Parametric differentiation Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Parametric differentiation questions. See exactly how to solve problems on parametric-differentiation, chain-rule.

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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.

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=2tx=2 t and y=t2−1y=t^{2} - 1. Find dydx\dfrac{dy}{dx} in terms of tt.

Worked solution

  1. Differentiate x and y with respect to t

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

    Differentiate each parametric equation separately.

  2. Divide to find dy/dx

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

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

  3. State dy/dx in terms of t

    dydx=t\frac{dy}{dx}=t

    This is the gradient of the curve at parameter t.

Answer
t
Question 3
2 markseasy
A curve is defined parametrically by x=t2x=t^{2} and y=2ty=2 t. Find dydx\dfrac{dy}{dx} in terms of tt.

Worked solution

  1. Differentiate x and y with respect to t

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

    Differentiate each parametric equation separately.

  2. Divide to find dy/dx

    dydx=22t=1t\frac{dy}{dx}=\frac{2}{2 t}=\frac{1}{t}

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

  3. State dy/dx in terms of t

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

    This is the gradient of the curve at parameter t.

Answer
1t\frac{1}{t}
Question 4
2 markseasy
A curve is defined parametrically by x=t3x=t^{3} and y=t2y=t^{2}. Find dydx\dfrac{dy}{dx} in terms of tt.

Worked solution

  1. Differentiate x and y with respect to t

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

    Differentiate each parametric equation separately.

  2. Divide to find dy/dx

    dydx=2t3t2=23t\frac{dy}{dx}=\frac{2 t}{3 t^{2}}=\frac{2}{3 t}

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

  3. State dy/dx in terms of t

    dydx=23t\frac{dy}{dx}=\frac{2}{3 t}

    This is the gradient of the curve at parameter t.

Answer
23t\frac{2}{3 t}
Question 5
2 markseasy
A curve is defined parametrically by x=t+1x=t + 1 and y=t2y=t^{2}. Find dydx\dfrac{dy}{dx} in terms of tt.

Worked solution

  1. Differentiate x and y with respect to t

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

    Differentiate each parametric equation separately.

  2. Divide to find dy/dx

    dydx=2t1=2t\frac{dy}{dx}=\frac{2 t}{1}=2 t

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

  3. State dy/dx in terms of t

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

    This is the gradient of the curve at parameter t.

Answer
2t2 t

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