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WMA11/01 Oct 2019, Question 5: Equation of the Tangent Line

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 1 (WMA11) · Calculator · 7 marks

The question, in short

Differentiate y = x³/6 + 4√x − 15 for x ≥ 0, then find the normal to the curve at P(4, 11/3) in the form ax + by + c = 0 with integer coefficients.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 3 marks

Find dy/dx of the given curve, giving the result in its simplest form.

  1. 1Write y = (1/6)x³ + 4x^(½) − 15
  2. 2d/dx of (1/6)x³ is ½x², and d/dx of 4x^(½) is 2x^(−½); the constant gives 0
  3. 3dy/dx = ½x² + 2/√x

Answerdy/dx = ½x² + 2x^(−½) (= x²/2 + 2/√x)

Part (b) · 4 marks

Using point P(4, 11/3) on the curve, find the normal's equation as ax + by + c = 0 with integers a, b, c.

  1. 1At x = 4: dy/dx = ½ × 16 + 2/2 = 9
  2. 2Gradient of normal = −1/9
  3. 3y − 11/3 = −(1/9)(x − 4)
  4. 49y − 33 = −x + 4, so x + 9y − 37 = 0

Answerx + 9y − 37 = 0

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Also searched as: WMA11 October 2019 · WMA11/01 · wma11 oct19 · IAS Maths Pure Mathematics 1 October 2019 · question 5.
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