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Related rates of increase

  1. Mar 5, 2006 #1
    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}{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]
  2. jcsd
  3. Mar 6, 2006 #2


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    Science Advisor

    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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