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Derivatives of implicit functions

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
 
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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?
 
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plugpoint said:
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
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
 
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apmcavoy said:
I don't see how you get the square root of 2.
He gave y', so I thought he plugged 0 into y, which he didn't give in his post.
 

mezarashi

Homework Helper
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apmcavoy said:
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?
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.
 
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mezarashi said:
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.
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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