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Consider the implicit function y(x) defined by the equation

  1. Nov 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider the implicit function y(x) defined by the equation
    cos(x-y)=xy for x>0.5
    a.find the smallest two positive values x>0.5 for which y(x)=0
    b.find the global minimum of y(x) for x>0.5
    c.call the numbers you found inpart(b) x1 and x2 find
    integral of y(x)dx from x1 to x2


    2. Relevant equations



    3. The attempt at a solution
    a.One of the two positive values is pi/2 and another one is 3*pi/2
    b.I could not figure this out.
    c.- 2*cos(y) – pi^2*y
     
  2. jcsd
  3. Nov 22, 2009 #2

    Mark44

    Staff: Mentor

    Are what you show in your attempts at a solution the answers in the back of the book? I don't see how you could get something for part c (which depends on part b), if you couldn't figure out part b.

    For part b, a maximum or minimum will occur at a point where y'(x) = 0 or is undefined or at an endpoint of the domain. Have you tried differentiating your equation implicitly to find y'(x)?
     
  4. Nov 22, 2009 #3
    I did part a. in matlab
    Here is the code:
    cos(x-y) = x*y
    y(x) = 0, cos(x) = 0
    solve('cos(x)=0')
    ans = pi/2
    Here is the code for part c.
    int('cos(x-y)-x*y',pi/2,3*pi/2)
    ans = - 2*cos(y) – pi^2*y
    So, did I do part a. and c. correctly?
    I still don't get how to do part b.
     
  5. Nov 22, 2009 #4

    Mark44

    Staff: Mentor

    What are you going to do in a test situation where you don't have access to matlab? This problem can be done without technology more advanced than pencil and paper.

    For part a, if y(x) = 0, substitute 0 for y in the equation cos(x - y) = xy, and find the two smallest x values that are larger than .5.

    For part b, this is what I said earlier.
    When you get y'(x), we can work on part c.
     
  6. Nov 22, 2009 #5
    I differentiate the equation and I got:
    y'(x)=sin(x-y)-x
    Substituting y=0 into the equation:
    y'(x)=sin(x)-x
    I got 0 for the answer but I think it is wrong because it has to be greater than 0.5
     
  7. Nov 22, 2009 #6

    Mark44

    Staff: Mentor

    This is not right. You are forgetting that you need to use the chain rule and the product rule.
     
  8. Nov 22, 2009 #7
    I redo the problem and I got
     
  9. Nov 22, 2009 #8
    I redo the problem and I got
    y'(x)=-y-sin(x-y)-sin(x-y)+x
    y'(x)=-sin(x)-sin(x)+x
    y'(x)=-2sin(x)+x
     
  10. Nov 22, 2009 #9
    Am I on the right track?
     
  11. Nov 22, 2009 #10

    Mark44

    Staff: Mentor

    Not even close. You should not be starting with y'(x) = <whatever>.

    Starting with cos(x - y) = xy, your next equation should have on the left, the derivative with respec to x of cos(x - y). On the right, this equation should have the derivative with respect to x of xy.

    To differentiate cos(x - y), you need to use the chain rule. To differentiate xy, you need the product rule.
     
  12. Nov 22, 2009 #11

    Mark44

    Staff: Mentor

    Also, you have posted the same problem in another thread. Don't do this.
     
  13. Nov 22, 2009 #12
    I think I did it right this time.
    -sin(x-y)+cos(x-y)=x+y
     
  14. Nov 22, 2009 #13

    Mark44

    Staff: Mentor

    No, both sides are incorrect.
    The derivative with respect to x of cos(x - y) is -sin(x - y) times the derivative with respect to x of x - y. (chain rule)

    For the derivative with respect to x of xy, you need the product rule.
    d/dx(xy) = x*d/dx(y) + y*d/dx(x). Can you finish this and the one above?
     
  15. Nov 22, 2009 #14
    The left hand side is -sin(x-y)*1
    The right hand side is y
     
  16. Nov 22, 2009 #15
    How can I solve this?

    Mod note: merging threads
    --Redbelly98


    1. The problem statement, all variables and given/known data
    The derivative of cos(x-y)=xy is -sin(x-y)*(1+dy/dx)=y+x*dy/dx


    2. Relevant equations



    3. The attempt at a solution
    I simplified a little and I got
    -sin(x-y)-dy/dxsin(x-y) = y+x*dy/dx
    I got stuck
     
    Last edited by a moderator: Nov 22, 2009
  17. Nov 22, 2009 #16

    Mark44

    Staff: Mentor

    No, both are wrong. Did you read what I wrote? Namely, d/dx(xy) = x*d/dx(y) + y*d/dx(x). That's not equal to y.

    Go back in your calculus book and review the section on implicit differentiation.
     
  18. Nov 22, 2009 #17
    This should be the right derivative:
    y'(x)=-y-sin(x-y)/sin(x-y)+x
    What do I do next?
     
  19. Nov 23, 2009 #18

    Mark44

    Staff: Mentor

    It's not right, but it's close.

    cos(x - y) = xy
    d/dx(cos(x - y)) = d/dx(xy)
    -sin(x - y)*(1 - y') = xy' + y
    y'sin(x - y) - xy' = y + sin(x -y)
    y'(sin(x - y) - x) = y + sin(x - y)
    y' = [y + sin(x - y)]/[sin(x - y) - x]

    A minimum or maximum can occur only at a number x at which the derivative is zero or undefined, or at an endpoint of the domain.

    Find all numbers x > .5 at which the derivative y' is zero or undefined.
     
  20. Nov 23, 2009 #19
    What did I do wrong?
     
  21. Nov 23, 2009 #20

    Mark44

    Staff: Mentor

    I don't know. I showed you how to do it. Compare what you did with what I did and see where they are different.
     
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