# I need help in this related rates problem?

A spherical balloon is inflated so that its volume is increasing at the rate of 5ft^3/min. How fast is the diameter of the balloon increasing when the radius is 3ft?

Rate of change of the diameter = ??? ft/min

Relevance
• Ian H
Lv 7
1 month ago

Without calculus:

V = (4π/3)(D/2)^3 = (π/6)D^3

When D = 6, V = 36π

In 1/1000 of a minute V becomes 36π + 5/1000

D^3 = 6V/π = 216 + 3/(100π) ~ 216.0095493

D ~ 6.000088418...

Change in diameter in 1/1000 of a minute ~ 0.000088418

Estimate of dD/dt is ~ 0.088 ft/min

dV/dt = dV/dD * dD/dt = 5 ft^3/min

V = (4π/3)r^3 = (π/6)D^3

dV/dD = πD^2/2

dD/dt = 5/(dV/dD) = 10/(πD^2), (in general)

When diameter D = 6 ft, dD/dt = 10/(36π)

dD/dt = 5/(18π) ft/min (answer in exact form)

dD/dt ~ 0.088419 ft/min, or, about 1.06 inches per minute

• 1 month ago

V = (4/3)πr³

so, V = (4/3)π(d/2)³ => (1/6)πd³

Hence, dV/dt = (1/2)πd².dd/dt

At the point when r = 3, d = 6 so,

5 = (1/2)π(6)².dd/dt

so, dd/dt = 5/18π => 0.088 feet/min

:)>

• 1 month ago

Sphere V = ⁴/₃πr³

A = 4πr²

d = 2r

V = ⁴/₃πr³ = ⁴/₃π(d/2)³ = (1/6)πd³

dV/dd = (1/2)πd²

dV/dt = 5 ft³/min

dd/dt = dV/dt / dV/dd = (1/2)πd² / 5 = (1/10)πd²

dd/dt  (d is 6) =  (1/10)π6² = 21.6π ft/min