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Rate of change (surface area/radius)

  1. Aug 30, 2011 #1
    The problem statement, all variables and given/known data
    I am 99% positive I did a) correctly, as the answer is correct according to the answer in the back of my text book. However, I cannot do b).
    The question: A spherical balloon is being inflated. Find the rate of change of the surface area with respect to the radius when:

    a) The radius is 5cm
    b) The surface area is 200cm2

    The attempt at a solution

    a) 8[itex]\pi[/itex]*5cm
    =125.66 cm2/cm (or 40[itex]\pi[/itex] cm2/cm)

    b) I am unsure of what to do for this. Is it something like:

    [itex]\sqrt{}200/4\pi[/itex]

    =3.9894

    And then do something else, which I don't know what that is.

    The answer to this question, according to the book, is 100.27cm2/cm

    Sorry if the equations don't look very clear. This is my first post and I will learn how to write it all up neatly when I am not so tired.
     
  2. jcsd
  3. Aug 30, 2011 #2

    Hootenanny

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    Welcome to Physics Forums.

    Since you have solved the first part, I assume that you have worked our that

    [tex]\frac{dS}{dr} = 8\pi r\;.[/tex]

    Now, what you need to ask yourself if: "How is the radius related to the surface area?". :wink:
     
  4. Aug 30, 2011 #3
    There's an excellent video here:
    http://press.princeton.edu/video/banner/

    which, I think, should give you what you need to know. It's 50 minutes in to video 5, and you shouldn't need to watch any of the others in order to follow the explanation (although they are all a good intuitive primer to calculus). Professor Banner covers related rates problems for about 20 minutes, and it should address your question.
     
  5. Aug 30, 2011 #4

    HallsofIvy

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    I presume you know that the surface area of a sphere of radius r is [itex]A= 4\pi r^2[/itex]. That gives, of course, [itex]dA/dr= 8\pi r[/itex] which is where you got "8π*5cm" for (a).

    For (b), instead of being given r, you are given A= 200. Solve [itex]4\pi r^2= 200[/itex] for r, then do it the same as (a).
     
  6. Aug 30, 2011 #5
    Unfortunately I won't be able to view the video as it will take a while to download and I've gotta get going to school for an excursion. Thank you though.

    So if I try:

    200*4pi

    = 2513.27

    Get the square root of that makes 50.1325

    Multiple by 2 for some reason? Makes 100.265. Rounded off gives 100.27. I am pretty sure I tried this method before, however I got 100.29 (rounding error) and I forgot how I did it all. Nevertheless, I don't really understand why I must multiple my answer by 2.
    Thanks for all your help so far.
     
  7. Aug 30, 2011 #6
    Try to solve for r again. Look at the equation [itex]4\pi r^2= 200[/itex], make sure you're correctly solving for r. Then, once you have r, you can use your derivative equation to get the answer.
     
  8. Aug 31, 2011 #7
    Ok I did it.

    Here is my working:

    4[itex]\pi[/itex]r2 = 200
    r2 = 200/4[itex]\pi[/itex]
    r = Square root of 50/[itex]\pi[/itex]

    dS/dr = 8[itex]\pi[/itex]r
    = 8[itex]\pi[/itex]Square root of 50/[itex]\pi[/itex]
    = 100.27 cm2/cm

    Thanks everyone for all your help :smile:
     
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