V=4 π r3 / 3
dV/dt = dV/dr . dr/dt
It's very similar to the triangle problem.
A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.4 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 10 cm. (Note the answer is a positive number).
3 answers
diameter is decreasing at a rate of 0.4 cm/min
---> dd/dt = -.4 cm/min or dr/dt = -.2 cm/min
V = (4/3)pi(r^3)
dV/dt = 4pi(r^2)dr/dt
so when d=10, r=5 and
dV/dt = 4pi(25)(-.2) cm^3/min
= -20pi cm^3/min
The negative rate for dV/dt tells me that the volume is decreasing, so if the word "decreasing" is used in the concluding sentence you would use 20pi.
---> dd/dt = -.4 cm/min or dr/dt = -.2 cm/min
V = (4/3)pi(r^3)
dV/dt = 4pi(r^2)dr/dt
so when d=10, r=5 and
dV/dt = 4pi(25)(-.2) cm^3/min
= -20pi cm^3/min
The negative rate for dV/dt tells me that the volume is decreasing, so if the word "decreasing" is used in the concluding sentence you would use 20pi.
I see it now.