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Connected rates of change and small changes Cambridge IGCSE Additional Mathematics revision

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In plain words

A balloon is being blown up. You know how fast its radius is growing, and you want to know how fast its volume is growing. The two rates are connected through the formula linking volume and radius, and the chain rule makes the connection.

5 things to know

  1. A rate of change is a derivative with respect to time: dr/dt is the rate at which r is changing.
  2. Connected rates: dy/dt = dy/dx × dx/dt. Differentiate the formula that links the two quantities, then multiply by the rate you know.
  3. Work out the derivative at the moment the question specifies, such as "when r = 4".
  4. Small changes: if x changes by a small amount δx, the change in y is approximately δy ≈ (dy/dx) × δx.
  5. The same relation turned round finds the small change in x that produces a given small change in y.

Worked example

The radius of a circle is increasing at 0.5 cm per second. Find the rate at which the area is increasing when the radius is 4 cm.

  1. A = πr², so dA/dr = 2πr. When r = 4, dA/dr = 8π.
  2. dA/dt = dA/dr × dr/dt = 8π × 0.5 = 4π.
  3. The area is increasing at 4π, about 12.6, cm² per second.

Worked example

y = x³. Find the approximate change in y when x increases from 2 to 2.01.

  1. dy/dx = 3x², which is 12 when x = 2.
  2. δy ≈ 12 × 0.01 = 0.12.

Tips and tricks

  • Write down what is given and what is wanted as derivatives, such as dr/dt = 0.5 and dA/dt = ?. The chain rule to use then becomes obvious.
  • For small changes, work out dy/dx at the starting value of x, not the final one.
5 questions, about 2 minutes.

It lands in your notebook with its questions as flashcards.

Connected rates of change and small changes: 5 questions and answers

These are the quiz’s questions. Do the quiz first, then come back here for the ones that got you.

  1. Which rule connects dy/dt, dy/dx and dx/dt?
    • dy/dt = dy/dx + dx/dt
    • dy/dt = dy/dx × dx/dt (the answer)
    • dy/dt = dy/dx ÷ dx/dt
    • dy/dt = dx/dt − dy/dx

    The chain rule.

  2. y = x² and x is increasing at 3 units per second. How fast is y increasing when x = 5?
    • 10
    • 15
    • 30 (the answer)
    • 75

    dy/dx = 10, and 10 × 3.

  3. What is the approximate change in y, given a small change δx in x?
    • δx ÷ (dy/dx)
    • (dy/dx) × δx (the answer)
    • dy/dx + δx
    • y × δx

    The gradient times the small step.

  4. y = 5x². What is the approximate change in y when x increases from 1 to 1.02?
    • 0.1
    • 0.2 (the answer)
    • 0.5
    • 10

    dy/dx = 10 at x = 1, and 10 × 0.02.

  5. What does dV/dt represent?
    • the volume
    • the rate at which the volume changes with time (the answer)
    • the time taken
    • the rate at which time changes

    A derivative with respect to time is a rate.

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