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Derivative as an Instantaneous rate of change

  1. Sep 15, 2006 #1
  2. jcsd
  3. Sep 15, 2006 #2

    Galileo

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    Mathematically, what is asked is [tex]V'(d)[/tex] (prime denotes differentiation), where V(d) is the volume as a function of the depth d. So find V(d) first.
     
  4. Sep 15, 2006 #3

    radou

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    Yes, mathematically, the result should be a constant, but there's no serious physical sense here.
     
  5. Sep 15, 2006 #4
    Thats the thing i am having trouble with, starting the equation? I know I need the Volume for a cylinder equation but how do I use it to figure it out?
     
  6. Sep 15, 2006 #5

    StatusX

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    The volume of a cone is 1/3 b*h, where b is the area of the base (pi r^2) and h is the height.
     
  7. Sep 15, 2006 #6
    So would the new equation be 1/3 4 - d * 4 - d (pi r^2)?
     
  8. Sep 15, 2006 #7

    radou

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    The equation is V = 1/3 * d^3 * Pi . Since d can change, it is a variable, which makes the equation a function, so you can write V(d) = 1/3 * d^3 * Pi , because the volume of the fluid V depends of it's depth d.
     
  9. Sep 15, 2006 #8
    Okay, Thanks a lot i didn't get the d^3.
     
  10. Sep 15, 2006 #9

    radou

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    The d^3 is implied by radius = height.
     
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