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

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)   (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)   (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)   (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 = . Since both the volume and the radius depend on time, the formula becomes V(t) = . The derivative over time is = Hence, = = . (1) The surface area of a sphere is A = , or A(t) = . The derivative over time is = = Substitute here the expression = from (1) and r(t) = 15 cm. You will get = = = = 2*8 = 16 cm^2/min.

Solved.

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