If the volume of a hemisphere is increasing at a steady rate of 18π m(^{3}) s(^{-1}), at what rate is its radius changing when its is 6m?
-
A.
2.50m/s -
B.
2.00 m/s -
C.
0.25 m/s -
D.
0.20 m/s
Correct Answer: Option C
Explanation
(V = frac{2}{3} pi r^{3})
Given: (frac{mathrm d V}{mathrm d t} = 18pi m^{3} s^{-1})
(frac{mathrm d V}{mathrm d t} = frac{mathrm d V}{mathrm d r} times frac{mathrm d r}{mathrm d t})
(frac{mathrm d V}{mathrm d r} = 2pi r^{2})
(18pi = 2pi r^{2} times frac{mathrm d r}{mathrm d t})
(frac{mathrm d r}{mathrm d t} = frac{18pi}{2pi r^{2}} = frac{9}{r^{2}})
The rate of change of the radius when r = 6m,
(frac{mathrm d r}{mathrm d t} = frac{9}{6^{2}} = frac{1}{4})
= (0.25 ms^{-1})