Rates of change and connected rates

A typical question: The radius of a circle is increasing at $0.5\ \text{cm s}^{-1}$. Find the rate of increase of its area when the radius is 6 cm.

Giveaway words: rate of change, rate of increase, per second, at the instant when.

• $t$ is the time in seconds. $\dfrac{dr}{dt}$ (read "d r by d t") is how fast $r$ changes as time passes. $\text{cm s}^{-1}$ means centimetres per second.
• A rate tells you how much a quantity changes each second. Here $\dfrac{dr}{dt}=0.5\ \text{cm s}^{-1}$ means the radius grows by 0.5 cm each second.
• $\dfrac{dA}{dt}$ means how many square centimetres the area changes by each second. Increasing means a positive rate; decreasing means a negative rate.

The same four steps every time:

1. Know and want: write the rate you're given and the rate you want. In the circle example, the radius grows by $0.5$ cm each second; we want the area change in square centimetres each second.
2. Differentiate the formula that links area and radius: $A = \pi r^2$ gives $\dfrac{dA}{dr} = 2\pi r$. This tells how much the area changes for each extra centimetre of radius.
3. Chain: write $\dfrac{dA}{dt} = \dfrac{dA}{dr} \times \dfrac{dr}{dt}$ with your letters, before any numbers. Why: (cm² of area per cm of radius) × (cm of radius per second) = cm² of area per second. It's the chain rule from the last lesson, with time as the variable. Don't think of it as cancelling fractions.
4. Numbers: put the values in and solve for the rate you want.