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Continuously differentiable

  1. Mar 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Given:
    w=9(cos([tex]\theta[/tex])+sin([tex]\theta[/tex])) and u=r*w.
    1) Is u continuously differentiable?
    2)Is it possible to get u continuously differentiable with a different w?

    2. Relevant equations



    3. The attempt at a solution

    1) u is continuously differentiable since w is in terms of sines and cosines which are continuously differentiable functions. (Is that right?)
    2) I'm stuck on this one.

    Thank you
     
  2. jcsd
  3. Mar 15, 2009 #2

    matt grime

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    What's r? A constant? A function of theta? A variable? I'm inclined to think that r and theta are polar co-ordinates, is that correct? If so what are you differentiating with respect to?

    What level of proof is required? I.e. do you have to do things from first principles? Can you appeal to known results: sin and cos are certainly continuously differentiable, but can you use that?
     
  4. Mar 15, 2009 #3

    Dick

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    As a function of two real variables r and theta, u is continuously differentiable. But if you think of u as a function on the x-y plane described by the polar coordinates (r,theta) you MAY have a problem. Think about what happens at r=0. Do you?
     
  5. Mar 15, 2009 #4
    u(x,y)=rw(theta) where r and theta are polar coordinates.When we differentiate, it will be with respect to x and y.
    When r=0 then u=0 which is continuously differentiable..??
     
  6. Mar 15, 2009 #5

    Dick

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    u=0 is the value of the function at r=0. I would be more worried about the values of the derivatives of u near r=0. Try writing it as a function of x and y.
     
  7. Mar 15, 2009 #6
    Do you mean I should write w in terms of x and y?
     
  8. Mar 15, 2009 #7

    Dick

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    Sure. And then write u as a function of x and y.
     
  9. Mar 15, 2009 #8
    w=9cos(theta)+9sin(theta)
    =9(x/r)+9(y/r)
    so u=rw=9(x+y)
    the derivative (whether its with respect to x or y) is 9 which is continuously differentiable.
     
  10. Mar 15, 2009 #9

    Dick

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    Seems fine to me.
     
  11. Mar 15, 2009 #10
    Thanks.
    What about the next one: Is it possible to get u continuously differentiable with a different w?
     
  12. Mar 15, 2009 #11

    Dick

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    You shouldn't have any trouble coming up with others that are. It's more interesting to think about ones that aren't. How about u=r?
     
  13. Mar 16, 2009 #12
    u=r is continuously differentiable.
    and we can have other functions like u=rsin(2theta) etc.. right?
     
  14. Mar 16, 2009 #13

    Dick

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    u=r is not even differentiable at 0. Think harder about it. It's a lot like absolute value, f(x)=|x|.
     
  15. Mar 16, 2009 #14
    u=r=sqrt(x^2+y^2)
    the derivative is u'=(1/2)(2x or 2y)(x^2+y^2)^(-1/2)
    the derivative at 0 is 0.


    But can we not have other functions for u that are continuously differentiable?
     
  16. Mar 16, 2009 #15

    Dick

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    The derivative at zero is not zero. It's undefined. To see whether it's continuously differentiable you have to look at limits. Try limit du/dx as x->0+ and x->0- and y=0. To find other functions that ARE continuously differentiable, write any function that is continuously differentiable at 0 in x-y coordinates and convert to polar coordinates. Like u=x^2+y^2=r^2.
     
  17. Mar 16, 2009 #16
    oh right i see, we have to take the limit of the derivative.
    ok so say u=r^2=x^2+y^2
    then du/dx=2x and the limit of 2x as x tends to 0 is 0 so it is continuously differentiable...right?
     
  18. Mar 16, 2009 #17

    Dick

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    Right. u=r^2 is continuously differentiable. u=r is not.
     
  19. Mar 16, 2009 #18
    thank you.
     
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