Hard A-Level Parametric modelling Questions

Challenging, exam-style A-Level Parametric modelling questions with worked solutions. Stretch yourself on the hardest parametric, tangent, normal, motion problems.

parametrictangentnormalmotionspeedmaximum
A-Level34 questionsStep-by-step solutions
Question 1
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
Question 2
8 markschallenging
The gradient dydx\frac{dy}{dx} of a curve given by x(t)x(t), y(t)y(t) is found by:
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
dividing dydt\frac{dy}{dt} by dxdt\frac{dx}{dt}
Question 3
8 markschallenging
A curve has parametric equations x=cos⁡(t)x=\cos{\left(t \right)} and y=sin⁡(t)y=\sin{\left(t \right)}. Which of the following is dydx\frac{dy}{dx}?
Show worked solution

Worked solution

  1. Differentiate both parametric equations

    dxdt=−sin⁡(t),dydt=cos⁡(t)\frac{dx}{dt}=- \sin{\left(t \right)},\quad \frac{dy}{dt}=\cos{\left(t \right)}

    Differentiate x and y with respect to the parameter.

  2. Apply the parametric differentiation rule

    dydx=dy/dtdx/dt=cos⁡(t)−sin⁡(t)\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{\cos{\left(t \right)}}{- \sin{\left(t \right)}}

    Divide dy/dt by dx/dt.

  3. Differentiate x with respect to t

    dxdt=−sin⁡(t)\frac{dx}{dt}=- \sin{\left(t \right)}

    Differentiate the x-equation term by term.

  4. Differentiate y with respect to t

    dydt=cos⁡(t)\frac{dy}{dt}=\cos{\left(t \right)}

    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. Substitute the derivatives into the rule

    dydx=cos⁡(t)−sin⁡(t)\frac{dy}{dx}=\frac{\cos{\left(t \right)}}{- \sin{\left(t \right)}}

    Divide dy/dt by dx/dt to get the gradient in terms of t.

  7. Simplify the gradient expression

    dydx=−1tan⁡(t)\frac{dy}{dx}=- \frac{1}{\tan{\left(t \right)}}

    Simplify to a single expression in the parameter t.

  8. Interpret the derivatives as velocity components

    x˙=−sin⁡(t),y˙=cos⁡(t)\dot{x}=- \sin{\left(t \right)},\quad \dot{y}=\cos{\left(t \right)}

    In a motion model dx/dt and dy/dt are the velocity components.

  9. Write the speed as the magnitude of velocity

    v=(dxdt)2+(dydt)2v=\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}

    Speed is the magnitude of the velocity vector.

  10. Recall the equation of a straight line

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

    A line through a known point with a known gradient.

  11. Recall the gradient of the normal

    mn=−1mm_{n}=-\frac{1}{m}

    The normal is perpendicular to the tangent.

  12. Note the condition for a horizontal tangent

    dydt=0\frac{dy}{dt}=0

    The path has a horizontal tangent where dy/dt vanishes.

  13. Note the condition for a vertical tangent

    dxdt=0\frac{dx}{dt}=0

    The path has a vertical tangent where dx/dt vanishes.

  14. Keep every quantity in terms of the parameter

    Work in t until the final substitution\text{Work in } t \text{ until the final substitution}

    Parametric methods stay in terms of t until the last step.

  15. Simplify to identify the correct expression

    dydx=−1tan⁡(t)\frac{dy}{dx}=- \frac{1}{\tan{\left(t \right)}}

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

Answer
−1tan⁡(t)- \frac{1}{\tan{\left(t \right)}}
Question 4
8 markschallenging
A curve has parametric equations x=sin⁡(t)x=\sin{\left(t \right)} and y=sin⁡(2t)y=\sin{\left(2 t \right)}. Find dydx\frac{dy}{dx} in terms of tt.
Show worked solution

Worked solution

  1. Differentiate both parametric equations

    dxdt=cos⁡(t),dydt=2cos⁡(2t)\frac{dx}{dt}=\cos{\left(t \right)},\quad \frac{dy}{dt}=2 \cos{\left(2 t \right)}

    Differentiate x and y separately with respect to t.

  2. Divide the derivatives

    dydx=2cos⁡(2t)cos⁡(t)\frac{dy}{dx}=\frac{2 \cos{\left(2 t \right)}}{\cos{\left(t \right)}}

    Apply the parametric differentiation rule.

  3. Differentiate x with respect to t

    dxdt=cos⁡(t)\frac{dx}{dt}=\cos{\left(t \right)}

    Differentiate the x-equation term by term.

  4. Differentiate y with respect to t

    dydt=2cos⁡(2t)\frac{dy}{dt}=2 \cos{\left(2 t \right)}

    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. Substitute the derivatives into the rule

    dydx=2cos⁡(2t)cos⁡(t)\frac{dy}{dx}=\frac{2 \cos{\left(2 t \right)}}{\cos{\left(t \right)}}

    Divide dy/dt by dx/dt to get the gradient in terms of t.

