- #1
jimbo007
- 41
- 2
hey guys,
just wondering if i did this correctly
a spherical balloon is to be filled with water so that there is a constant increase in the rate of its surface area of 3cm2/sec .
a) Find the rate of increase in the radius when the radius is 3cm.
b) Find the volume when the volume is increasing at a rate of 10cm3=sec.
A=surface area, V=volume
a) i thought was fairly easy and got 1/(8Pi) cm/sec
b) was a bit more trickier here is what i did
(1) [tex]\frac{dV}{dt} = \frac{dV}{dr} \frac{dr}{dt} [/tex]
(2) [tex]\frac{dA}{dt} = \frac{dA}{dr} \frac{dr}{dt} [/tex]
from the question we have
[tex]\frac{dV}{dt}=10[/tex]
[tex]\frac{dA}{dt}=3[/tex]
[tex]\frac{dV}{dr}=4 \pi r^2[/tex]
[tex]\frac{dA}{dr}=8 \pi r[/tex]
from (2)
[tex]3=8 \pi r \frac{dr}{dt}[/tex]
so [tex]\frac{dr}{dt}=\frac{3}{8 \pi r}[/tex]
sub into (1) we get that r=20/3
so [tex]V=\frac{4}{3}\pi r^3=\frac{4 \pi}{3}(20/3)^3[/tex]
just wondering if i did this correctly
a spherical balloon is to be filled with water so that there is a constant increase in the rate of its surface area of 3cm2/sec .
a) Find the rate of increase in the radius when the radius is 3cm.
b) Find the volume when the volume is increasing at a rate of 10cm3=sec.
A=surface area, V=volume
a) i thought was fairly easy and got 1/(8Pi) cm/sec
b) was a bit more trickier here is what i did
(1) [tex]\frac{dV}{dt} = \frac{dV}{dr} \frac{dr}{dt} [/tex]
(2) [tex]\frac{dA}{dt} = \frac{dA}{dr} \frac{dr}{dt} [/tex]
from the question we have
[tex]\frac{dV}{dt}=10[/tex]
[tex]\frac{dA}{dt}=3[/tex]
[tex]\frac{dV}{dr}=4 \pi r^2[/tex]
[tex]\frac{dA}{dr}=8 \pi r[/tex]
from (2)
[tex]3=8 \pi r \frac{dr}{dt}[/tex]
so [tex]\frac{dr}{dt}=\frac{3}{8 \pi r}[/tex]
sub into (1) we get that r=20/3
so [tex]V=\frac{4}{3}\pi r^3=\frac{4 \pi}{3}(20/3)^3[/tex]