The radius of a circle increases at a rate of 0.5(cms^{-1}). Find the rate of change in the area of the circle with radius 7cm. ([pi = frac{22}{7}])
-
A.
11(cm^{2}s^{-1}) -
B.
22(cm^{2}s^{-1}) -
C.
33(cm^{2}s^{-1}) -
D.
44(cm^{2}s^{-1})
Correct Answer: Option B
Explanation
With radius = 7cm, (Area = pi r^{2} = frac{22}{7} times 7^{2})
= (154cm^{2})
The next second, radius = 7.5cm, (Area = pi r^{2} = frac{22}{7} times 7.5^{2})
= (176cm^{2})
Change in area = ((176 – 154)cm^{2} = 22cm^{2})
(therefore) The rate of increase = (22cm^{2}s^{-1})
OR
(Area (A) = pi r^{2} implies frac{mathrm d A}{mathrm d r} = 2pi r)
Given (frac{mathrm d r}{mathrm d t} = 0.5)
(frac{mathrm d A}{mathrm d r} times frac{mathrm d r}{mathrm d t} = frac{mathrm d A}{mathrm d t})
(frac{mathrm d A}{mathrm d t} = 2pi r times 0.5 = 2 times frac{22}{7} times 7 times 0.5)
= (22cm^{2}s^{-1})