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Average rate of change of a sphere

  1. Oct 23, 2006 #1
    the Volume of a sphere is given by [tex]V(r) = \frac{4}{3}\pi(r)^3[/tex]

    Find the average rate of change of volume with respect to radius as the radius changes from 10 cm to 15 cm.

    Ok what I tried is: [tex]\lim_{r\rightarrow 15} \frac{V(r)-V(a)}{r-a}[/tex]

    [tex]P(a,V(a))=(15,14 137)\\ Let\ r=10 \lim_{r\rightarrow 15} \frac{\frac{4}{3}\pi(10)^3-14,137}{10-15} = 1989[/tex]

    and I would do that for 10 to 15 and get the average rate but somethign is wrong here....
    Last edited: Oct 23, 2006
  2. jcsd
  3. Oct 23, 2006 #2
    if r goes to 15 wouldn't it be wherever there is an r a 15 would go... I think you did it backwards... it looks as if you put "a" where "r" was supposed to go and "r" where "a" is supposed to go.

    P(a,V(a))=(15,14 137)\\ Let\ r=10 \lim_{r\rightarrow 15} \frac{\frac{4}{3}\pi(15)^3-frac{4}{3}\pi(10)^3}{15-10} = 1989.675

    I am not sure exactly about this... although I do remember doing something similar to this last semester. Is this the whole problem? Because if it wants the rate it would usually be for ex. the sphere is increasing by 1989 cm^3 per min.

    I don't know if the formula will show up because it is in code right now and I am not 100% sure how this forum works with formulas. Hope this helped some. If it still seems to be wrong send me a message and I'll look at it again.
  4. Oct 23, 2006 #3
    well I guess the formula didn't work lol... how did you get yours up there?
  5. Oct 23, 2006 #4


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    The average rate of change as the radius changes from 10 cm to 15 cm is simply


    No limit, no hocus pocus involved, just plain soustraction and division of numbers.
  6. Oct 23, 2006 #5
    answer says [tex]\frac{1900}{3}\pi[/tex]
  7. Oct 23, 2006 #6
    that is the same as 1989.675, if you want to get an answer like that just keep pie separate.
  8. Oct 23, 2006 #7
    second part is: Find the rate of change of volume when the radius is 8cm...

    So that means the instantaneous rgight?
  9. Oct 23, 2006 #8
    hmm... do you have the answer for this one as well? and do you know how to do derivatives? This dose look like it means instantaneous.
  10. Oct 23, 2006 #9
    i got it its [tex]256\pi[/tex]

    Used the instantaneous formula thing, i.e. the long way of derivative
  11. Oct 23, 2006 #10
    yeah you will learn how to get it easier just by deriving the equation later. Is this a college course or high school? Just wondering because the part you posted in was precalculus. If you get to where you need help again, I wouldn't mind helping. I need a refresher now and then to make sure I don't forget things.
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