A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.3 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 16 cm.

Question

anonymous · Answer

120.6371579 possibly

anonymous · Answer

Bring back memories huh, mnad1:)

anonymous · Answer

haha yeah, deja vu

bahrom7893 · Answer

V = (4/3)pir^3
dv/dt = 4pir^2(dr/dt)

bahrom7893 · Answer

d = 16, so r = 8
dv/dt = 4pi * 8^2 * 0.3

anonymous · Answer

differntiate the formular for volume of a sphere, then use 8 for the radius and 0.15 for dd/dt(meaing the rate of change of the diameter)

bahrom7893 · Answer

dv/dt = 76.8pi

bahrom7893 · Answer

dv/dt = 241.152...

bahrom7893 · Answer

someone confirm/reject my answer.. i am coming back from a math exam.. my brain is dead tired.. and im doin this off the top of my head..

bahrom7893 · Answer

oh wait I AM SOO WRONG!

bahrom7893 · Answer

A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.3 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 16 cm.

anonymous · Answer

you're not. your ans/2 =correct.

anonymous · Answer

the answer 120.637 is correct

anonymous · Answer

okay...

V = 4/3*pi*r^3 = 4/3*pi*(d/2)^3

= 4/3*pi*d^3/8

= 1/6*pi*d^3

dV/dt = 1/2*pi*d^2*dd/dt

dV/dt = -0.3, when d = 16:

-0.3 = 0.5*pi*256*dd/dt

so dd/dt = -0.3/(128*pi)

so decreasing at 0.3/(128*pi)

anonymous · Answer

dv/dt=(4)(pi)(8^2)(0.15)=120.637

anonymous · Answer

oh...i got the diameter decrease rate... lol

dv/dt = 0.5*pi*256*-0.3

= -120.64

so decreasing at 120.64

bahrom7893 · Answer

Okay I got -38.4pi, which is 120.64 I confirm..

bahrom7893 · Answer

TOGETHER WE SHALL DEFEAT RELATED RATES!

bahrom7893 · Answer

sorry meant which is -120.64 lol

bahrom7893 · Answer

All of yall.. good answers everyone xD!

anonymous · Answer

good effort