Review of Related Rates

Expert reviewed • 21 July 2024 • 4 minute read

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Video coming soon!

Related Rates

Going back to the year 11 course, we learnt about the concept of related rates, and using the chain rule to solve problems involving rates and geometry.

To solve a related rates question, it is easiest to first identify all relevant formulas. It is then important to draw a diagram of the problem if possible, to visually conceptualise and identify the problem. The next step would be to derive relevant formulas, to apply them to the chain rule.

\frac{dy}{dt}=\frac{dy}{dx}\times\frac{dx}{dt}

Thus, by manipulation of the chain rule, and substituting in any known variables, we can find the solution to any related rates problem. Don’t forget to include units!

Practice Question 1

A spherical balloon is inflating and increasing in size. Find the rate at which the volume is increasing when the radius is 8 cm, if the radius is increasing at 2 cm/s.

First we must derive a formula for the volume of the spherical balloon and derive it. In the question we are told that the balloon is a sphere and thus the formula for its volume is:

V=\frac{4}{3}\pi r^3

Differentiating to find $\frac{dV}{dr}$ :

\frac{dV}{dr}=3\times\frac{4}{3}\pi r^2\\=4\pi r^2

Substituting $r=8$ , given by the question, we find $\frac{dV}{dr}$ to be:

\frac{dV}{dt}=4\pi(8)^2\\=4\pi(64)\\=256\pi

Now, adapting the chain rule to this question we get:

\frac{dV}{dt}=\frac{dV}{dr}\times \frac{dr}{dt}

Substituting known values found above and in the question, we can see that $\frac{dV}{dt}$ equals to:

\frac{dV}{dt}=256\pi\times 2\\=512\pi

$\therefore$ The volume of the balloon is increasing at a rate of $512\pi\: cm^3/s$

Integrating with respect to Time

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Rates and Differentiation

Return to Module 7: Motion and Rates