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Finding the point on the graph where the tangent is parallel

  1. Dec 16, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the point on the graph y = sqrt 2x-1 where the tangent line is parallel to x - 3y = 16


    2. Relevant equations



    3. The attempt at a solution

    So what I started off was was trying to find the slope of x - 3y = 16. I rearranged the equation to a y = equation and got 16-x /-3 which I got a slope of -3.

    Then I went to find the deriviative of y = sqrt 2x-1 which I get (x - 1/2)exp-1/2

    From there i get a little lost. The derivative = the slope of the tangent which would get me my x and y points, but my x values get a lot wonky. Would somebody be able to lead me in the right direction of finding my x point?

    Thanks!
     
  2. jcsd
  3. Dec 16, 2008 #2

    gabbagabbahey

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    Be careful with your negative signs!

    [tex]y=16-\frac{x}{-3}=16+\frac{x}{3}[/tex]

    so your slope is ___?:wink:

    You should double check your derivative too!:wink:
     
  4. Dec 16, 2008 #3
    OKay, with that then I get a slope of 1/3, im still not clear on the derivative though? If its
    (2x-1)exp 1/2, wouldnt that be (1/2)(2x-1)exp-1/2?
     
  5. Dec 16, 2008 #4

    gabbagabbahey

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    You need to use the chain rule:

    [tex]\frac{d}{dx} (2x-1)^{1/2}=\frac{1}{2}(2x-1)^{-1/2}\cdot\frac{d}{dx}(2x-1)[/tex]
     
  6. Dec 16, 2008 #5
    Okay, this is what ive got from your equation.

    1/2(2x-1)-1/2 x 2

    (2x-1)-1/2

    So with that then I have the deriviative equaling the slope of the tangent which ive figured to be 1/3.

    How do I get rid of that negative exponent?
     
  7. Dec 16, 2008 #6
    To answer my own question, I took the reciprocal of the 2x-1 line so I got

    1/(2x-1) = 1/9 A little cross multiplication and I landed x = 5?

    Throw that into the equation and we get a y value of 3, therefore the answer is (5,3)?


    haha, sorry bud im editting this as your trying to reply, it dawned on me in like 5 seconds, what did I do with the square root??
     
    Last edited: Dec 16, 2008
  8. Dec 16, 2008 #7

    gabbagabbahey

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    Looks good to me :smile:
     
  9. Dec 16, 2008 #8
     
  10. Dec 16, 2008 #9

    HallsofIvy

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    He was assuming you meant what you said!

    16- x/-3 means 16- (x/-3). If you mean (16- x)/-3 you should have used parentheses to make that clear.
     
  11. Dec 16, 2008 #10
    So which is it that I go from?

    If you have x - 3y = 16 how do you adjust that to make it a y = equation?
     
  12. Dec 16, 2008 #11

    Dick

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    x-3y=16 -> x-16=3y -> (x-16)/3=y -> x/3-16/3=y. This shouldn't be the hardest part of the problem. gabba didn't check the derivation from the start.
     
  13. Dec 17, 2008 #12
    If you wouldnt mind confirming for me, but I dont think that changes anything, my answer of (5,10) is still correct?
     
  14. Dec 17, 2008 #13

    Dick

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    x=5 is good. But if y=sqrt(2x-1) what does that make y?
     
  15. Dec 17, 2008 #14
    Yup, sorry I misread my answer, my answer was (5,3)
     
  16. Dec 17, 2008 #15

    Dick

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    That's fine.
     
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