Related rates of increase

  • Thread starter jimbo007
  • Start date
  • #1
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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]
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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I did it slightly differently, using
[tex]\frac{dV}{dA}= \frac{\frac{dV}{dt}}{\frac{dA}{dt}}= \frac{r}{2}= \frac{10}{3}[/tex]
to find r and then V but I got exactly the same answer!
 

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