A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.1 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).
We are given the rate of decrease of the diameter `(dd)/(dt)=.1"cm"/"min" ` . We are asked to find the rate of the decrease of the volume when the diameter is 10cm:
The formula for the volume of a sphere is `V=4/3pir^3 ` .
The derivative is ` (dV)/(dt)=4pir^2(dr)/(dt) `
If the diameter is 10cm, the radius is 5cm. Also, if `(dd)/(dt)=.1 ` then `(dr)/(dt)=.05 `
Then substituting the given values:
(The entire section contains 2 answers and 240 words.)
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You can also use the equation:
V = (1/6)pi d^3
dV/dt = (1/2)pi d^2 dd/dt
Since dd/dt = .1 cm/min, and d = 10 cm, we only have to plug in and find that the answer is 15.7 cm^3/min.