A-Level Rates of change Practice Questions

Free A-Level Rates of change practice questions with full step-by-step worked solutions. Covers rates of change, differentiation, connected rates, chain rule. Practise exam-style problems and check your method.

rates of changedifferentiationconnected rateschain ruleexponential growthdifferential equation
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A sphere of radius rr has volume V=4πr33V=\frac{4 \pi r^{3}}{3}. Find dVdr\frac{dV}{dr}, the rate of change of the quantity with respect to rr.
Show worked solution

Worked solution

  1. Write down the given formula

    V=4πr33V=\frac{4 \pi r^{3}}{3}

    State the quantity as a function of the variable.

  2. Differentiate with respect to r

    ddr(4πr33)=4πr2\frac{d}{dr}\left(\frac{4 \pi r^{3}}{3}\right)=4 \pi r^{2}

    Apply the power rule to each term.

  3. State the derivative

    dVdr=4πr2\boxed{\frac{dV}{dr}=4 \pi r^{2}}

    This is the required rate of change.

Answer
4πr24 \pi r^{2}
Question 2
2 markseasy
A cube's side xx increases with time tt, so its volume VV changes. Which of the following correctly links the rates of change using the chain rule?
Show worked solution

Worked solution

  1. Recall the chain rule for connected rates

    dVdt=dVdx×dxdt\frac{dV}{dt}=\frac{dV}{dx}\times\frac{dx}{dt}

    The two rates are linked by multiplying through the shared variable.

  2. Compare with the five options

    find the option that multiplies the correct derivatives\text{find the option that multiplies the correct derivatives}

    The correct link multiplies x-derivatives, not divides or adds.

  3. Select the correct linking

    dVdt=dVdx×dxdt\frac{dV}{dt}=\frac{dV}{dx}\times\frac{dx}{dt}

    This is the valid chain-rule statement.

Answer
dVdt=dVdx×dxdt\frac{dV}{dt}=\frac{dV}{dx}\times\frac{dx}{dt}
Question 3
3 marksintermediate
A cube's side xx increases with time tt, so its surface area AA changes. Which of the following correctly links the rates of change using the chain rule?
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Worked solution

  1. Identify the quantities that change with time

    A and x both depend on tA\text{ and }x\text{ both depend on }t

    Both quantities vary with time, so their rates are connected.

  2. Recall the chain rule for connected rates

    dAdt=dAdx×dxdt\frac{dA}{dt}=\frac{dA}{dx}\times\frac{dx}{dt}

    Multiply the derivatives through the shared variable x.

  3. Check the derivative factors multiply, not add or divide

    dAdx×dxdt\frac{dA}{dx}\times\frac{dx}{dt}

    The linking operation in the chain rule is multiplication.

  4. Compare each option with the correct structure

    match the multiplication of the two derivatives\text{match the multiplication of the two derivatives}

    Only one option multiplies the correct pair of derivatives.

  5. Reject options that divide or add the derivatives

    ÷ and + are incorrect here\div\ \text{and}\ +\ \text{are incorrect here}

    Dividing or adding the derivatives does not follow from the chain rule.

  6. Select the correct linking

    dAdt=dAdx×dxdt\frac{dA}{dt}=\frac{dA}{dx}\times\frac{dx}{dt}

    This is the valid chain-rule statement.

Answer
dAdt=dAdx×dxdt\frac{dA}{dt}=\frac{dA}{dx}\times\frac{dx}{dt}
Question 4
5 markshard
An ice cube of side xx shrinks with time tt, changing its volume VV. Which of the following correctly links the rates of change using the chain rule?
Show worked solution

Worked solution

  1. Identify the quantities that change with time

    V and x both depend on tV\text{ and }x\text{ both depend on }t

    Both quantities vary with time, so their rates are connected.

  2. Recall the chain rule for connected rates

    dVdt=dVdx×dxdt\frac{dV}{dt}=\frac{dV}{dx}\times\frac{dx}{dt}

    Multiply the derivatives through the shared variable x.

  3. Check the derivative factors multiply, not add or divide

    dVdx×dxdt\frac{dV}{dx}\times\frac{dx}{dt}

    The linking operation in the chain rule is multiplication.

  4. Recall that a rate of change means differentiate with respect to time

    rate=ddt(quantity)\text{rate}=\frac{d}{dt}\left(\text{quantity}\right)

    Whenever a problem asks how fast something changes, differentiate it with respect to t.

