Parametric curves: 7 exam questions solved step by step

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What this lesson covers

  • For a curve given by x(t) and y(t), dy/dx is dy/dt divided by dx/dt.
  • The second derivative is the derivative of dy/dx with respect to t, divided by dx/dt again.
Key idea

7 questions, one method

  • 1. dy/dx = dy/dt ÷ dx/dt
  • 2. Stationary: dy/dt = 0
  • 3. Vertical tangent: dx/dt = 0
  • 4. d²y/dx² = d/dt(dy/dx) ÷ dx/dt
Your turn

Question 1 · Achieved

  • A curve has x = t³ + 1, y = t² − 4t.
  • Find dy/dx when t = 1.
Worked example

Q1: dy/dx at t = 1

  • Differentiate both in t.
  • dy/dt on top.
  1. dx/dt = 3t²differentiate x
  2. dy/dt = 2t − 4differentiate y
  3. dy/dx = (dy/dt) ÷ (dx/dt)chain rule
  4. = (2t − 4)/(3t²)
  5. t = 1: (2 − 4)/(3 × 1)
  6. = −2/3
  7. 3/(−2) = −3/2upside down
Your turn

Question 2 · Achieved

  • x = t² + 2t, y = t³ − 2
  • Find the equation of the tangent
  • at the point where t = 1.
Worked example

Q2: tangent at t = 1

  • The point needs x and y.
  • Then the gradient.
  1. t = 1: x = 1 + 2 = 3the point
  2. y = 1 − 2 = −1
  3. dx/dt = 2t + 2 = 4at t = 1
  4. dy/dt = 3t² = 3at t = 1
  5. m = 3/4dy/dt ÷ dx/dt
  6. y + 1 = (3/4)(x − 3)through (3, −1)
  7. 4y + 4 = 3x − 9× 4
  8. 3x − 4y − 13 = 0general form
Your turn

Question 3 · Merit

  • x = e²ᵗ, y = 2eᵗ + t
  • Find the equation of the normal
  • at the point where t = 0.
Worked example

Q3: normal at t = 0

  • e⁰ = 1
  • Normal: −1 ÷ m.
  1. t = 0: x = e⁰ = 1, y = 2e⁰ + 0 = 2
  2. dx/dt = 2e²ᵗchain rule
  3. dy/dt = 2eᵗ + 1
  4. t = 0: m = (2 + 1)/2 = 3/2dy/dt ÷ dx/dt
  5. normal m = −2/3negative reciprocal
  6. y − 2 = −(2/3)(x − 1)through (1, 2)
  7. 3y − 6 = −2x + 2× 3
  8. 2x + 3y − 8 = 0general form
Your turn

Question 4 · Merit

  • x = t² − 2t, y = t³ − 12t
  • Find the stationary points, and
  • the point where the tangent is vertical.
Worked example

Q4: flat and vertical tangents

  • Flat: top is 0.
  • Vertical: bottom is 0.
  1. dx/dt = 2t − 2, dy/dt = 3t² − 12
  2. 3t² − 12 = 0stationary: dy/dt = 0
  3. t² = 4, so t = ±2
  4. t = 2: x = 4 − 4 = 0, y = −16
  5. t = −2: x = 4 + 4 = 8, y = 16
  6. (0, −16) and (8, 16)stationary points
  7. 2t − 2 = 0, so t = 1vertical: dx/dt = 0
  8. t = 1: x = −1, y = −11dy/dt = −9, not 0
  9. vertical tangent at (−1, −11)
Your turn

Question 5 · Merit

  • The same curve: x = t² − 2t, y = t³ − 12t
  • Use d²y/dx² to decide whether each
  • stationary point is a max or a min.
Worked example

Q5: the second derivative

  • Quotient rule in t.
  • Then ÷ dx/dt again.
  1. dy/dx = (3t² − 12)/(2t − 2)
  2. u = 3t² − 12, v = 2t − 2quotient rule
  3. vu′ = (2t − 2)(6t) = 12t² − 12t
  4. uv′ = (3t² − 12)(2) = 6t² − 24
  5. d/dt(dy/dx) = (vu′ − uv′)/v²
  6. = (6t² − 12t + 24)/(2t − 2)²
  7. d²y/dx² = d/dt(dy/dx) ÷ dx/dt
  8. = (6t² − 12t + 24)/(2t − 2)³
Worked example

Q5: classify each point

  • Positive: minimum.
  • Negative: maximum.
  1. d²y/dx² = (6t² − 12t + 24)/(2t − 2)³
  2. t = 2: (24 − 24 + 24)/2³
  3. = 24/8 = 3
  4. 3 > 0: (0, −16) is a minimum
  5. t = −2: (24 + 24 + 24)/(−6)³
  6. = 72/(−216) = −1/3
  7. −1/3 < 0: (8, 16) is a maximum
  8. not ÷ dx/dt: 72/36 = 2the wrong sign
Your turn

Question 6 · Excellence

  • x = t³ + 3t, y = t²
  • Find the points of inflection,
  • and justify that they are inflections.
Worked example

Q6: the second derivative

  • Quotient rule in t.
  • Then ÷ dx/dt again.
  1. dx/dt = 3t² + 3, dy/dt = 2t
  2. dy/dx = 2t/(3t² + 3)dy/dt ÷ dx/dt
  3. d/dt(dy/dx) = (vu′ − uv′)/v²
  4. top: 2(3t² + 3) − 2t(6t)u = 2t, v = 3t² + 3
  5. = 6t² + 6 − 12t² = 6 − 6t²
  6. d/dt(dy/dx) = (6 − 6t²)/(3t² + 3)²
  7. d²y/dx² = (6 − 6t²)/(3t² + 3)³
Worked example

Q6: zero, and a change of sign

  • d²y/dx² = 0 is not enough.
  • Show the sign changes.
  1. d²y/dx² = (6 − 6t²)/(3t² + 3)³
  2. 6 − 6t² = 0d²y/dx² = 0
  3. t² = 1, so t = ±1
  4. (3t² + 3)³ > 0 for all tsign comes from the top
  5. −1 < t < 1: 6 − 6t² > 0concave up
  6. t > 1 or t < −1: 6 − 6t² < 0
  7. t = 1: x = 1 + 3 = 4, y = 1
  8. t = −1: x = −1 − 3 = −4, y = 1
  9. (4, 1) and (−4, 1)concavity changes
Your turn

Question 7 · Excellence

  • x = 2sin t, y = cos 2t
  • Show that dy/dx = −x.
  • Hence find d²y/dx², and describe
  • the concavity of the curve.
Worked example

Q7: show that dy/dx = −x

  • sin 2t = 2sin t cos t
  • Get back to x.
  1. dx/dt = 2cos t
  2. dy/dt = −2sin 2tchain rule
  3. dy/dx = −2sin 2t ÷ 2cos t
  4. = −2(2sin t cos t) ÷ 2cos t
  5. = −2sin tcancel 2cos t
  6. = −xx = 2sin t
  7. d/dt(−2sin t) = −2cos t
  8. d²y/dx² = −2cos t ÷ 2cos t = −1
  9. −1 < 0: concave down everywhere
Summary

The method that solved all seven

  • Differentiate in t, divide, and divide again.
StepHow
dy/dxdy/dt ÷ dx/dt
Pointput t into x and y
Stationarydy/dt = 0
Verticaldx/dt = 0
d²y/dx²d/dt(dy/dx) ÷ dx/dt
Inflectiond²y/dx² = 0, sign changes

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