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  1. Aug 31, 2014 #1
    Here is a good question, been trying to work it out all evening, and were all engineering students, and struggling!!!

    Points (p,q) lie on the curve √x + √y = 1

    Rearranging to make y the subject we get y=(√x +1)2

    We than take the derivative which gives us 1-[itex]\frac{1}{√x}[/itex]

    Than since we know x=p and y=q

    Plugging in p into y' we get 1-[itex]\frac{1}{√p}[/itex]

    Than into tangent equation y-y1=m(x-x1)

    Giving us y - q = 1-[itex]\frac{1}{√p}[/itex] (x-p)

    Now compared to the answer which says its √pq or x√q + y√p

    I am so lost?

    I have attached a picture as well to help! :confused:
     

    Attached Files:

  2. jcsd
  3. Aug 31, 2014 #2
    For starters, check that you have rearranged & taken the derivative correctly. Maybe doing dx/dy, as well, will get you somewhere. Just guessing!
     
  4. Aug 31, 2014 #3

    PeroK

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    You could try Implicit Differentiation.
     
  5. Aug 31, 2014 #4

    PeroK

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    Something else that might help is:

    [tex]p\sqrt{q} + q\sqrt{p} = \sqrt{pq}(\sqrt{p}+\sqrt{q})[/tex]
     
  6. Aug 31, 2014 #5
    So I did implict differention of [itex]\sqrt{x}[/itex] + [itex]\sqrt{y}[/itex]= 1

    and get [itex]\frac{dy}{dx}[/itex]=[itex]\frac{-\sqrt{y}}{\sqrt{x}}[/itex]

    Knowing x=p and y=q

    I sub in p (as x) to get m (the gradient of the tangent)

    When substituting this into the tangent equation, I still don't get a similar answer.

    Am I going wrong somewhere?
     
    Last edited: Aug 31, 2014
  7. Aug 31, 2014 #6

    vela

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    Remember that ##\sqrt{p} + \sqrt{q} = 1##.
     
  8. Aug 31, 2014 #7
    Okay so I have no idea anymore.

    I have got the implicit derivative as [itex]\frac{-\sqrt{y}}{x}[/itex].

    Therefore the gradient of the tangent to the curve ([itex]\sqrt{x}[/itex]+[itex]\sqrt{y}[/itex]=1 at points p,q is:

    Now as I said earlier, x1=p and y1=q

    m= [itex]\frac{-\sqrt{p}}{\sqrt{q}}[/itex]

    Substituting all known information into the tangent equation of (y-y1)=m(x-x1)

    We get y - q = [itex]\frac{-\sqrt{p}}{\sqrt{q}}[/itex] * (x - p)

    From here, is where I got no idea?
     
  9. Aug 31, 2014 #8
    ImageUploadedByPhysics Forums1409547175.334908.jpg

    I think this is correct?
     
    Last edited: Sep 1, 2014
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