Question

The radius of a soap bubble is increasing at the rate of 0.2 cm/sec. If its radius is 5 cm, find the rate of increase of its volume.

Answer 1

Answer:

Explanation. is the answer.

Step-by-step explanation :If r is the radius and V is the volume of a soap bubble at any time t, then V=43πr3.

Differentiating w.r.t. t, we get,

dV/dt=4π/3×3r2dr/dt

∴dV/dt=4πr2dr/dt ...(1)

Now, dr/dt=0.2 cm/sec and r = 5 cm

∴ (1) gives, dV/dt=4π(5)2×0.2=20π cc/sec

∴ the rate of increase of volume =20π cc/sec.

Answer 2

Answer:

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