Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Proving an equation related to order of derivatives

  1. Jul 20, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove that for all y(x)=ax^2+bx+c where a is a constant !=0 and x is a real number that [tex]\frac{y'(x2)^2- y'(x1)^2}{(x2-x1)}[/tex] = 2y''(x)


    2. Relevant equations
    I don't know what to put here in mathematics but here...
    y(x)=ax^2+bx+c
    y'(x)=2ax+b
    y''(x)=2a


    3. The attempt at a solution
    I made a lot of attempts and tried many thing but here is the one that looks like it works.
    y''(x)=dy'(x)/dx
    y''(x) * dy/dx = y'(x)dy'/dx
    now here I don't know if its ok to take the dx es of the fraction
    y''(x) dy= y'(x)dy'
    integral from x1 to x2 (y''(x) dy)= integral from x1 to x2 (y'(x)dy')
    and since y''(x)=2a so its a constant
    y''(x)* (x2-x1)= 1/2 * (y'(x2)^2-y'(x1)^2)
    2y''(x)= (y'(x2)^2-y'(x1)^2)/ (x2-x1)
     
  2. jcsd
  3. Jul 20, 2010 #2

    Mark44

    Staff: Mentor

    This is ambiguous. What does y'(x2)2 mean? Is it the derivative evaluated at (x2)2 or is it the square of the derivative evaluated at x2?
    You already know that y''(x) = 2a, so I don't see that this or the next line are of any help.
    I'm not sure that what you have above is correct. In any case, it's a lot more convoluted than it needs to be.
     
  4. Jul 20, 2010 #3
     
  5. Jul 20, 2010 #4

    Mark44

    Staff: Mentor

    Then to be clear, you should write (y'(x2)2.

    Evaluate the left side:
    [tex]\frac{(y'(x_2))^2 - (y'(x_1))^2}{x_2 - x_1}[/tex]

    What does it take for the simplified left side to equal 4a (= 2y''(x))? Aside from taking the first and second derivatives of the given function, this problem seems to me to be an algebra exercise.
     
  6. Jul 20, 2010 #5
    But is the solution I already wrote incorrect?
     
  7. Jul 20, 2010 #6

    Mark44

    Staff: Mentor

    It took me awhile to figure out what you were doing, but it seems to be OK.
     
  8. Jul 20, 2010 #7
    I don't understand how you solved it with algebra. since my solution is ok it would be ok for you to post the algebriac solution right? it won't be cheating for me since I already made my own solution right?
     
  9. Jul 20, 2010 #8

    Mark44

    Staff: Mentor

    Right now, I like your approach better than mine. Let me think about it a bit.
     
  10. Jul 20, 2010 #9

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You're integrating the lefthand side with respect to y, so you should get 2a(y2-y1), not 2a(x2-x1).

    The original statement is false. If you have y=x2 for instance, you get

    [tex]\frac{[y'(x_2)]^2-[y'(x_1)]^2}{x_2-x_1} = 4(x_2+x_1)[/tex]

    You're not going to be able to prove it.
     
  11. Jul 20, 2010 #10
    oh sorry the denominator was y(x2)-y(x1) I was translating it from a different language with different variables so...
    Edit:
    also I think this is proven
    in the dimensional motion
    (V(t))^2=V(0))^2 + 2a (x2-x1) I just noticed lol think about it
    v= y'
    and a=y'' and x=y then it would all make sense lol
     
  12. Jul 20, 2010 #11

    Mark44

    Staff: Mentor

    Good catch, vela!
     
  13. Jul 20, 2010 #12
    I mistyped the question sorry but I thinks its fixed now
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook