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Is differentiation a possible approach?

  1. Apr 22, 2015 #1
    Question:
    I have a function of time. Its expression has a constant 'b' in it. I am asked to ascertain how changing 'b' affects the function.

    Specifically, I have velocity as a function of time which accounts for drag forces; 'b' is the drag coefficient. I am asked to ascertain how changing 'b' affects how quickly terminal velocity is attained.

    Attempt 1:
    I understand that Calculus is the study of change. I am tempted to employ it. Hence, I obtained the derivative of the velocity function [tex] \frac{dv}{db}[/tex]

    Trouble:
    My limited understanding also tells me 'derivatives' help me determine how quickly a function changes given a change in one of its variables. I doubt if that's what my velocity-b context is demanding.

    Attempt 2:
    I have also labored through studying [tex]v(t)[/tex] graphs for different values of 'b'.

    Trouble:
    I am not in favor of the method from Attempt 2 since it's labor-intensive and, perhaps, crude (do you agree?).


    Any guidance as to how to address this question would be greatly appreciated.

    Best regards,
    wirefree
     
  2. jcsd
  3. Apr 22, 2015 #2

    HallsofIvy

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    The derivative of v with respect to b ([itex]\partial v/\partial b[/itex]) while holding other variables or parameters constant is the "rate of change of v as b changes".
     
  4. Apr 23, 2015 #3
    Appreciate the response, HallsofIvy. It has prompted me to consider the situation.

    The derivative of v with respect to b while holding other variables or parameters constant, or the rate of change of v as b changes, if positive, will indicate that v increases as b increases. But that's not what the question concerns itself with. To restate it: for different values of b, does v change over time faster?

    How do I address such a situation, please?

    Regards,
    wirefree
     
  5. Apr 24, 2015 #4
    rbxf0w.gif

    Would appreciate some guidance on how to interpret the above expression.

    Regards,
    wirefree
     
  6. Apr 24, 2015 #5

    HallsofIvy

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    Faster than what? Or do you mean that the rate of change itself is increasing? That will be true when the second derivative is positive.

     
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