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HSC/VCE-style Calculus Practice with Worked Solutions — Answers

Answer sheet with worked solutions for the Year 12 calculus worked examples.

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Year 12 · Mathematics · Calculus

HSC/VCE-style Calculus Practice with Worked Solutions — Answers

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Section A: Differentiation

1.Differentiate y = x2e3x. Give your answer in factorised form.[2 marks]

Answer: = xe3x(2 + 3x)

  1. Product rule with u = x2 and v = e3x.
  2. u′ = 2x and v′ = 3e3x (chain rule: the derivative of 3x is 3).
  3. = u′v + uv′ = 2xe3x + 3x2e3x
  4. Factorise the common factor xe3x: = xe3x(2 + 3x)
2.Differentiate y = ln(x2 + 1).[2 marks]

Answer: =

  1. The derivative of ln(f(x)) is .
  2. Here f(x) = x2 + 1 and f′(x) = 2x.
  3. =
3.Differentiate y = .[3 marks]

Answer: =

  1. Quotient rule with u = sin x and v = x.
  2. u′ = cos x and v′ = 1.
  3. = =
4.Find the equation of the tangent to the curve y = e2x at the point where x = 0.[3 marks]

Answer: y = 2x + 1

  1. Point: when x = 0, y = e0 = 1, so the point is (0, 1).
  2. Gradient: = 2e2x, so at x = 0 the gradient is 2e0 = 2.
  3. Equation: y − 1 = 2(x − 0)
  4. y = 2x + 1

Section B: Integration and area

5.Evaluate ∫ from 1 to 4 of 3√x dx.[3 marks]

Answer: 14

  1. Write √x as x1/2. Raise the power to and divide by it: 3 ÷ = 2, so ∫ 3x1/2 dx = 2x3/2.
  2. Evaluate [2x3/2] from 1 to 4.
  3. At x = 4: 2 × 43/2 = 2 × 8 = 16. At x = 1: 2 × 1 = 2.
  4. 16 − 2 = 14
6.Evaluate ∫ from 0 to of cos 2x dx. Give an exact answer.[3 marks]

Answer:

  1. ∫ cos 2x dx = sin 2x + C (divide by the coefficient of x).
  2. Evaluate [ sin 2x] from 0 to .
  3. At x = : sin = × = . At x = 0: sin 0 = 0.
  4. ∫ = − 0 =
7.Find the area of the region enclosed between the parabola y = x2 and the line y = 2x.[4 marks]

Answer: square units

  1. Intersections: x2 = 2x, so x2 − 2x = 0, x(x − 2) = 0, giving x = 0 and x = 2.
  2. Between x = 0 and x = 2 the line is above the parabola (test x = 1: line gives 2, parabola gives 1).
  3. Area = ∫ from 0 to 2 of (2x − x2) dx = [x2 − ] from 0 to 2
  4. = (4 − ) − 0 = − =
  5. Area = square units

Section C: Applications

8.Find the coordinates of the stationary points of y = x3 − 6x2 + 9x + 1 and determine their nature.[4 marks]

Answer: Maximum at (1, 5); minimum at (3, 1)

  1. y′ = 3x2 − 12x + 9 = 3(x2 − 4x + 3) = 3(x − 1)(x − 3)
  2. y′ = 0 at x = 1 and x = 3.
  3. y(1) = 1 − 6 + 9 + 1 = 5 and y(3) = 27 − 54 + 27 + 1 = 1.
  4. y″ = 6x − 12. At x = 1: y″ = −6 < 0, so (1, 5) is a maximum.
  5. At x = 3: y″ = 6 > 0, so (3, 1) is a minimum.
9.An open box has a square base of side x cm and height h cm. Its volume is 32 cm3. Show that its external surface area is S = x2 + , and find the dimensions that minimise S.[5 marks]

Answer: x = 4 cm, h = 2 cm, minimum S = 48 cm2

  1. Volume: x2h = 32, so h = .
  2. Surface area (base + 4 sides, no lid): S = x2 + 4xh = x2 + 4x × = x2 + , as required.
  3. S′ = 2x − . Set S′ = 0: 2x = , so x3 = 64 and x = 4.
  4. S″ = 2 + > 0 for x > 0, so x = 4 gives a minimum.
  5. h = = 2. Minimum S = 16 + = 16 + 32 = 48 cm2.
  6. Dimensions: base 4 cm × 4 cm, height 2 cm.
10.A particle moves in a straight line so that its displacement from the origin after t seconds is x = t3 − 6t2 + 9t metres. (a) Find the velocity when t = 2. (b) Find when the particle is at rest. (c) Find the acceleration when t = 2.[4 marks]

Answer: (a) −3 m/s (b) t = 1 s and t = 3 s (c) 0 m/s2

  1. Velocity v = = 3t2 − 12t + 9.
  2. (a) v(2) = 3(4) − 24 + 9 = −3 m/s (the particle is moving towards the origin, in the negative direction).
  3. (b) At rest means v = 0: 3t2 − 12t + 9 = 3(t − 1)(t − 3) = 0, so t = 1 s and t = 3 s.
  4. Acceleration a = = 6t − 12.
  5. (c) a(2) = 12 − 12 = 0 m/s2.
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