Question

the volume v=4/3πr³ of a spherical balloon changes with the radius

a) at what rate (ft³/ft) does the volume change with respect to the radius when r=2ft?

(type an exact answer in terms of π)

b)using the rate from part a, by approximately how much does the volume increase when the radius changes from 2 to 2.4ft?

Answer 1

a)

When , .

b)

.

Answer 2

Differentiate V with respect to r:

When r = 2ft

dV/dr = 4πx4 = 16π ft³/ft.

when the radius changes from 2 to 2.4, dr = 2.4 - 2 = 0.4

The change in volume is obtained by puttind dr = 0.24

Answer 3

a)V=4/3π r^3

dV/dr =d/dr(4/3π r^3)

=4/3 π d/dr(r^3)

=4/3 π 3r^2

dv/dr =4πr^2 ft^3/ft

dV/dr at r= 2ft

is dv/dr|r=2ft =4π2^2 ft^3/ft

=16π ft^3/ft

b) given dr=2.4-2=.4

=>dV= 4π r^2 dr

r=2ft,dr=.4ft

=> dV =4π2^2 *.4

=>dV=6.4π ft^3

so increase in volume is 6.4π ft^3