7ARelatedRates


 * Related Rates**

A **__rate of change__** is a measure of how quickly something is changing.

Rate of change is equivalent to gradient or derivative.

In descriptions, look for the word **__rate__** **or** look for the word **__per__**.

Example ..... A bucket is filling at a **__rate__** of 5 litres **__per__** minute.

..... This is a change in Volume (V) with respect to time (t), so this would be written as:

math . \qquad \dfrac{dV}{dt}=5 \; \text{ litres/min} math

Rates of Change

Most rates of change are with respect to **__time__**(t). If nothing else is stated, assume time.
 * An __increasing__ rate of change is __positive__.
 * A __decreasing__ rate of change is __negative__.

Example 1 ... ... The area of a stain is __growing__ at 7cm 2 **__per__** hour.

math . \qquad \dfrac{dA}{dt}=7 \; \text{ cm}^2\text{/hour} math

Example 2 ... ... A water tank is found to be __leaking__ at a **__rate__** proportional to the depth of water in the tank.

... ... So the Volume of water (V) in the tank can be described as: math . \qquad \dfrac{dV}{dt}=-kh \, \text{ litres/min} math


 * k the constant of proportionality
 * h is the depth of water in the tank

Solving Problems with Related Rates

This topic uses the Chain Rule math . \qquad \dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx} math

Example ... ... The __**radius**__ of a __circular__ oil spill is __increasing__ at a constant of 3m/min.

... ... Find the rate at which the **__area__** of the spill is increasing when the radius is 8m.

From the question, we get that: math . \qquad \dfrac{dr}{dt}=3 \qquad \dfrac{dA}{dt}= \; ? math

We know the oil spill is circular, so: math , \qquad A=\pi r^2 \;\; \Rightarrow \;\; \dfrac{dA}{dr}=2\pi r math

Now write a chain rule that connects: math . \qquad \dfrac{dr}{dt}, \;\; \dfrac{dA}{dr} \; \text{ and } \; \dfrac{dA}{dt} math

We get the rule: math . \qquad \dfrac{dA}{dt}=\dfrac{dA}{dr} \times \dfrac{dr}{dt} math

substitute the values we know: math \\ . \qquad \dfrac{dA}{dt}=2\pi r \times 3 \\. \\ . \qquad \qquad = 6\pi r math

Now answer the last part of the question: When **r = 8** math \\ . \qquad \dfrac{dA}{dt} = 6\pi \times 8 \\. \\ . \qquad \qquad = 48\pi \;\; m^2/min math

... .... {Units for Area = m 2, units for time = min, so units for rate = m 2 /min}

Reciprocal of Derivatives

Sometimes you will need to take the reciprocal of a derivative to solve problems involving related rates.

Example math . \qquad \dfrac{dA}{dr}=5 \quad \Rightarrow \quad \dfrac{dr}{dA}=\dfrac{1}{5} math

**See also:** www.mathsonline.com.au Y12Advanced --> Calculus --> Calculus & the Physical World --> Lesson 3 Y12Extension --> Calculus --> Physical Applications --> Lesson 2 .