A-Level Parametric differentiation Practice Questions

Free A-Level Parametric differentiation practice questions with full step-by-step worked solutions. Covers parametric-differentiation, chain-rule. Practise exam-style problems and check your method.

parametric-differentiationchain-rule
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A curve is defined parametrically by x=t2x=t^{2} and y=t3y=t^{3}. Find dydx\dfrac{dy}{dx} in terms of tt.
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Worked solution

  1. Differentiate x and y with respect to t

    dxdt=2t,dydt=3t2\frac{dx}{dt}=2 t,\quad \frac{dy}{dt}=3 t^{2}

    Differentiate each parametric equation separately.

  2. Divide to find dy/dx

    dydx=3t22t=3t2\frac{dy}{dx}=\frac{3 t^{2}}{2 t}=\frac{3 t}{2}

    Use dy/dx = (dy/dt)/(dx/dt) and simplify.

  3. State dy/dx in terms of t

    dydx=3t2\frac{dy}{dx}=\frac{3 t}{2}

    This is the gradient of the curve at parameter t.

Answer
3t2\frac{3 t}{2}
Question 2
2 markseasy
A curve is defined parametrically by x=t2x=t^{2} and y=ty=t. Describe the tangent to the curve at the point where t=0t=0.
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Worked solution

  1. Differentiate x and y with respect to t

    dxdt=2t,dydt=1\frac{dx}{dt}=2 t,\quad \frac{dy}{dt}=1

    Differentiate each parametric equation separately.

  2. Divide to find dy/dx

    dydx=12t=12t\frac{dy}{dx}=\frac{1}{2 t}=\frac{1}{2 t}

    Use dy/dx = (dy/dt)/(dx/dt) and simplify.

  3. Compare dx/dt and dy/dt to classify the tangent

    dxdt=0,dydt=1\frac{dx}{dt}=0,\quad \frac{dy}{dt}=1

    Zero dy/dt gives a horizontal tangent; zero dx/dt gives a vertical tangent.

Answer
Vertical
Question 3
3 marksintermediate
A curve is defined parametrically by x=t2x=t^{2} and y=t3−3ty=t^{3} - 3 t. For which value of tt is the tangent to the curve horizontal?
Show worked solution

Worked solution

  1. Write down the parametric equations

    x=t2,y=t3−3tx=t^{2},\quad y=t^{3} - 3 t

    Both coordinates are expressed in terms of the parameter t.

  2. Recall the parametric chain rule

    dydx=dy/dtdx/dt\frac{dy}{dx}=\frac{dy/dt}{dx/dt}

    The gradient is the ratio of the two rates of change with respect to t.

  3. Differentiate x with respect to t

    dxdt=2t\frac{dx}{dt}=2 t

    Differentiate the expression for x term by term.

  4. Differentiate y with respect to t

    dydt=3t2−3\frac{dy}{dt}=3 t^{2} - 3

    Differentiate the expression for y term by term.

  5. Form and simplify dy/dx

    dydx=3t2−32t=3(t2−1)2t\frac{dy}{dx}=\frac{3 t^{2} - 3}{2 t}=\frac{3 \left(t^{2} - 1\right)}{2 t}

    Divide dy/dt by dx/dt and cancel any common factors.

  6. Identify the parameter giving a horizontal tangent

    dydt=3t2−3=0 ⇒ t=−1\frac{dy}{dt}=3 t^{2} - 3=0\ \Rightarrow\ t=-1

    Horizontal tangents occur where dy/dt = 0.

Answer
t=−1t=-1
Question 4
5 markshard
A curve is defined parametrically by x=t2x=t^{2} and y=t3−12ty=t^{3} - 12 t. For which value of tt is the tangent to the curve horizontal?
Show worked solution

Worked solution

  1. Write down the parametric equations

    x=t2,y=t3−12tx=t^{2},\quad y=t^{3} - 12 t

    Both coordinates are expressed in terms of the parameter t.

  2. Recall the parametric chain rule

    dydx=dy/dtdx/dt\frac{dy}{dx}=\frac{dy/dt}{dx/dt}

    The gradient is the ratio of the two rates of change with respect to t.

  3. Differentiate x with respect to t

    dxdt=2t\frac{dx}{dt}=2 t

    Differentiate the expression for x term by term.

  4. Differentiate y with respect to t

    dydt=3t2−12\frac{dy}{dt}=3 t^{2} - 12

    Differentiate the expression for y term by term.

  5. Form and simplify dy/dx

    dydx=3t2−122t=3t2−6t\frac{dy}{dx}=\frac{3 t^{2} - 12}{2 t}=\frac{3 t}{2} - \frac{6}{t}

    Divide dy/dt by dx/dt and cancel any common factors.

