Proportionality Problems

Index card: 12ab

This lesson draws on and extends the concepts from the preceding lesson: Direct and Inverse Variation.

A Typical Proportionality Problem: Finding the Constant of Proportionality

Question: Write an equation for

‘$A\,$ is proportional to the square of $\,t\,,$ and inversely proportional to the cube of $\,x\,.$’

If $\,A = 3\,$ when $\,t = 1\,$ and $\,x = 2\,,$ find the constant of proportionality.

What is the value of $\,A\,$ when $\, t = -1\,$ and $\,x = 4\,$?

Solution:

$\displaystyle A = k\cdot \frac{t^2}{x^3}$

Write the equation that describes the relationship between the variables, using the information from Direct and Inverse Variation. Don't forget the constant of proportionality!

$\displaystyle 3 = k\cdot\frac{1^2}{2^3}\,,$ $\displaystyle 3 = \frac{k}{8}$

Substitute the known values of $\,A\,,$ $\,t\,$ and $\,x\,$; simplify.

$k = 24$

Solve for the constant of proportionality, $\,k\,.$

$\displaystyle A = 24\cdot\frac{t^2}{x^3}\,,$ $\displaystyle A= \frac{24t^2}{x^3}$

The final equation can be written in slightly different ways.

$\displaystyle A = \frac{24\cdot (-1)^2}{4^3} = \frac{24}{64} = \frac{3}{8}$

Now, whenever any two of the three variables are known, the remaining variable can be determined.