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Question

Mathematics Question on Application of derivatives

A spherical balloon is expanding. If the radius is increasing at the rate of 22 centimeters per minute, the rate at which the volume increases (in cubic centimeters per minute) when the radius is 55 centimetres is

A

10π\pi

B

100π\pi

C

200π\pi

D

50π\pi

Answer

200π\pi

Explanation

Solution

Let r and V be the respectively radius and volume of the balloon. Let t represents the time. The rate of increament in radius is drdt=2\frac{dr}{dt}=2 cm/minute. The volume of the balloon is given by
V=43πr3V =\frac{4}{3}\pi r^{3}
Differentiating w.r. to t, we get
dVdt=43π(3r2drdt)\frac{dV}{dt}=\frac{4}{3}\pi\left(3r^{2}\frac{dr}{dt}\right)
Substituting the values of and drdt\frac{dr}{dt} , we get
dVdt=43π(3×52×2)=200πcm3/minute\frac{dV}{dt}=\frac{4}{3}\pi\left(3\times5^{2}\times2\right)=200\pi cm^{3} / minute