A spherical snowball melts so that its radius decreases at a rate of 4 in/sec. At what rate is the volume of the snowball changing when the radius is 8 in?

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A spherical snowball melts so that its radius decreases at a rate of 4 in/sec. At what rate is the volume of the snowball changing when the radius is 8 in?

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Answer 1 · 2016-09-01T15:04:47

The formula for volume of a sphere is $V = \frac{4}{3} r^{3} π$ $V = 4/3r^3pi$ .

Differentiating with respect to $t$ $t$ , time.

$\frac{d V}{d t} = 4 r^{2} (\frac{d r}{d t})$ $(dV)/dt = 4r^2((dr)/dt)$

The rate of change of the snowball is given by $\frac{d V}{d t}$ $(dV)/dt$ . We know $\frac{d r}{d t} = - 4$ $(dr)/dt = -4$ . We want to find the rate of change when $r = 8$ $r= 8$ . Hence,

$\frac{d V}{d t} = 4 {(8)}^{2} (- 4)$ $(dV)/dt = 4(8)^2(-4)$

$\frac{d V}{d t} = 4 (64) (- 4)$ $(dV)/dt = 4(64)(-4)$

$\frac{d V}{d t} = - 1024$ $(dV)/dt = -1024$

Thus, the volume of the snowball is decreasing at a rate of $- 1024 \frac{{in}^{3}}{sec}$ $-1024" in"^3/"sec"$ .

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A spherical snowball melts so that its radius decreases at a rate of 4 in/sec. At what rate is the volume of the snowball changing when the radius is 8 in?

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