Answer to Question #336373 in Calculus for John

Question

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

Answer 1

$$V=\dfrac{4}{3}\pi R^3=\dfrac{\pi}{6}D^3$$

Differentiate both sides with respect to $t$

$$\dfrac{dV}{dt}=\dfrac{\pi}{6}(3D^2)(\dfrac{dD}{dt})$$

Then

$$\dfrac{dD}{dt}=\dfrac{2}{\pi D^2}(\dfrac{dV}{dt})$$

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

$$\dfrac{dD}{dt}=\dfrac{2}{\pi (12ft)^2}(5{ft}^3/min)=\dfrac{5}{72\pi}ft/min$$

$$\approx0.0221 ft/min$$

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

is 12 feet.