SPM Additional Mathematics 2019, Paper 2 (Question 7)

Quesion 7:
A metal solid is formed by combining a cone and a cylinder with common radius, r cm. The total
surface area of the solid, A cm², is given by

A = 2 π (\frac{18}{r} + \frac{r^{2}}{3}) .

$A = 2 π (\frac{18}{r} + \frac{r^{2}}{3}) .$

(a)(i) The solid expands when heated. It is given that the surface area of the solid changes at the
rate of 1.4π cm²s^-1. Find the rate of change of its radius in, cm s^-1, when its radius is 6 cm.

(ii) Find the approximate change in the surface area of the solid, in terms of π, when its radius
increases from 6 cm to 6.02 cm.
[6 marks]

(b) If a solid of a same shape is to be formed in such that the total surface area is minimum, find
the minimum total surface area of the solid, in terms of π. [4 marks]

Solution:
(a)(i)

\begin{array}{l} A = 2 π (\frac{18}{r} + \frac{r^{2}}{3}) \\ = \frac{36 π}{r} + \frac{2 π r^{2}}{3} \\ = 36 π r^{- 1} + \frac{2 π}{3} r^{2} \\ \frac{d A}{d r} = - 36 π r^{- 2} + \frac{4 π}{3} r \\ = - \frac{36 π}{r^{2}} + \frac{4 π}{3} r \\ When r = 6, \\ \frac{d A}{d r} = - \frac{36 π}{{(6)}^{2}} + \frac{4 π}{3} (6) \\ = - π + 8 π \\ = 7 π \\ \frac{d A}{d t} = \frac{d A}{d r} \times \frac{d r}{d t} \\ 1.4 π = 7 π \times \frac{d r}{d t} \\ \frac{d r}{d t} = \frac{1.4 π}{7 π} \\ = 0.2 cm/s \end{array}

$\begin{array}{l} A = 2 π (\frac{18}{r} + \frac{r^{2}}{3}) \\ = \frac{36 π}{r} + \frac{2 π r^{2}}{3} \\ = 36 π r^{- 1} + \frac{2 π}{3} r^{2} \\ \frac{d A}{d r} = - 36 π r^{- 2} + \frac{4 π}{3} r \\ = - \frac{36 π}{r^{2}} + \frac{4 π}{3} r \\ When r = 6, \\ \frac{d A}{d r} = - \frac{36 π}{{(6)}^{2}} + \frac{4 π}{3} (6) \\ = - π + 8 π \\ = 7 π \\ \frac{d A}{d t} = \frac{d A}{d r} \times \frac{d r}{d t} \\ 1.4 π = 7 π \times \frac{d r}{d t} \\ \frac{d r}{d t} = \frac{1.4 π}{7 π} \\ = 0.2 cm/s \end{array}$

(a)(ii)

\begin{array}{l} δ r = 6.02 - 6 \\ = 0.02 cm \\ δ A = \frac{d A}{d r} \times δ r \\ = 8 π \times 0.02 \\ = 1.6 π {cm}^{2} \end{array}

$\begin{array}{l} δ r = 6.02 - 6 \\ = 0.02 cm \\ δ A = \frac{d A}{d r} \times δ r \\ = 8 π \times 0.02 \\ = 1.6 π {cm}^{2} \end{array}$

(b)

\begin{array}{l} \frac{d A}{d r} = 0, \\ - \frac{36 π}{r^{2}} + \frac{4 π r}{3} = 0 \\ \frac{4 r}{3} = \frac{36}{r^{2}} \\ r^{3} = 27 \\ r^{3} = 3^{3} \\ r = 3 \\ \frac{d^{2} A}{d r^{2}} = 72 π r^{- 3} + \frac{4 π}{3} \\ When r = 3, \\ \frac{d^{2} A}{d r^{2}} = \frac{72 π}{27} + \frac{4 π}{3} \\ = 4 π > 0 \\ ∴ A is minimum \\ A_{minimum} = 2 π (\frac{18}{3} + \frac{9}{3}) \\ = 18 π {cm}^{2} \end{array}

$\begin{array}{l} \frac{d A}{d r} = 0, \\ - \frac{36 π}{r^{2}} + \frac{4 π r}{3} = 0 \\ \frac{4 r}{3} = \frac{36}{r^{2}} \\ r^{3} = 27 \\ r^{3} = 3^{3} \\ r = 3 \\ \frac{d^{2} A}{d r^{2}} = 72 π r^{- 3} + \frac{4 π}{3} \\ When r = 3, \\ \frac{d^{2} A}{d r^{2}} = \frac{72 π}{27} + \frac{4 π}{3} \\ = 4 π > 0 \\ ∴ A is minimum \\ A_{minimum} = 2 π (\frac{18}{3} + \frac{9}{3}) \\ = 18 π {cm}^{2} \end{array}$

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