SOLUTION: A spherical snow ball is melting at 120cm3/min. At what rate is the surface area decreasing at when the radius is 15cm?

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Question 1163173: A spherical snow ball is melting at 120cm3/min. At what rate is the surface area decreasing at when the radius is 15cm?
Found 3 solutions by solver91311, dkppathak, ikleyn:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!














You can do this arithmetic (leave in terms of )



Plug in 15 for and the value for calculated above to find at

Note: is in units of and is in units of


John

My calculator said it, I believe it, that settles it

Answer by dkppathak(439) About Me  (Show Source):
You can put this solution on YOUR website!
A spherical snow ball is melting at 120cm3/min. At what rate is the surface area decreasing at when the radius is 15cm?
solution
dv/dt=120cm^3/min given
v=4/3(πr3)
dv/dt=4πr2 dr/dt than dr/dt=120/4πr2
Surface area of sphere A=4πr2
da/dt=A=4*2πr dr/dt=8πrdr/dt
da/dt=8πrx120/4πr2=240/r
putting r=15
da/dt=240/15=16
rate of change of decreasing surface are 16cm^2/min

Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.

            After reading the posts of the two other respectful tutors,  I have a feeling
            that another solution should be written and presented in more clear form. 


The volume of a sphere is  V = %284%2F3%29%2Api%2Ar%5E3.


Since both the volume and the radius depend on time, the formula becomes  V(t) = %284%2F3%29%2Api%2Ar%5E3%28t%29.


The derivative over time is  %28dV%29%2F%28dt%29 = 4%2Api%2Ar%5E2%28t%29%2A%28%28dr%29%2F%28dt%29%29


Hence,  %28dr%29%2F%28dt%29 = %28%28dV%29%2F%28dt%29%29%2F%284%2Api%2Ar%5E2%28t%29%29 = 120%2F%284%2Api%2A15%5E2%29.   (1)


The surface area of a sphere is  A = 4%2Api%2Ar%5E2,  or  A(t) = 4%2Api%2Ar%5E2%28t%29.


The derivative over time is  %28%28dA%29%2F%28dt%29%29 = 4%2Api%2A2%2Ar%28t%29%2A%28%28dr%29%2F%28dt%29%29 = 8%2Api%2Ar%28t%29%2A%28%28dr%29%2F%28dt%29%29


Substitute here  the expression  %28dr%29%2F%28dt%29 = 120%2F%284%2Api%2A15%5E2%29  from (1)  and r(t) = 15 cm.  You will get


    %28%28dA%29%2F%28dt%29%29 = 8%2Api%2A15%2A%28120%2F%284%2Api%2A15%5E2%29%29 = %288%2A15%2A120%29%2F%284%2A15%5E2%29 = %282%2A120%29%2F15 = 2*8 = 16 cm^2/min.

Solved.