Related Rates of Change

Related rates of change involve problems where two or more quantities are changing with respect to time or another variable, and these rates of change are related through a given equation. These problems typically arise in physical situations where multiple quantities depend on each other.

In other words, we can find a rate of change from two related rates of change :

\frac{dy}{dx} = \frac{dz}{dx} \cdot \frac{dy}{dz}

Notice that the above expression is just a fraction multiplication, so the term $dz$ will cancel out.

\frac{dy}{dx} = \frac{\cancel{dz}}{dx} \cdot \frac{dy}{\cancel{dz}}

Example

infoNote

A circle's radius is increasing at a rate of 2 cm per second. How fast is the area of the circle increasing when the radius is 5 cm?

First, identify the quantities involved.

Let $r(t)$ be the radius of the circle at time $t$ .
Let $A(t)$ be the area of the circle at time $t$ . We are given :

\frac{dr}{dt}=2cm/s.

What we want to find, is the rate of change in area with respect to time.

\frac{dA}{dt}

Let's write a related rate of change expression, using the radius $r$ .

\frac{dA}{dt} = \frac{dr}{dt} \cdot \frac{dA}{dr}

The term we need to find is $\frac{dA}{dr}$ . We can find this from the area of a circle equation.

A=\pi r^2

\frac{dA}{dr} = 2\pi r

Substitute known values into the related rate of change equation.

\frac{dA}{dt} = \frac{dr}{dt} \cdot \frac{dA}{dr}

\frac{dA}{dt} = 2 \cdot 2\pi r

= 4\pi r

When the radius is 5 cm ?

\frac{dA}{dt}=4 \pi (5)=20 \pi

The area is increasing at a rate of 20π cm²/s.

Related Rates of Change Simplified Revision Notes for Leaving Cert Mathematics

Related Rates of Change

A circle's radius is increasing at a rate of 2 cm per second. How fast is the area of the circle increasing when the radius is 5 cm?

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