Parametric modelling Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Parametric modelling questions. See exactly how to solve problems on parametric, gradient, motion, speed.

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A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A curve has parametric equations x=t2x=t^{2} and y=t3y=t^{3}. Find dydx\frac{dy}{dx} at the point where t=2t=2.

Worked solution

  1. Differentiate both parametric equations

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

    Differentiate x and y separately with respect to t.

  2. Form and simplify the gradient

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

    Divide dy/dt by dx/dt and simplify.

  3. Substitute the given parameter value

    dydx∣t=2=3\left.\frac{dy}{dx}\right|_{t=2}=3

    Substitute t to obtain the numerical gradient.

Answer
33
Question 2
2 markseasy
A curve has parametric equations x=2tx=2 t and y=t2y=t^{2}. Find dydx\frac{dy}{dx} at the point where t=3t=3.

Worked solution

  1. Differentiate both parametric equations

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

    Differentiate x and y separately with respect to t.

  2. Form and simplify the gradient

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

    Divide dy/dt by dx/dt and simplify.

  3. Substitute the given parameter value

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

    Substitute t to obtain the numerical gradient.

Answer
33
Question 3
2 markseasy
A curve has parametric equations x=t2x=t^{2} and y=2ty=2 t. Find dydx\frac{dy}{dx} at the point where t=2t=2.

Worked solution

  1. Differentiate both parametric equations

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

    Differentiate x and y separately with respect to t.

  2. Form and simplify the gradient

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

    Divide dy/dt by dx/dt and simplify.

  3. Substitute the given parameter value

    dydx∣t=2=12\left.\frac{dy}{dx}\right|_{t=2}=\frac{1}{2}

    Substitute t to obtain the numerical gradient.

Answer
12\frac{1}{2}
Question 4
2 markseasy
A curve has parametric equations x=t3x=t^{3} and y=t2y=t^{2}. Find dydx\frac{dy}{dx} at the point where t=1t=1.

Worked solution

  1. Differentiate both parametric equations

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

    Differentiate x and y separately with respect to t.

  2. Form and simplify the gradient

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

    Divide dy/dt by dx/dt and simplify.

  3. Substitute the given parameter value

    dydx∣t=1=23\left.\frac{dy}{dx}\right|_{t=1}=\frac{2}{3}

    Substitute t to obtain the numerical gradient.

Answer
23\frac{2}{3}
Question 5
2 markseasy
A curve has parametric equations x=4tx=4 t and y=t2−3y=t^{2} - 3. Find dydx\frac{dy}{dx} at the point where t=2t=2.

Worked solution

  1. Differentiate both parametric equations

    dxdt=4,dydt=2t\frac{dx}{dt}=4,\quad \frac{dy}{dt}=2 t

    Differentiate x and y separately with respect to t.

  2. Form and simplify the gradient

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

    Divide dy/dt by dx/dt and simplify.

  3. Substitute the given parameter value

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

    Substitute t to obtain the numerical gradient.

Answer
11

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