1. Identify the quantities involved: Determine which quantities are changing with respect to time (or another variable).
2. Write down the relationship: Express the relationship between the quantities using an equation.
3. Differentiate the equation with respect to time: Use implicit differentiation to find the rates of change of the quantities with respect to time (or the other variable).
4. Substitute known values: Plug in the given values (including the known rate of change) and solve for the unknown rate of change.
5. Interpret the result: Make sure the solution makes sense in the context of the problem.

Example 1: Expanding Circle

Problem: 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?

• Step 1: 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 $t$ .
• Given: $\frac{dr}{dt}$ = 2 cm/s.
• Find: $\frac{dA}{dt}$ when $r = 5 cm.$

• Step 2: Write down the relationship:
• The area of a circle is given by $A = \pi r^2 .$

• Step 3: Differentiate the equation with respect to time: $\frac{dA}{dt} = \frac{d}{dt} \left(\pi r^2\right) = 2\pi r \frac{dr}{dt}$

• Step 4: Substitute known values:
• When $r = 5 cm\ and\ \frac{dr}{dt} = 2 cm/s:$ $\frac{dA}{dt} = 2\pi(5)(2) = 20\pi \text{ cm}^2/\text{s}$

• Step 5: Interpret the result:
• The area of the circle is increasing at a rate of 20π square centimetres per second when the radius is 5 cm.

Example 2: Draining Water from a Conical Tank

Problem: Water is draining from a conical tank at a rate of 3 cubic meters per minute. If the tank has a height of 12 meters and a base radius of 6 meters, how fast is the water level falling when the water is 4 meters deep?

• Step 1: Identify the quantities involved:
• Let V(t) be the volume of water in the tank at time t .
• Let h(t) be the height of the water in the tank at time t .
• Given: $\frac{dV}{dt}$= -3 m³/min (negative because the volume is decreasing).
• Find: $\frac{dh}{dt}$ when h = 4 meters.
• Step 2: Write down the relationship:
• The volume of a cone is $V = \frac{1}{3} \pi r^2 h .$
• The radius $r$ of the water surface and the height h of the water are related by similar triangles: $\frac{r}{h} = \frac{6}{12} = \frac{1}{2} .$
• So, $r = \frac{h}{2} .$
• Step 3: Substitute $r = \frac{h}{2}$ into the volume formula:

$\ V = \frac{1}{3} \pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3} \pi \frac{h^3}{4} = \frac{\pi h^3}{12}$

• Step 4: Differentiate the equation with respect to time: $\frac{dV}{dt} = \frac{d}{dt} \left(\frac{\pi h^3}{12}\right) = \frac{\pi}{4} h^2 \frac{dh}{dt}$

• Step 5: Substitute known values:

• Given $\frac{dV}{dt} = -3 m^3/min\ and\ h = 4$ meters:

• $-3 = \frac{\pi}{4} (4)^2 \frac{dh}{dt} \\$ $-3 = \frac{\pi}{4} \times 16 \times \frac{dh}{dt} \\$ $-3 = 4\pi \frac{dh}{dt} \\$ $\frac{dh}{dt} = -\frac{3}{4\pi} \text{ meters/minute}$

• Step 6: Interpret the result:

• The water level is falling at a rate of 3/(4π) meters per minute when the water is 4 meters deep.

• Always start by identifying the relationship between the quantities involved.
• Use implicit differentiation when necessary, and always differentiate with respect to the independent variable, typically time.
• Pay attention to the signs of the rates (positive for increasing quantities, negative for decreasing quantities).

Connected Rates of Change

The radius of a circle is increasing at a constant rate of 0.2 cm s⁻¹.

a) Show that the perimeter of the circle is increasing at the rate of 0.4π cm s⁻¹.

b) Find the rate at which the area of the circle is increasing when the radius is 10 cm.

c) Find the radius of the circle when its area is increasing at the rate of 20 cm² s⁻¹.

The area of a circle is decreasing at a constant rate of 0.5 cm² s⁻¹. a) Find the rate at which the radius of the circle is decreasing when the radius is 8 cm.

b) Find the rate at which the perimeter of the circle is decreasing when the radius is 8 cm.