  6. Find where the tangent is horizontal

    dydt=0 ⇒ t=−2, t=2\frac{dy}{dt}=0\ \Rightarrow\ t=-2,\ t=2

    A horizontal tangent occurs where dy/dt = 0 (with dx/dt non-zero).

  7. Find where the tangent is vertical

    dxdt=0 ⇒ t=0\frac{dx}{dt}=0\ \Rightarrow\ t=0

    A vertical tangent occurs where dx/dt = 0 (with dy/dt non-zero).

  8. Find the second derivative

    d2ydx2=ddt ⁣(dydx)÷dxdt=34t+3t3\frac{d^2y}{dx^2}=\frac{d}{dt}\!\left(\frac{dy}{dx}\right)\div\frac{dx}{dt}=\frac{3}{4 t} + \frac{3}{t^{3}}

    Differentiate dy/dx with respect to t, then divide by dx/dt again.

  9. State the gradient at t=1

    dydx∣t=1=−92\left.\frac{dy}{dx}\right|_{t=1}=- \frac{9}{2}

    Substitute t=1 into the gradient function.

  10. Identify the parameter giving a horizontal tangent

    dydt=3t2−12=0 ⇒ t=−2\frac{dy}{dt}=3 t^{2} - 12=0\ \Rightarrow\ t=-2

    Horizontal tangents occur where dy/dt = 0.

Answer
t=−2t=-2
Question 5
8 markschallenging
A curve is defined parametrically by x=t2x=t^{2} and y=t3−3ty=t^{3} - 3 t. For which value of tt is the tangent to the curve horizontal?
Show worked solution

Worked solution

  1. Write down the parametric equations

    x=t2,y=t3−3tx=t^{2},\quad y=t^{3} - 3 t

    Both coordinates are expressed in terms of the parameter t.

  2. Recall the parametric chain rule

    dydx=dy/dtdx/dt\frac{dy}{dx}=\frac{dy/dt}{dx/dt}

    The gradient is the ratio of the two rates of change with respect to t.

  3. Differentiate x with respect to t

    dxdt=2t\frac{dx}{dt}=2 t

    Differentiate the expression for x term by term.

  4. Differentiate y with respect to t

    dydt=3t2−3\frac{dy}{dt}=3 t^{2} - 3

    Differentiate the expression for y term by term.

  5. Form and simplify dy/dx

    dydx=3t2−32t=3(t2−1)2t\frac{dy}{dx}=\frac{3 t^{2} - 3}{2 t}=\frac{3 \left(t^{2} - 1\right)}{2 t}

    Divide dy/dt by dx/dt and cancel any common factors.

  6. Find where the tangent is horizontal

    dydt=0 ⇒ t=−1, t=1\frac{dy}{dt}=0\ \Rightarrow\ t=-1,\ t=1

    A horizontal tangent occurs where dy/dt = 0 (with dx/dt non-zero).

  7. Find where the tangent is vertical

    dxdt=0 ⇒ t=0\frac{dx}{dt}=0\ \Rightarrow\ t=0

    A vertical tangent occurs where dx/dt = 0 (with dy/dt non-zero).

  8. Find the second derivative

    d2ydx2=ddt ⁣(dydx)÷dxdt=3(t2+1)4t3\frac{d^2y}{dx^2}=\frac{d}{dt}\!\left(\frac{dy}{dx}\right)\div\frac{dx}{dt}=\frac{3 \left(t^{2} + 1\right)}{4 t^{3}}

    Differentiate dy/dx with respect to t, then divide by dx/dt again.

  9. State the gradient at t=1

    dydx∣t=1=0\left.\frac{dy}{dx}\right|_{t=1}=0

    Substitute t=1 into the gradient function.

  10. State the gradient at t=2

    dydx∣t=2=94\left.\frac{dy}{dx}\right|_{t=2}=\frac{9}{4}

    Substitute t=2 into the gradient function.

  11. State the gradient at t=3

    dydx∣t=3=4\left.\frac{dy}{dx}\right|_{t=3}=4

    Substitute t=3 into the gradient function.

  12. State the gradient at t=-1

    dydx∣t=−1=0\left.\frac{dy}{dx}\right|_{t=-1}=0

    Substitute t=-1 into the gradient function.

  13. State the gradient at t=-2

    dydx∣t=−2=−94\left.\frac{dy}{dx}\right|_{t=-2}=- \frac{9}{4}

    Substitute t=-2 into the gradient function.

  14. State the gradient at t=4

    dydx∣t=4=458\left.\frac{dy}{dx}\right|_{t=4}=\frac{45}{8}

    Substitute t=4 into the gradient function.

  15. Identify the parameter giving a horizontal tangent

    dydt=3t2−3=0 ⇒ t=−1\frac{dy}{dt}=3 t^{2} - 3=0\ \Rightarrow\ t=-1

    Horizontal tangents occur where dy/dt = 0.

Answer
t=−1t=-1

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