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Gradient of stationary points

  1. Feb 8, 2010 #1
    1. The problem statement, all variables and given/known data
    gradient of the stationary points of y=1-2sinx domain 0<x<2pi


    2. Relevant equations



    3. The attempt at a solution
    dy/dx = -2cosx
    -2cosx=0......?
     
    Last edited: Feb 8, 2010
  2. jcsd
  3. Feb 8, 2010 #2

    rock.freak667

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    Gradient of the stationary point? Do you need to find that or the gradient? Because when you put dy/dx=0 it sort of implies what the gradient will be.
     
  4. Feb 8, 2010 #3
    Sorry. the qu reads 'the curve y=1-2sinx has domain 0<x<pi. find the gradients of the curve at the points where the curve crosses the x-axis
     
  5. Feb 8, 2010 #4

    rock.freak667

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    Ah that is different.

    Well you know the gradient function is dy/dx.

    You need to find when the curve crosses the x-axis, or when y=0.
     
  6. Feb 8, 2010 #5
    so do i differentiate it and make it = 0?

    so i get -2cosx=0
     
  7. Feb 8, 2010 #6
    how to i rearrange this to work it out?
     
  8. Feb 8, 2010 #7

    rock.freak667

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    No if y=1-2sinx, the curve crosses, the x-axis when y=0

    so solve 1-2sinx=0 for x. Can you do that?
     
  9. Feb 8, 2010 #8
    -2sinx=1
    sinx=-1/2
     
  10. Feb 8, 2010 #9

    rock.freak667

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    which gives x as ? ( in the given domain that is)
     
  11. Feb 8, 2010 #10
    principle value of -0.523 (but out of range) the pi-(-0.523) = 3.6651

    therfore value = 3.6651 and 6.2832?????
     
  12. Feb 8, 2010 #11

    rock.freak667

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    this should be sinx=1/2.


    But how did you get that value for x?
     
  13. Feb 8, 2010 #12
    ohhh... did sin^-1... so ....

    x principle = 0.479425
    so pi - 0.4879425 = 2.66216
     
  14. Feb 8, 2010 #13

    rock.freak667

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    you should get x= π/6 and another one, best to use the exact values, I don't have a calculator with me to divide out pi.
     
  15. Feb 8, 2010 #14
    ok. i'll have a play and post back if i cant do it.. Thanks for the inspiration!
     
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