Hard A-Level Rates of change Questions

Challenging, exam-style A-Level Rates of change questions with worked solutions. Stretch yourself on the hardest rates of change, connected rates, chain rule, exponential growth problems.

rates of changeconnected rateschain ruleexponential growthdifferential equationdifferentiation
A-Level34 questionsStep-by-step solutions
Question 1
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}
Question 2
8 markschallenging
A circular ripple of radius rr expands over time tt, changing its area AA. 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

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

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

  2. Recall the chain rule for connected rates

    dAdt=dAdr×drdt\frac{dA}{dt}=\frac{dA}{dr}\times\frac{dr}{dt}

    Multiply the derivatives through the shared variable r.

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

    dAdr×drdt\frac{dA}{dr}\times\frac{dr}{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

    dAdt=dAdr×drdt\frac{dA}{dt}=\frac{dA}{dr}\times\frac{dr}{dt}

    This is the valid chain-rule statement.

Answer
dAdt=dAdr×drdt\frac{dA}{dt}=\frac{dA}{dr}\times\frac{dr}{dt}
Question 3
8 markschallenging
A spherical balloon of radius rr is inflated over 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 r both depend on tV\text{ and }r\text{ both depend on }t

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

  2. Recall the chain rule for connected rates

    dVdt=dVdr×drdt\frac{dV}{dt}=\frac{dV}{dr}\times\frac{dr}{dt}

    Multiply the derivatives through the shared variable r.

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

    dVdr×drdt\frac{dV}{dr}\times\frac{dr}{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=dVdr×drdt\frac{dV}{dt}=\frac{dV}{dr}\times\frac{dr}{dt}

    This is the valid chain-rule statement.

Answer
dVdt=dVdr×drdt\frac{dV}{dt}=\frac{dV}{dr}\times\frac{dr}{dt}
Question 4
8 markschallenging
A cone of base radius rr (height twice the radius) has volume V=2πr33V=\frac{2 \pi r^{3}}{3}. Which of the following is dVdr\frac{dV}{dr}?
Show worked solution

Worked solution

  1. Write down the formula

    V=2πr33V=\frac{2 \pi r^{3}}{3}

    State clearly what is being differentiated.

  2. Recall the power rule for differentiation

    ddr(a rn)=a n rn−1\frac{d}{dr}\left(a\,r^{n}\right)=a\,n\,r^{n-1}

    Multiply by the power and reduce the power by one.

  3. Differentiate with respect to r

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

    Differentiate the formula term by term.

  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 the derivative with each option

    choose the option equal to 2πr2\text{choose the option equal to }2 \pi r^{2}

    Choose the option equal to this expression.

  14. Eliminate the incorrect options

    reject expressions not equal to 2πr2\text{reject expressions not equal to }2 \pi r^{2}

    Options that differ (for example by integrating instead) are wrong.

  15. Select the correct option

    2πr22 \pi r^{2}

    This matches the derivative found above.

Answer
2πr22 \pi r^{2}
Question 5
8 markschallenging
A cube of side xx has volume V=x3V=x^{3}. Which of the following is dVdx\frac{dV}{dx}?
Show worked solution

Worked solution

  1. Write down the formula

    V=x3V=x^{3}

    State clearly what is being differentiated.

  2. Recall the power rule for differentiation

    ddx(a xn)=a n xn−1\frac{d}{dx}\left(a\,x^{n}\right)=a\,n\,x^{n-1}

    Multiply by the power and reduce the power by one.

  3. Differentiate with respect to x

    dVdx=3x2\frac{dV}{dx}=3 x^{2}

    Differentiate the formula term by term.

  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 the derivative with each option

    choose the option equal to 3x2\text{choose the option equal to }3 x^{2}

    Choose the option equal to this expression.

  14. Eliminate the incorrect options

    reject expressions not equal to 3x2\text{reject expressions not equal to }3 x^{2}

    Options that differ (for example by integrating instead) are wrong.

  15. Select the correct option

    3x23 x^{2}

    This matches the derivative found above.

Answer
3x23 x^{2}

Unlock 29 more Rates of change questions

Create a free account to work through every A-Level Rates of change question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Rates of change practice

Practice QuestionsWorked Solutions

Related Pure Maths topics

Areas and further applicationsMathematical argument and notationArithmetic sequences and seriesBinomial expansion (rational powers)Binomial expansionCircles