A-Level Parametric modelling Practice Questions

Free A-Level Parametric modelling practice questions with full step-by-step worked solutions. Covers parametric, gradient, motion, speed. Practise exam-style problems and check your method.

parametricgradientmotionspeedreasoningtangent
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.
Show worked solution

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 is given parametrically. At a point where dxdt=0\frac{dx}{dt}=0 and dydt≠0\frac{dy}{dt}\ne 0, the tangent to the curve is:
Show worked solution

Worked solution

  1. Identify what is being asked

    Interpret the parametric condition\text{Interpret the parametric condition}

    Read the condition carefully and relate it to the geometry.

  2. Apply the relevant parametric principle

    dydx=y˙x˙\frac{dy}{dx}=\frac{\dot{y}}{\dot{x}}

    Use the parametric gradient and motion relationships.

  3. State the correct conclusion

    See the selected option\text{See the selected option}

    This follows directly from the parametric definition.

Answer
vertical
Question 3
3 marksintermediate
A curve has parametric equations x=t2x=t^{2} and y=t4y=t^{4}. Which of the following is dydx\frac{dy}{dx}?
Show worked solution

Worked solution

  1. Differentiate both parametric equations

    dxdt=2t,dydt=4t3\frac{dx}{dt}=2 t,\quad \frac{dy}{dt}=4 t^{3}

    Differentiate x and y with respect to the parameter.

  2. Apply the parametric differentiation rule

    dydx=dy/dtdx/dt=4t32t\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{4 t^{3}}{2 t}

    Divide dy/dt by dx/dt.

  3. Differentiate x with respect to t

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

    Differentiate the x-equation term by term.

  4. Differentiate y with respect to t

    dydt=4t3\frac{dy}{dt}=4 t^{3}

    Differentiate the y-equation term by term.

  5. State the parametric differentiation rule

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

    The chain rule links the two parametric derivatives.

  6. Simplify to identify the correct expression

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

    This is the gradient of the curve in terms of t.

Answer
2t22 t^{2}
Question 4
5 markshard
In a motion model with position x(t)x(t), y(t)y(t), the speed of the particle at time tt is equal to:
Show worked solution

Worked solution

  1. Identify what is being asked

    Interpret the parametric condition\text{Interpret the parametric condition}

    Read the condition carefully and relate it to the geometry.

  2. Apply the relevant parametric principle

    dydx=y˙x˙\frac{dy}{dx}=\frac{\dot{y}}{\dot{x}}

    Use the parametric gradient and motion relationships.

  3. Recall the chain rule for parametric curves

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

    The gradient is built from the two parametric derivatives.

  4. Write the standard parametric gradient formula

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

    Divide the rate of change of y by the rate of change of x.

  5. Condition for a horizontal tangent

    dydt=0, dxdt≠0\frac{dy}{dt}=0,\ \frac{dx}{dt}\ne 0

    Zero vertical rate with nonzero horizontal rate.

  6. Condition for a vertical tangent

    dxdt=0, dydt≠0\frac{dx}{dt}=0,\ \frac{dy}{dt}\ne 0

    Zero horizontal rate with nonzero vertical rate.

  7. Velocity components in a motion model

    x˙=dxdt,y˙=dydt\dot{x}=\frac{dx}{dt},\quad \dot{y}=\frac{dy}{dt}

    Differentiating each coordinate gives a velocity component.

  8. Speed is the magnitude of velocity

    v=x˙2+y˙2v=\sqrt{\dot{x}^{2}+\dot{y}^{2}}

    Combine the components with Pythagoras.

  9. Tangent line at a point on the path

    y−y1=m(x−x1)y-y_{1}=m\left(x-x_{1}\right)

    Use the point on the curve and the gradient there.

  10. State the correct conclusion

    See the selected option\text{See the selected option}

    This follows directly from the parametric definition.

Answer
the magnitude of the velocity vector
Question 5
8 markschallenging
A particle following a parametric path is momentarily at rest when:
Show worked solution

Worked solution

  1. Identify what is being asked

    Interpret the parametric condition\text{Interpret the parametric condition}

    Read the condition carefully and relate it to the geometry.

  2. Apply the relevant parametric principle

    dydx=y˙x˙\frac{dy}{dx}=\frac{\dot{y}}{\dot{x}}

    Use the parametric gradient and motion relationships.

  3. Recall the chain rule for parametric curves

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

    The gradient is built from the two parametric derivatives.

  4. Write the standard parametric gradient formula

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

    Divide the rate of change of y by the rate of change of x.

  5. Condition for a horizontal tangent

    dydt=0, dxdt≠0\frac{dy}{dt}=0,\ \frac{dx}{dt}\ne 0

    Zero vertical rate with nonzero horizontal rate.

  6. Condition for a vertical tangent

    dxdt=0, dydt≠0\frac{dx}{dt}=0,\ \frac{dy}{dt}\ne 0

    Zero horizontal rate with nonzero vertical rate.

  7. Velocity components in a motion model

    x˙=dxdt,y˙=dydt\dot{x}=\frac{dx}{dt},\quad \dot{y}=\frac{dy}{dt}

    Differentiating each coordinate gives a velocity component.

  8. Speed is the magnitude of velocity

    v=x˙2+y˙2v=\sqrt{\dot{x}^{2}+\dot{y}^{2}}

    Combine the components with Pythagoras.

  9. Tangent line at a point on the path

    y−y1=m(x−x1)y-y_{1}=m\left(x-x_{1}\right)

    Use the point on the curve and the gradient there.

  10. Normal gradient is the negative reciprocal

    mn=−1mm_{n}=-\frac{1}{m}

    The normal is perpendicular to the tangent.

  11. Eliminating the parameter gives the Cartesian form

    Solve for t and substitute\text{Solve for } t \text{ and substitute}

    Removing t recovers a relation between x and y.

  12. Stationary points of the path

    dydx=0\frac{dy}{dx}=0

    These occur where dy/dt=0 while dx/dt is nonzero.

  13. A point is momentarily at rest when velocity vanishes

    x˙=0 and y˙=0\dot{x}=0 \text{ and } \dot{y}=0

    Both velocity components are zero simultaneously.

  14. The sign of dy/dx describes the direction of travel

    dydx>0⇒rising path\frac{dy}{dx}>0 \Rightarrow \text{rising path}

    A positive gradient means y increases with x.

  15. State the correct conclusion

    See the selected option\text{See the selected option}

    This follows directly from the parametric definition.

Answer
both dxdt=0\frac{dx}{dt}=0 and dydt=0\frac{dy}{dt}=0

Unlock 65 more Parametric modelling questions

Create a free account to work through every A-Level Parametric modelling 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 Parametric modelling practice

Hard QuestionsWorked Solutions

Related Pure Maths topics

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