WAEC 2017 · Paper 2 · Q3

(a)
Given that $\log_{10} p = a$ , $\log_{10} q = b$ and $\log_{10} s = c$ , express $\log_{10}\left(\dfrac{p^{\frac12}q^4}{s^2}\right)$ in terms of $a$ , $b$ and $c$ .
(b)
The radius of a circle is $6\text{ cm}$ . If the area of the circle is increasing at the rate of $20\text{ cm}^2\text{ s}^{-1}$ , find, leaving the answer in terms of $\pi$ , the rate at which the radius is increasing.

Worked solution (try it first)

(a)

Split the log of the fraction: $\log_{10}\dfrac{p^{\frac12}q^4}{s^2} = \log_{10} p^{\frac12} + \log_{10} q^4 - \log_{10} s^2$ .
Bring each power down in front: $\frac12\log_{10} p + 4\log_{10} q - 2\log_{10} s$ .
Replace the logs with $a$ , $b$ and $c$ : $\dfrac a2 + 4b - 2c$ .

(b)

The area is $A = \pi r^2$ , so $\dfrac{dA}{dr} = 2\pi r$ .
By the chain rule, $\dfrac{dA}{dt} = \dfrac{dA}{dr} \times \dfrac{dr}{dt}$ : $20 = 2\pi(6) \times \dfrac{dr}{dt}$ .
So $\dfrac{dr}{dt} = \dfrac{20}{12\pi}$
$= \dfrac{5}{3\pi}\text{ cm s}^{-1}$ .

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