Understanding Related Rates

In many real-world situations, two or more quantities change with time. If these quantities are related by an equation, then their rates of change are also related. The chain rule connects these rates.

dy/dt = dy/dx \times dx/dt

The chain rule links the rate of change of y with respect to time to the rate of change of x with respect to time.

For example, if a balloon is being inflated, both the radius and the volume change with time. Since volume depends on radius (V = 4/3 πr³), we can find how fast the volume changes if we know how fast the radius changes.

The General Strategy

Identify all quantities that change with time.
Write an equation relating those quantities.
Differentiate both sides with respect to time t.
Substitute known values and solve for the unknown rate.

Common Geometric Relationships

Many related rates problems use standard geometric formulas. Here are the most common ones you will encounter in HSC exams:

Circle

A = πr², C = 2πr

Sphere

V = 4/3πr³, S = 4πr²

Cone

V = 1/3πr²h

Cylinder

V = πr²h

Tip: In cone problems, the ratio r/h is often constant. Use similar triangles to express r in terms of h (or vice versa) before differentiating.

Solving a Related Rates Problem

Let us work through a classic example step by step.

Example: Expanding Circular Oil Slick

An oil slick spreads in a circle. The radius increases at 2 m/s. How fast is the area increasing when r = 10 m?

Step 1: A = πr², and dr/dt = 2 m/s.

Step 2: Differentiate with respect to t: dA/dt = 2πr · dr/dt.

Step 3: Substitute r = 10 and dr/dt = 2:

dA/dt = 2π(10)(2) = 40π

Answer: The area is increasing at 40π ≈ 125.7 m²/s.

Key Vocabulary

Rate of Change

How quickly a quantity changes with respect to another variable, often time (e.g. dr/dt).

Chain Rule

The rule that connects rates: dy/dt = dy/dx × dx/dt.

Similar Triangles

A geometric relationship often used to eliminate a variable in cone and shadow problems.

Implicit Differentiation

Differentiating both sides of an equation with respect to t, treating all variables as functions of t.

Worked Examples

1

A spherical balloon is inflated so that its volume increases at 100 cm³/s. Find the rate of increase of the radius when r = 5 cm.

Step 1: V = (4/3)πr³, dV/dt = 100 cm³/s.

Step 2: dV/dt = 4πr² · dr/dt

Step 3: 100 = 4π(25) · dr/dt = 100π · dr/dt

Answer: dr/dt = 1/π ≈ 0.318 cm/s

2

A 5 m ladder slides down a wall. The base moves away at 1 m/s. How fast is the top sliding down when the base is 3 m from the wall?

Step 1: x² + y² = 25, dx/dt = 1 m/s.

Step 2: Differentiate: 2x · dx/dt + 2y · dy/dt = 0

Step 3: When x = 3: y = 4. So 2(3)(1) + 2(4)(dy/dt) = 0

Answer: dy/dt = −3/4 m/s (the top is sliding down at 0.75 m/s).

3

Water flows into a conical tank (radius 3 m, height 6 m) at 2 m³/min. How fast is the water level rising when h = 4 m?

Step 1: By similar triangles: r/h = 3/6 = 1/2, so r = h/2.

Step 2: V = (1/3)π(h/2)²h = πh³/12

Step 3: dV/dt = πh²/4 · dh/dt. With dV/dt = 2 and h = 4:

2 = π(16)/4 · dh/dt = 4π · dh/dt

Answer: dh/dt = 1/(2π) ≈ 0.159 m/min

Knowledge Check

Select the correct answer for each question. Click "Check Answer" to see if you are right.

Question 1

A circle has area A = πr². If dr/dt = 3 cm/s, what is dA/dt when r = 5 cm?

15π cm²/s 30π cm²/s 10π cm²/s 25π cm²/s

Question 2

If y = x³ and dx/dt = 2, what is dy/dt when x = 3?

27 18 54 6

Question 3

For the ladder problem (x² + y² = 25), if the base is 4 m from the wall and moving at 0.5 m/s, how fast is the top sliding down?

−1/3 m/s −3/4 m/s −4/3 m/s −2/3 m/s

Question 4

The volume of a sphere increases at 36π cm³/s. What is dr/dt when r = 3 cm?

1 cm/s 3 cm/s 1/π cm/s 4 cm/s

Question 5

What is the first step in solving a related rates problem?

Differentiate with respect to t immediately Identify the variables and write an equation relating them Substitute the given values Solve for dy/dt

Related Rates