Asked by Branden
A spherical balloon is inflated at a rate of 10 cubic feet per minute. How fast is the radius of the balloon changing at the instant the radius is 4 feet?
And
The radius of a circle is decreasing at a rate of 2 ft/minute. Find the rate at which the area is decreasing with respect to time when the radius is 4 feet?
And
The radius of a circle is decreasing at a rate of 2 ft/minute. Find the rate at which the area is decreasing with respect to time when the radius is 4 feet?
Answers
Answered by
Reiny
V = (4/3)πr^3
dV/dt = 4πr^2 dr/dt
given dV/dt = 10 ft^3/min
so at r = 4
10 = 4π(16) dr/dt
dr/dt = 10/(64π) or appr. .05 ft/min
If A = 4πr^2, do the second question the same way.
dV/dt = 4πr^2 dr/dt
given dV/dt = 10 ft^3/min
so at r = 4
10 = 4π(16) dr/dt
dr/dt = 10/(64π) or appr. .05 ft/min
If A = 4πr^2, do the second question the same way.
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