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Connected rate of change & Calculus

  1. Jul 26, 2009 #1
    1. The problem statement, all variables and given/known data
    quesaa.jpg

    3. The attempt at a solution

    Managed part a) just fine, but the trouble comes with part b. Erm, the thing is i've got the answer just fine using the chain rule;

    i worked out dh/dt = 1/(dV/dh) * dV/dt

    Giving me 18 * [tex]\frac{1}{\frac{pi}{3}(18h-3h^2)}[/tex]

    Now, I thought, "Okay, it says something about 3 so i'll insert that" and i did, giving me the correct answer, but when I was explaining to my friend about it and I thought i could try to explain it to him geometrically by drawing out the semi-circle and the line x = 3 - h I got stuck myself. I just can't justify it to myself geometrically because;

    you have the semi-circle passing through (-3 , 0) (0, 3) & (3 , 0) and the vertical line x = 3 - h, now when it's 'full' surely h will equal 6? So the line goes back to (-3,0)...

    I think i'm generally confused.

    I know it's a long problem but i'd really appreciate any help
     
  2. jcsd
  3. Jul 26, 2009 #2

    HallsofIvy

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    This is a hemisphere of radius 3. When it is full, the height of the water equals the radius, 3.
     
  4. Jul 26, 2009 #3
    Yeah, which is the same reasoning that I used when answering it, however when i sketch the actual graph of y = (9 - x²)^0.5 & x = 3 - h. You'll see what I mean, sureley only when h is 6 will the whole volume of the hemisphere be given when you integrate the volume...
     
  5. Jul 26, 2009 #4
    never minddddd :PP i got it!

    Is it that when you integrate you would only need the cross-section for 1/4 of the hemisphere? (Because it rotates it around)

    Thank you!! :P
     
    Last edited: Jul 26, 2009
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