SPM 2020 Add Maths Paper 2, Question 5Differentiation

Solved on video in English and Bahasa Melayu · Form 5, Differentiation · 5:41 video

The question

(a)

Diagram 2 shows a metal solid with a uniform cross section PQR in the shape of a right-angled triangle (right angle at Q). It is given that PQ = x cm, PQ:QR = 2:5, and the area of the cross section is A cm². (a) Express A in terms of x.

(b)(i)

02 cm s⁻¹. Find the rate of change of the area, in cm² s⁻¹, of the cross section when x = 8 cm.

(b)(ii)

05 cm.

SPM 2020 Add Maths Paper 2, Question 5: the question as drawn in Eduly's video

5:41

Part (a)

Step 1

From PQ:QR = 2:5 with PQ = x: QR/PQ = 5/2, so QR = (5/2)x.

Why this step

QR is the longer side, so QR = (5/2)x. Flipping the ratio to (2/5)x is the common slip. Always check QR comes out longer than PQ.

2 more steps, each with the reason for it, in the full solution

Final answer

A = (5/4)x²

Where students lose marks

Watch out: PQ:QR = 2:5 means QR is longer, so QR = (5/2)x, not (2/5)x. Don't use x for both legs.

Part (b)(i)

4 more steps, each with the reason for it, in the full solution

Final answer

dA/dt = 0.4 cm² s⁻¹

Where students lose marks

Watch out: the question asks for the rate dA/dt, so use the chain rule with dx/dt. Don't use the small change δA.

Part (b)(ii)

5 more steps, each with the reason for it, in the full solution

Final answer

δV ≈ 2.4 cm³

Where students lose marks

Watch out: for an approximate change use δV ≈ (dV/dx)(δx). Don't multiply by a rate dx/dt.

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SPM 2020 Add Maths Paper 2, Question 5Differentiation

The question

Part (a)

Part (b)(i)

Part (b)(ii)

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