  7. Interpret the derivatives as velocity components

    x˙=cos⁡(t),y˙=2cos⁡(2t)\dot{x}=\cos{\left(t \right)},\quad \dot{y}=2 \cos{\left(2 t \right)}

    In a motion model dx/dt and dy/dt are the velocity components.

  8. Write the speed as the magnitude of velocity

    v=(dxdt)2+(dydt)2v=\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}

    Speed is the magnitude of the velocity vector.

  9. Recall the equation of a straight line

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

    A line through a known point with a known gradient.

  10. Recall the gradient of the normal

    mn=−1mm_{n}=-\frac{1}{m}

    The normal is perpendicular to the tangent.

  11. Note the condition for a horizontal tangent

    dydt=0\frac{dy}{dt}=0

    The path has a horizontal tangent where dy/dt vanishes.

  12. Note the condition for a vertical tangent

    dxdt=0\frac{dx}{dt}=0

    The path has a vertical tangent where dx/dt vanishes.

  13. Keep every quantity in terms of the parameter

    Work in t until the final substitution\text{Work in } t \text{ until the final substitution}

    Parametric methods stay in terms of t until the last step.

  14. Check the parameter value lies in the model's range

    Confirm t is admissible\text{Confirm } t \text{ is admissible}

    Only parameter values in the model's domain are meaningful.

  15. Simplify to give dy/dx in terms of t

    dydx=2cos⁡(2t)cos⁡(t)\frac{dy}{dx}=\frac{2 \cos{\left(2 t \right)}}{\cos{\left(t \right)}}

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

Answer
2cos⁡(2t)cos⁡(t)\frac{2 \cos{\left(2 t \right)}}{\cos{\left(t \right)}}
Question 5
8 markschallenging
A curve has parametric equations x=t+1tx=t + \frac{1}{t} and y=t−1ty=t - \frac{1}{t}. Find dydx\frac{dy}{dx} in terms of tt.
Show worked solution

Worked solution

  1. Differentiate both parametric equations

    dxdt=1−1t2,dydt=1+1t2\frac{dx}{dt}=1 - \frac{1}{t^{2}},\quad \frac{dy}{dt}=1 + \frac{1}{t^{2}}

    Differentiate x and y separately with respect to t.

  2. Divide the derivatives

    dydx=1+1t21−1t2\frac{dy}{dx}=\frac{1 + \frac{1}{t^{2}}}{1 - \frac{1}{t^{2}}}

    Apply the parametric differentiation rule.

  3. Differentiate x with respect to t

    dxdt=1−1t2\frac{dx}{dt}=1 - \frac{1}{t^{2}}

    Differentiate the x-equation term by term.

  4. Differentiate y with respect to t

    dydt=1+1t2\frac{dy}{dt}=1 + \frac{1}{t^{2}}

    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. Substitute the derivatives into the rule

    dydx=1+1t21−1t2\frac{dy}{dx}=\frac{1 + \frac{1}{t^{2}}}{1 - \frac{1}{t^{2}}}

    Divide dy/dt by dx/dt to get the gradient in terms of t.

  7. Simplify the gradient expression

    dydx=t2+1t2−1\frac{dy}{dx}=\frac{t^{2} + 1}{t^{2} - 1}

    Simplify to a single expression in the parameter t.

  8. Interpret the derivatives as velocity components

    x˙=1−1t2,y˙=1+1t2\dot{x}=1 - \frac{1}{t^{2}},\quad \dot{y}=1 + \frac{1}{t^{2}}

    In a motion model dx/dt and dy/dt are the velocity components.

  9. Write the speed as the magnitude of velocity

    v=(dxdt)2+(dydt)2v=\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}

    Speed is the magnitude of the velocity vector.

  10. Recall the equation of a straight line

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

    A line through a known point with a known gradient.

  11. Recall the gradient of the normal

    mn=−1mm_{n}=-\frac{1}{m}

    The normal is perpendicular to the tangent.

  12. Note the condition for a horizontal tangent

    dydt=0\frac{dy}{dt}=0

    The path has a horizontal tangent where dy/dt vanishes.

  13. Note the condition for a vertical tangent

    dxdt=0\frac{dx}{dt}=0

    The path has a vertical tangent where dx/dt vanishes.

  14. Keep every quantity in terms of the parameter

    Work in t until the final substitution\text{Work in } t \text{ until the final substitution}

    Parametric methods stay in terms of t until the last step.

  15. Simplify to give dy/dx in terms of t

    dydx=t2+1t2−1\frac{dy}{dx}=\frac{t^{2} + 1}{t^{2} - 1}

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

Answer
t2+1t2−1\frac{t^{2} + 1}{t^{2} - 1}

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