Question

From a spherical balloon of radius r, gas is escaping at the rate of 500cm^3/s. Find how fast the surface area A is shrinking at the instant r=250cm.

Answer 1

let V be the volume then V=4/3pi r^3
dV/dt=4 pi r^2dr/dt=500cm^/s
dr/dt=500/4 pi r^2
let A bethe area = 4pi r^2
dA/dt=8 pi r dr/dt
= 8 pi r * 500 /4 pi r^2
= 1000/r
when r = 250
dA/dt= 4cm^2/s

Answer 2

From a spherical balloon of radius r,
gas is escaping at the rate of
500cm^3/s.

Find how fast the surface area A is shrinking at the instant
r=250cm.

\[\frac{dr}{dt}=\frac{dr}{dV}\frac{dV}{dt}\ ;\ \frac{dV}{dt}=500\]
\[V = \frac{4}{3}pi\ r^3\ ;\ derive\ with\ repect\ to\ "V"\]
\[\frac{dV}{dV} = \frac{4(3)}{3}pi\ r^2*\frac{dr}{dV}\]
\[1 = 4pi\ r^2*\frac{dr}{dV}\]
\[\frac{1}{4pi\ r^2}=\frac{dr}{dV}\]
find the value of "r" and use it to determine dr/dV

Answer 3

-500 might be more accurate, but just keep in mind that the radius is "shrinking" and apply the right sign afterwards :)

Answer 4

..... and now that it reads surface area ... doh!!

Answer 5

im going for a coffee :)