  5. State the general chain rule that links two connected rates

    dydt=dydx×dxdt\frac{dy}{dt}=\frac{dy}{dx}\times\frac{dx}{dt}

    Connected rates are joined by multiplying derivatives through a shared variable.

  6. Note that the shared variable links the two quantities

    x=x(t)x=x(t)

    Because both quantities depend on t, their rates of change are connected.

  7. Remember the chain rule can be rearranged for an unknown rate

    dxdt=dy/dtdy/dx\frac{dx}{dt}=\frac{dy/dt}{dy/dx}

    If a different derivative is required, divide by the known one.

  8. Compare each option with the correct structure

    match the multiplication of the two derivatives\text{match the multiplication of the two derivatives}

    Only one option multiplies the correct pair of derivatives.

  9. Reject options that divide or add the derivatives

    ÷ and + are incorrect here\div\ \text{and}\ +\ \text{are incorrect here}

    Dividing or adding the derivatives does not follow from the chain rule.

  10. Select the correct linking

    dVdt=dVdx×dxdt\frac{dV}{dt}=\frac{dV}{dx}\times\frac{dx}{dt}

    This is the valid chain-rule statement.

Answer
dVdt=dVdx×dxdt\frac{dV}{dt}=\frac{dV}{dx}\times\frac{dx}{dt}
Question 5
8 markschallenging
A cylinder fills so its water depth hh rises over time tt, changing the volume VV. Which of the following correctly links the rates of change using the chain rule?
Show worked solution

Worked solution

  1. Identify the quantities that change with time

    V and h both depend on tV\text{ and }h\text{ both depend on }t

    Both quantities vary with time, so their rates are connected.

  2. Recall the chain rule for connected rates

    dVdt=dVdh×dhdt\frac{dV}{dt}=\frac{dV}{dh}\times\frac{dh}{dt}

    Multiply the derivatives through the shared variable h.

  3. Check the derivative factors multiply, not add or divide

    dVdh×dhdt\frac{dV}{dh}\times\frac{dh}{dt}

    The linking operation in the chain rule is multiplication.

  4. Recall that a rate of change means differentiate with respect to time

    rate=ddt(quantity)\text{rate}=\frac{d}{dt}\left(\text{quantity}\right)

    Whenever a problem asks how fast something changes, differentiate it with respect to t.

  5. State the general chain rule that links two connected rates

    dydt=dydx×dxdt\frac{dy}{dt}=\frac{dy}{dx}\times\frac{dx}{dt}

    Connected rates are joined by multiplying derivatives through a shared variable.

  6. Note that the shared variable links the two quantities

    x=x(t)x=x(t)

    Because both quantities depend on t, their rates of change are connected.

  7. Remember the chain rule can be rearranged for an unknown rate

    dxdt=dy/dtdy/dx\frac{dx}{dt}=\frac{dy/dt}{dy/dx}

    If a different derivative is required, divide by the known one.

  8. Keep any factor of pi exact throughout the working

    π≈3.14159\pi\approx 3.14159

    Working in terms of pi avoids rounding errors in the final rate.

  9. Track the units of each rate

    [dydt]=[y][t]\left[\frac{dy}{dt}\right]=\frac{\left[y\right]}{\left[t\right]}

    The units of a rate are the units of the quantity divided by time.

  10. Set up the algebra before substituting numbers

    dydt=dydx⋅dxdt\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}

    Arrange the symbols first so the substitution at the end is clean.

  11. Recall the standard volume and surface-area formulae

    V=43πr3, A=4πr2V=\tfrac{4}{3}\pi r^{3},\ A=4\pi r^{2}

    These mensuration results are needed to differentiate geometric quantities.

  12. Check that a positive rate corresponds to an increasing quantity

    dxdt>0 ⇒ increasing\frac{dx}{dt}>0\ \Rightarrow\ \text{increasing}

    A positive derivative means the quantity is growing with time.

  13. Compare each option with the correct structure

    match the multiplication of the two derivatives\text{match the multiplication of the two derivatives}

    Only one option multiplies the correct pair of derivatives.

  14. Reject options that divide or add the derivatives

    ÷ and + are incorrect here\div\ \text{and}\ +\ \text{are incorrect here}

    Dividing or adding the derivatives does not follow from the chain rule.

  15. Select the correct linking

    dVdt=dVdh×dhdt\frac{dV}{dt}=\frac{dV}{dh}\times\frac{dh}{dt}

    This is the valid chain-rule statement.

Answer
dVdt=dVdh×dhdt\frac{dV}{dt}=\frac{dV}{dh}\times\frac{dh}{dt}

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