(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 + 4log_{10} q – 2log_{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} = 2pi r)
(frac{mathrm d A}{mathrm d t} = frac{mathrm d A}{mathrm d r} times frac{mathrm d r}{mathrm d t})
(20 = 2pi (6) times frac{mathrm d r}{mathrm d t})
(frac{mathrm d r}{mathrm d t} = frac{20}{12pi})
= (frac{5}{3pi})