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Derivative with constant variables

  1. Oct 12, 2012 #1
    Derivative with "constant" variables

    Hi, I need some help figuring out this one situation that's got me thrown for a loop... not the first time, since my grasp of calculus is slippery at best.

    1. The problem statement, all variables and given/known data
    a and b are both positive constants
    Drag(D) = av^2+b/v^2
    I'm trying to find the value of V that would represent a min for D

    2. Relevant equations

    D=av^2+b/v^2
    [tex]\frac{dD}{dv} = 2av-\frac{2b}{v^{3}}[/tex]

    [tex]\frac{d^{2}D}{dv^{2}}= 2a+\frac{6b}{v^{4}}[/tex]

    3. The attempt at a solution

    If I learned this proper, the derivative = 0 at Max / Min points... so;
    [tex]0=2av-\frac{2b}{v^{3}}[/tex]

    [tex]\frac{2b}{v^{3}}= 2a[/tex]
    [tex]v=\left(\frac{b}{a}\right)^{-1/4}[/tex]

    I figured that with the second derivative that since all positive values of v will have a positive value of d^2D/dv^2, and so any point of the derivative =0 will be a minimum point for D.

    Now, since this is homework, I don't necessarily need the correct answer, but if I could be pointed to where I'm gong wrong (if I am)... I don't know why, but having these variables as constants is really throwing me for a loop, and any help / advice here would be appreciated.
     
    Last edited: Oct 12, 2012
  2. jcsd
  3. Oct 12, 2012 #2

    Mark44

    Staff: Mentor

    Re: Derivative with "constant" variables

    You have a mistake above - it should be
    $$ \frac{2b}{v^{3}}= 2av$$
    What you have above is OK, though.
    I'm not sure what your question is.
     
    Last edited: Oct 12, 2012
  4. Oct 12, 2012 #3
    I gotta find the values of v where d would be at a min.

    I think I was going wrong by isolating v, instead I think I should isolate a and b and plug those into the equation of d= av^2+b/v^2.

    This gave me d is at min when d = 2b/v^2 and/ or d=2av^2

    Otherwise, the min for d=[tex]\frac{ab +ba}{\sqrt{ab}}[/tex]
    (got that by plugging in the values for v=(b/a)^1/4)

    Ultimately, my question is: does this look like im on the right track?
     
  5. Oct 12, 2012 #4

    Mark44

    Staff: Mentor

    Re: Derivative with "constant" variables

    Without more information, there's no way you can solve for a and b.
    I'm not sure what your reasoning here is, but it looks like you're saying that if a = 0, then D = 2b/v2, and when b = 0, D = 2av2.
    In post 1, you had v = (b/a)-1/4, which is the same as (a/b)1/4.
     
  6. Oct 12, 2012 #5

    Mute

    User Avatar
    Homework Helper

    Re: Derivative with "constant" variables

    No, a and b are just constants - they could be any number, the number just isn't important for what you're trying to do. The velocity v is the variable, so that what you need to solve for, and you found the velocity which minimizes the drag. You can then plug that value of the velocity into your drag equation,

    $$D(v_{min}) = a\left(\frac{a}{b}\right)^{-1/2} + b\left(\frac{a}{b}\right)^{+1/2}.$$

    Then you simplify. There's no need to set a or b to zero.

    The expression you derived, ##(ba+ab)/\sqrt{ba}## is correct, but you can simplify it further.

    You will then have an expression for the minimum drag that involves the unspecified constants a and b. You can now choose any values for those constants and your formula will tell you what the minimum drag on the system is (which it attains when moving at the minimum velocity you calculated in post 3 - the expression you gave in post 1 appears to have been flipped).
     
  7. Oct 13, 2012 #6
    Thanks, I had just finished solving the problem and it, in spite of my lack of confidence in the answer, was exactly what you described here.

    Most of all, not only did I finally work this out, I figured out what I was doing wrong in my approach at first... As evidence of not needing help on the next, equally challenging, question, for that I thank you all.

    The last point you raised of the exponent -1/4, that was an error that I made and partially edited out.

    Forgot that the fourth square of a function is the same as 1/4, and not -4 as I had mistakenly used at first, and only partially corrected, so in post 1 I flipped it by mistake, it wasn't until I looked closer that I realized there was no way the two were equivalent.

    Either way, thanks again for the help, like I said, better to be shown how to do it then just gettin the answer as you guys do. Appreciate it.
     
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