2.11.3 Differentiation (Paper 2), Question 4

Question 4:

(a) A worker is pumping air into a spherical shape balloon at the rate of 25 cm³ s^-1.

$[Volume of a sphere, V = \frac{4}{3} π r^{3}]$

Leaving answer in terms of π, find,
(i) rate of change of radius of the balloon when its radius is 10 cm. [3 marks]
(ii) approximate change of volume when radius of the balloon decrease from 10 cm to 9.95 cm. [2 marks]

(b) A curve

$y = h x^{3} - \frac{3}{x}$ has turning point x = 1, find value of h. [3 marks]

Solution:

(a)(i)

$\begin{array}{l} V = \frac{4}{3} π r^{3} \\ \frac{d V}{d r} = 4 π r^{2} \\ \frac{d r}{d t} = \frac{d r}{d V} \times \frac{d V}{d t} \\ = \frac{1}{4 π r^{2}} \times 25 \\ = \frac{1}{4 π {(10)}^{2}} \times 25 \\ = \frac{1}{16 π} \end{array}$

(a)(ii)

$\begin{array}{l} \frac{δ V}{δ r} \approx \frac{d V}{d r} \\ δ V = \frac{d V}{d r} \times δ r \\ = 4 π r^{2} \times (9.95 - 10) \\ = 4 π {(10)}^{2} \times (- 0.05) \\ = - 20 π \end{array}$

(b)

$\begin{array}{l} y = h x^{3} - \frac{3}{x} \\ y = h x^{3} - 3 x^{- 1} \\ \frac{d y}{d x} = 3 h x^{2} + 3 x^{- 2} \\ \frac{d y}{d x} = 3 h x^{2} + \frac{3}{x^{2}} \\ At turning points, \frac{d y}{d x} = 0 \\ 0 = 3 h x^{2} + \frac{3}{x^{2}} \\ 0 = 3 h {(1)}^{2} + \frac{3}{1} \to given x = 1 \\ 3 h = - 3 \\ h = - 1 \end{array}$

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