Asked by Cathy
a spherical hailstone is losing mass by melting uniformly over its surface as it falls. At a certain time, its radius is 2 cm and its volume is decreasing at the rate of o.1 cm^3/s. How fast is the radius decreasing at the time?
Answers
Answered by
saiko
V=pi*r^3
dV/dt=3*pi*(r^2)*(dr/dt)
dV/dt=.1
r=2
pi=3.14
dr/dt=?
dV/dt=3*pi*(r^2)*(dr/dt)
dV/dt=.1
r=2
pi=3.14
dr/dt=?
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