My 4. A spherical snowball is melting in such a way that its volume is decreasing at a rate of 2 cm /min . At what rate is the radius changing when the radius is 7 cm? [The volume of a sphere is given by v - mr^3.] A. 1 cm/min 771
B. 1 cm/min 497 1
C. cm/min 497
D. 1 cm/min 1961
E. 1 cm/min 987

E 1 cm/min 98π

Step-by-step explanation:

We know the volume of a sphere V = 4πr³/3. Now the rate of change of volume with respect to time dV/dt = dV/dr × dr/dt where dV/dr is the rate of change of volume with respect to the radius and dr/dt is the rate of change of radius with respect to time.

Now, dV/dt = -2 cm³/min since it is decreasing

dV/dr = 4πr²

So, dV/dt = dV/dr × dr/dt

-2 cm³/min =  4πr²dr/dt

dr/dt = -2 cm³/min ÷ 4πr²

when r = 7 cm,

dr/dt = -2 cm³/min ÷ 4π(7 cm)²

dr/dt = -1 cm³/min (2π × 49 cm²)

dr/dt = -1 cm/min 98π

So, the radius is changing (decreasing) at a rate of 1 cm/min 98π

⊂⊃⊂⊃ω

step-by-step explanation:

i hope this : )