(a) Given that \(\log_{10} p = a, \log_{10} q = b\) and \(\log_{10} s = c\), express \(\log_{10} (\frac{p^{\frac{1}{3}}q^{4}}{s^{2}}\) in terms of a, b and c.
(b) The radius of a circle is 6cm. If the area is increasing at the rate of 20\(cm^{2}s^{-1}\), find, leaving the answer in terms of \(\pi\), the rate at which the radius is increasing.
Explanation
(a) \(\log_{10} p = a ; \log_{10} q = b ; \log_{10} s = c\)
\(\log_{10} (\frac{p^{\frac{1}{3}}q^{4}}{s^{2}} = \log_{10} p^{\frac{1}{3}} + \log_{10} q^{4} - \log_{10} s^{2}\)
= \(\frac{1}{3}\log_{10} p + 4\log_{10} q - 2\log_{10} s\)
= \(\frac{1}{3}a + 4b - 2c\)
(b) Area of circle = \(\pi r^{2}\)
Given r = 6cm.
\(\frac{\mathrm d A}{\mathrm d r} = 2\pi r\)
\(\frac{\mathrm d A}{\mathrm d t} = \frac{\mathrm d A}{\mathrm d r} \times \frac{\mathrm d r}{\mathrm d t}\)
\(20 = 2\pi (6) \times \frac{\mathrm d r}{\mathrm d t}\)
\(\frac{\mathrm d r}{\mathrm d t} = \frac{20}{12\pi}\)
= \(\frac{5}{3\pi}\)