{9} & \left[\text{Now, put in the value of m and s } \right]\\ &= \frac{3600}{15} \\ &= 240\\ \end{align} {/eq}

Therefore, the value of {eq}\displaystyle \boxed{\color{blue} { r=240 }} {/eq} 