Asked by Barbara
This is my last question, when I usually get problems like this the volume would be there and I would be able to solve it but this question is very confusing.
A spherical iron ball is coated with a layer of ice of uniform thickness. If the ice melts at the rate of 8mL/min, how fast is the outer surface of the ice decreasing when the outer diameter (ball plus ice) is 20cm?
A spherical iron ball is coated with a layer of ice of uniform thickness. If the ice melts at the rate of 8mL/min, how fast is the outer surface of the ice decreasing when the outer diameter (ball plus ice) is 20cm?
Answers
Answered by
Damon
Area of sphere = 4 pi r^2 (here r = 10 cm)
d area/dr = 8 pi r
d area/dt = d area /dr * dr/dt (chain rule)
NOW
dVolume/dt = area * dr/dt
we are given d Volume/dt = 8 mL/min which is 8 cm^3/min
from that and area = 4 pi r^2 get dr/dt
then use that dr/dt up above to get d area/dt
d area/dr = 8 pi r
d area/dt = d area /dr * dr/dt (chain rule)
NOW
dVolume/dt = area * dr/dt
we are given d Volume/dt = 8 mL/min which is 8 cm^3/min
from that and area = 4 pi r^2 get dr/dt
then use that dr/dt up above to get d area/dt
Answered by
Barbara
Okay Thank You
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