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Real quick easy question: hori and vert tan lines

  1. Sep 14, 2008 #1
    sorry, i hit enter too soon

    1. The problem statement, all variables and given/known data
    Y=x(sqrt(4-x^2))

    how do I find all pts on the graph where the tan line is vert and hori?

    2. Relevant equations

    Y=x(sqrt(4-x^2))

    3. The attempt at a solution

    y'=(x)(.5(4-x^2)^(-1/2)(-2x))+(sqrt(4-x^2))(1)
     
    Last edited: Sep 14, 2008
  2. jcsd
  3. Sep 14, 2008 #2

    Dick

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    You have a horizontal tangent where y'=0. You have a vertical tangent where y' goes to + or - infinity.
     
  4. Sep 14, 2008 #3

    tiny-tim

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    Welcome to PF!

    Hi ClaudeFrollo! Welcome to PF! :smile:

    (have a square-root: √ and a squared: ² :smile:)

    You have y' = -x²/√(4 - x²) + √(4 - x²),

    and you want to solve y' = 0.

    Hint: simplify the expression for y' by putting the whole thing as a fraction with √(4 - x²) on the bottom. :wink:
     
  5. Sep 14, 2008 #4
    (1-x)/[tex]\sqrt{4-x^{2}}[/tex]=y'

    y'= 1/2

    Hmm... so does that mean that that's the slope of the horizontal tangent line?
     
  6. Sep 14, 2008 #5

    Dick

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    That's not y'. Try the algebra you used to simplify the derivative again. Then try and set y'=0 and figure out x and then the x values where y' goes to infinity.
     
  7. Sep 14, 2008 #6
    Whoops. I think I meant:

    x([tex]\sqrt{4-x^{2}}[/tex])/([tex]\sqrt{4-x^{2}}[/tex])

    so... y'=-x? So x=0 when y'=0?
     
  8. Sep 14, 2008 #7

    Dick

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    No. I don't think you meant that either. Can you put the two terms from the product rule over a common denominator and add them?
     
  9. Sep 15, 2008 #8

    tiny-tim

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    hmm :cry: … I think you need a Big Hint:

    [tex]\sqrt{4\ -\ x^2}\ =\ \frac{4\ -\ x^2}{\sqrt{4\ -\ x^2}}[/tex]​
     
  10. Sep 15, 2008 #9
    Ohhhhh yeah, so I guess

    ((1-x)+(4-x[tex]^{2}[/tex]))/([tex]\sqrt{4-x^{2}}[/tex])

    so f'(0)= 5/4?
     
  11. Sep 15, 2008 #10

    tiny-tim

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    I'm confused … where did 1-x come from? :confused:
     
  12. Sep 15, 2008 #11
    I meant -x^2, not 1-x D:
     
  13. Sep 15, 2008 #12

    tiny-tim

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    That's better! :biggrin:

    So y' = 0 when … ? :smile:
     
  14. Sep 15, 2008 #13
    when x= -2, 2?
     
  15. Sep 15, 2008 #14

    Dick

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    No. What did you finally get for the numerator of f'?
     
  16. Sep 15, 2008 #15

    tiny-tim

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    ClaudeFrollo, you'll make less mistakes if you always give a full answer, and don't try to do things in your head. :rolleyes:

    hmm … now seems as good a time as any to remark that your "Real quick easy question" was a clear breach of the Trade Descriptions Act :wink:
    Well? :smile:
     
  17. Sep 16, 2008 #16
    I thought I answered this when I said

    ((-x2(4-x2))/([tex]\sqrt{4-x^{2}}[/tex])

    ...wouldn't the numerator just be ((-x2(4-x2))?
     
  18. Sep 16, 2008 #17

    Dick

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    Don't you mean -x^2+(4-x^2)? While do you keep twisting things up?
     
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