A spherical ice ball is melting at the rate of 100 π cm³/min. The rate at which its radius is decreasing when its radius is 15 cm, is

The ice ball is melting at a rate of 100π cm³/min. The radius at this moment is 15 cm.

For a sphere, the volume is:

$$V = \frac{4}{3}\pi r^3$$V=34​πr3

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Differentiating both sides with respect to time:

$$\frac{dV}{dt} = \frac{4}{3}\pi \times 3r^2 \times \frac{dr}{dt}$$dtdV​=34​π×3r2×dtdr​

$$\frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}$$dtdV​=4πr2⋅dtdr​

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The volume is decreasing, so $\frac{dV}{dt} = -100\pi$dtdV​=−100π cm³/min.

At $r = 15$r=15 cm:

$$-100\pi = 4\pi (15)^2 \cdot \frac{dr}{dt}$$−100π=4π(15)2⋅dtdr​

$$-100\pi = 4\pi \times 225 \cdot \frac{dr}{dt}$$−100π=4π×225⋅dtdr​

$$-100\pi = 900\pi \cdot \frac{dr}{dt}$$−100π=900π⋅dtdr​

$$\frac{dr}{dt} = \frac{-100\pi}{900\pi}$$dtdr​=900π−100π​

$$\frac{dr}{dt} = -\frac{1}{9}$$dtdr​=−91​

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The negative sign indicates the radius is decreasing.

The radius is decreasing at a rate of $\frac{1}{9}$91​ cm/min.