Derivatives of implicit functions

  1. Oct 7, 2005 #1
    If you are given the derivative of an implicit function as [tex] y' = \frac{y}{2y+x} [/tex] how would you find all points (x,y) such that the slope at those points is 1/2? Ok so I did: [tex] \frac{y}{2y+x} = \frac{1}{2} [/tex] and got x = 0. So if I substitute x = 0 back into the original equation I get [tex] (0, \sqrt{2}) [/tex]. Would this be the correct method to solve this question? Also show that the slope is never equaled to 0. So I did [tex] \frac{y}{2y+x} = 0 [/tex]. If y = 0, then the original equation wouldnt make sense. Is this a valid response?

    Any help is appreciated

    Thanks
     
  2. jcsd
  3. Oct 7, 2005 #2
    Well, I wouldn't word it "the original equation wouldn't make sense". Can you word this more mathematically or demonstrate why cannot equal 0?
     
  4. Oct 7, 2005 #3
    When you solve:

    [tex]\frac{y}{2y+x}=\frac{1}{2}\quad (1)[/tex]

    and come up with x=0, I don't see how you get the square root of 2. Try plugging x=0 into equation (1). The y variable will cancel out and you get an identity. What does that tell you about the possible values of y?

    Alex
     
  5. Oct 7, 2005 #4
    He gave y', so I thought he plugged 0 into y, which he didn't give in his post.
     
  6. Oct 7, 2005 #5

    mezarashi

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    I think you forgot to note that equation one is the derivative of the function. I think plugpoint may have forgot to post the original equation to where he got the square root 2. The current differential is not separable.
     
  7. Oct 7, 2005 #6
    You don't need the original equation. If you plug x=0 into y', you get 1/2 no matter what value of y you have. This means that along the vertical line x=0 the slope is the same for all values of y.

    Alex
     
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