1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: This will get you!

  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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted