A spherical balloon is inflated with gas at the rate of 20 ft^3/min. How fast is the radius of the balloon changing at the instant that r=2ft. FIND dr/dt

Question

pls help

Answer 1

Tenzin,

One way of solving here is using logarithmic derivative

The volume of a sphere

V=(4/3)Pi R^3=Const x R^3 Const=(4/3) Pi] (1)

LnV= Ln(Const) + 3Ln R. (2)

Therefore, the logarithmic derivative of V

d(Ln V)/dt=(1/V)dV/dt=3dLnR/dt=(3/R) dR/dt (3)

From (3) dR/dt=(R/3V)dV/dt (4)

where V =(4/3)Pi R^3 (5)

Therefore, by substituting (5) into (4) we get

dR/dt=(R/4PiR^3)dV/dt=(1/4Pi R^2)dV/dt (6)

Note: The solution (6) could also be obtained by observing that a volume dV of a spherical layer of thickness dR between two concentric spheres with radii R and (R+dR) equals to area 4PiR^2 times dR

Hope it was helpful

Dr.Ariel B.

Answer 2

copy:

\frac{dV}{dt}&=20ft^3/\text{min}\\

V&=\frac{4}{3}\pi r^3\\

\frac{dV}{dt}&=\frac{4}{3}\pi 3r^2 \frac{dr}{dt}\\

20\cdot\frac{1}{4\pi\cdot 10^2}&=\frac{dr}{dt}\\

.1571\,ft/\text{min}&\approx \frac{dr}{dt}\\

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