Chapter SH, Problem 8.15EP

To determine

To find: the value of $\text{a}$ if $\text{a}$ varies jointly as $\text{b}$ $\text{and}$ $\text{c}$ .

Answer to Problem 8.15EP

  $\text{a=-}\frac{9}{2}$ .

Explanation of Solution

Given:

  ${\text{a}}_1=-36,{\text{b}}_1=12{\text{ and c}}_1=-4$ .

Concept used:

Joint variation problems are solved using the equation

  $\text{a=kbc}$ .

From the first given equation the value of $\text{k}$ can be calculated using above relation.

Second the Value of $\text{a}$ using same value of constant $\text{k}$ with same relations having different value of $\text{b}$ and $\text{c}$ .

Calculation:

Joint variation problems are solved using the equation

  $\text{a=kbc}$ .

Let the given expression be

  ${\text{a}}_1{\text{=kb}}_1{\text{c}}_1$ .

  ${\text{a}}_1=-36,{\text{b}}_1=12{\text{ and c}}_1=-4$ .

  $\text{k=}\frac{{\text{a}}_1}{{\text{b}}_1{\text{c}}_1}$ .

  $\Rightarrow \text{k=}\frac{\text{-36}}{\left(12\right)\left(-4\right)}\text{=}\frac{\text{-36}}{-48}\text{=}\frac{3}{4}$ .

  $\text{k=}\frac{3}{4}$ .

Here according to the given ${\text{b}}_{\text{2}}\text{=2}$ and ${\text{c}}_{\text{2}}\text{=-3}$ .

  ${\text{a}}_2{\text{=kb}}_2{\text{c}}_2$ .

  ${\text{a}}_2\text{=}\left(\frac{3}{4}\right)\left(2\right)\left(-3\right)$ .

  $\Rightarrow {\text{a}}_2\text{=}\left(\frac{3}{4}\right)\left(2\right)\left(-3\right)=-\frac{9}{2}$ .

  ${\text{a}}_2\text{=-}\frac{9}{2}$ .

Hence, ${\text{a}}_2\text{=-}\frac{9}{2}$