Question #336373

Related Rates: Problem Solving


Direction: Solve each word problem involving a related rates. Make a sketch of your work

if necessary, and then apply differentiation.


A spherical balloon is being inflated so that its volume is increasing at rate of 5

cubic feet per minute. At what rate is the diameter increasing when the diameter

is 12 feet?


PLEASE ANSWER MY QUESTION QUICKLY!!

DEADLINE : 05/03/2022 11 : 00 PM


1
Expert's answer
2022-05-06T13:12:51-0400
V=43πR3=π6D3V=\dfrac{4}{3}\pi R^3=\dfrac{\pi}{6}D^3

Differentiate both sides with respect to tt


dVdt=π6(3D2)(dDdt)\dfrac{dV}{dt}=\dfrac{\pi}{6}(3D^2)(\dfrac{dD}{dt})

Then


dDdt=2πD2(dVdt)\dfrac{dD}{dt}=\dfrac{2}{\pi D^2}(\dfrac{dV}{dt})

Given dVdt=5ft3/min,D=12ft.\dfrac{dV}{dt}=5{ft}^3/min, D=12ft.


dDdt=2π(12ft)2(5ft3/min)=572πft/min\dfrac{dD}{dt}=\dfrac{2}{\pi (12ft)^2}(5{ft}^3/min)=\dfrac{5}{72\pi}ft/min


0.0221ft/min\approx0.0221 ft/min

The diameter is increasing at rate of 0.0221feet per minute when the diameter

is 12 feet.



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