Rate of change with respect to a variable
A speedometer shows how fast the distance is changing at this moment. In the same way dy/dx tells how fast y changes when x changes.
If y = f(x), then dy/dx is the rate of change of y with respect to x. It is positive if y increases with x and negative if y decreases. At a given point, substitute that value of x.
Formula: Rate of change of y with respect to x = dy/dx
Rule 1: Differentiate and substitute — Find the derivative first, then put in the value.
- Easy: Area of a circle A = πr²: dA/dr = 2πr, which is 6π at r = 3.
- Medium: Volume of a cube V = x³: dV/dx = 3x², which is 12 at x = 2.
- Exam-type: Volume of a sphere V = (4/3)πr³: dV/dr = 4πr², which is 16π at r = 2.
Rule 2: Velocity as a rate — Position s(t) gives velocity s′(t).
- Easy: s = t³ − 2t: velocity v = 3t² − 2, which is 10 at t = 2.
- Medium: y = x² − 6x + 5 has rate zero where 2x − 6 = 0, i.e. at x = 3.
- Exam-type: Circumference C = 2πr: dC/dr = 2π, a constant.
Substitute the numbers only after differentiating.
Putting r = 3 into A = πr² before differentiating.
First differentiate, then plug in.
Practice questions
- 1Find the rate of change of the area of a circle with respect to its radius at r = 3.
- 2Find the rate of change of the volume of a cube with respect to its edge when x = 2.
- 3Find dy/dx at x = 2 if y = 3x² + 2.
- 4A particle has position s = t³ − 2t. Find its velocity at t = 2.
- 5Find the rate of change of the area of a square with respect to its side at side 7.
- 6Find the rate of change of the volume of a sphere with respect to its radius at r = 2.
- 7Find the rate of change of the circumference of a circle with respect to its radius.
- 8For y = x² − 6x + 5, find x where the rate of change is zero.
Show answers (try them yourself first)
- 12πr = 6π
- 23x² = 12
- 3dy/dx = 6x = 12
- 4v = 3t² − 2 = 10
- 5dA/dx = 2x = 14
- 64πr² = 16π
- 72π (a constant)
- 82x − 6 = 0, so x = 3
Rates of change with time (related rates)
When a balloon is blown up, its radius and its volume both change with time. The rate of one can be found from the rate of the other.
If a quantity depends on another which depends on time, use the chain rule: dA/dt = (dA/dr) · (dr/dt). Write the formula, differentiate with respect to t, then substitute the given values.
Formula: dA/dt = (dA/dr) × (dr/dt)
Rule 1: Find the rate in time — Differentiate the formula with respect to t.
- Easy: Circle radius grows at 2 cm/s: dA/dt = 2πr × 2, which is 20π cm²/s when r = 5.
- Medium: Cube edge grows at 3 cm/s: dV/dt = 3x² × 3, which is 144 cm³/s when x = 4.
- Exam-type: Balloon volume grows at 100 cm³/s: 4πr² dr/dt = 100, so at r = 5 the radius grows at dr/dt = 1/π cm/s.
Rule 2: Geometry link — Use a relation between the variables, then differentiate.
- Easy: A square's side grows at 2 cm/s: its perimeter grows at 8 cm/s and its area at 2x × 2 = 40 cm²/s when x = 10.
- Medium: A ladder: x² + y² = 25. Differentiating: 2x x′ + 2y y′ = 0.
- Exam-type: If x = 3, y = 4 and x′ = 1 m/s, then y′ = −3/4 m/s (the top slides down).
Write a relation first, then differentiate with respect to t, and only then substitute.
Substituting the numbers before differentiating.
Relate, differentiate in t, substitute.
Practice questions
- 1The radius of a circle grows at 3 cm/s. How fast is the area growing when r = 4?
- 2The edge of a cube grows at 2 cm/s. How fast is the surface area growing when the edge is 5 cm?
- 3The side of a square grows at 4 cm/s. How fast is its area growing when the side is 6 cm?
- 4The radius of a sphere grows at 0.5 cm/s. How fast is the volume growing at r = 4 cm?
- 5A balloon is filled at 200 cm³/s. How fast does its radius grow when r = 10 cm?
- 6A point moves on y = x² so that x grows at 3 units/s. How fast does y grow at x = 2?
- 7The radius of a circular oil spill grows at 2 cm/s. How fast does its circumference grow?
- 8A 5 m ladder leans on a wall. Its foot slides away at 1 m/s. How fast does the top slide down when the foot is 3 m from the wall?
Show answers (try them yourself first)
- 1dA/dt = 2πr × 3 = 24π cm²/s
- 2S = 6x², dS/dt = 12x × 2 = 120 cm²/s
- 3dA/dt = 2x × 4 = 48 cm²/s
- 4dV/dt = 4πr² × 0.5 = 32π cm³/s
- 54πr² dr/dt = 200, so dr/dt = 200 / (400π) = 1 / (2π) cm/s
- 6dy/dt = 2x × 3 = 12 units/s
- 7dC/dt = 2π × 2 = 4π cm/s
- 8x² + y² = 25: with x = 3, y = 4 the top slides down at 3/4 m/s