Hard A-Level Parametric differentiation Questions

Challenging, exam-style A-Level Parametric differentiation questions with worked solutions. Stretch yourself on the hardest parametric-differentiation, chain-rule problems.

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A-Level34 questionsStep-by-step solutions
Question 1
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
Question 2
8 markschallenging
A curve is defined parametrically by x=t2x=t^{2} and y=t3−3ty=t^{3} - 3 t. Compared with the gradient at t=1t=1, the gradient of the curve at t=2t=2 is:
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. Compare the two gradients

    dydx∣t=1=0,dydx∣t=2=94\left.\frac{dy}{dx}\right|_{t=1}=0,\quad \left.\frac{dy}{dx}\right|_{t=2}=\frac{9}{4}

    Evaluate dy/dx at both parameter values and compare them.

Answer
Greater
Question 3
8 markschallenging
A curve is defined parametrically by x=2cos⁡(t)x=2 \cos{\left(t \right)} and y=3sin⁡(t)y=3 \sin{\left(t \right)}. Describe the tangent to the curve at the point where t=π2t=\frac{\pi}{2}.
Show worked solution

Worked solution

  1. Write down the parametric equations

    x=2cos⁡(t),y=3sin⁡(t)x=2 \cos{\left(t \right)},\quad y=3 \sin{\left(t \right)}

    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=−2sin⁡(t)\frac{dx}{dt}=- 2 \sin{\left(t \right)}

    Differentiate the expression for x term by term.

  4. Differentiate y with respect to t

    dydt=3cos⁡(t)\frac{dy}{dt}=3 \cos{\left(t \right)}

    Differentiate the expression for y term by term.

  5. Form and simplify dy/dx

    dydx=3cos⁡(t)−2sin⁡(t)=−32tan⁡(t)\frac{dy}{dx}=\frac{3 \cos{\left(t \right)}}{- 2 \sin{\left(t \right)}}=- \frac{3}{2 \tan{\left(t \right)}}

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

  6. Find the x-coordinate at t=\frac{\pi}{2}

    x=0x=0

    Substitute the parameter value into the equation for x.

  7. Find the y-coordinate at t=\frac{\pi}{2}

    y=3y=3

    Substitute the parameter value into the equation for y.

  8. Evaluate the gradient at t=\frac{\pi}{2}

    dydx∣t=π2=0\left.\frac{dy}{dx}\right|_{t=\frac{\pi}{2}}=0

    Substitute the parameter value into dy/dx to get the tangent gradient.

  9. Use the point-slope form for the tangent

    y−3=0(x−0)y-3=0\left(x-0\right)

    The tangent passes through the point with the gradient found.

  10. Simplify the tangent equation

    y=3y=3

    Expand and tidy the tangent into the form y = mx + c.

  11. Find where the tangent is horizontal

    dydt=0 ⇒ t=π2, t=3π2\frac{dy}{dt}=0\ \Rightarrow\ t=\frac{\pi}{2},\ t=\frac{3 \pi}{2}

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

  12. Find where the tangent is vertical

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

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

  13. Find the second derivative

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

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

  14. State the gradient at t=\frac{\pi}{6}

    dydx∣t=π6=−332\left.\frac{dy}{dx}\right|_{t=\frac{\pi}{6}}=- \frac{3 \sqrt{3}}{2}

    Substitute t=\frac{\pi}{6} into the gradient function.

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

    dxdt=−2,dydt=0\frac{dx}{dt}=-2,\quad \frac{dy}{dt}=0

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

Answer
Horizontal
Question 4
8 markschallenging
A curve is defined parametrically by x=t3x=t^{3} and y=t2y=t^{2}. What is the gradient of the curve at t=−1t=-1?
Show worked solution

Worked solution

  1. Write down the parametric equations

    x=t3,y=t2x=t^{3},\quad y=t^{2}

    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=3t2\frac{dx}{dt}=3 t^{2}

    Differentiate the expression for x term by term.

  4. Differentiate y with respect to t

    dydt=2t\frac{dy}{dt}=2 t

    Differentiate the expression for y term by term.

  5. Form and simplify dy/dx

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

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

  6. Find the x-coordinate at t=-1

    x=−1x=-1

    Substitute the parameter value into the equation for x.

  7. Find the y-coordinate at t=-1

    y=1y=1

    Substitute the parameter value into the equation for y.

  8. Evaluate the gradient at t=-1

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

    Substitute the parameter value into dy/dx to get the tangent gradient.

  9. Use the point-slope form for the tangent

    y−1=−23(x−−1)y-1=- \frac{2}{3}\left(x--1\right)

    The tangent passes through the point with the gradient found.

  10. Simplify the tangent equation

    y=13−2x3y=\frac{1}{3} - \frac{2 x}{3}

    Expand and tidy the tangent into the form y = mx + c.

  11. Find the gradient of the normal

    mn=−1−23=32m_n=-\frac{1}{- \frac{2}{3}}=\frac{3}{2}

    The normal is perpendicular to the tangent, so its gradient is -1/m.

  12. Form the normal equation

    y=3x2+52y=\frac{3 x}{2} + \frac{5}{2}

    Use the same point together with the normal gradient.

  13. Find where the tangent is horizontal

    dydt=0 ⇒ t=0\frac{dy}{dt}=0\ \Rightarrow\ t=0

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

  14. 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).

  15. Select the correct gradient value

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

    Evaluate dy/dx at the given parameter value.

Answer
−23- \frac{2}{3}
Question 5
8 markschallenging
A curve is defined parametrically by x=t2x=t^{2} and y=t3−3ty=t^{3} - 3 t. What is the gradient of the curve at t=2t=2?
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 the x-coordinate at t=2

    x=4x=4

    Substitute the parameter value into the equation for x.

  7. Find the y-coordinate at t=2

    y=2y=2

    Substitute the parameter value into the equation for y.

  8. Evaluate the gradient at t=2

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

    Substitute the parameter value into dy/dx to get the tangent gradient.

  9. Use the point-slope form for the tangent

    y−2=94(x−4)y-2=\frac{9}{4}\left(x-4\right)

    The tangent passes through the point with the gradient found.

  10. Simplify the tangent equation

    y=9x4−7y=\frac{9 x}{4} - 7

    Expand and tidy the tangent into the form y = mx + c.

  11. Find the gradient of the normal

    mn=−194=−49m_n=-\frac{1}{\frac{9}{4}}=- \frac{4}{9}

    The normal is perpendicular to the tangent, so its gradient is -1/m.

  12. Form the normal equation

    y=349−4x9y=\frac{34}{9} - \frac{4 x}{9}

    Use the same point together with the normal gradient.

  13. 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).

  14. 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).

  15. Select the correct gradient value

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

    Evaluate dy/dx at the given parameter value.

Answer
94\frac{9}{4}

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