Find y' if (x-y)/(x+y)=(x/y)+1 and show that there are no points on that curve

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Homework Help Overview

The discussion revolves around using implicit differentiation to find the derivative \( y' \) from the equation \((x-y)/(x+y)=(x/y)+1\) and exploring whether there are any points on the corresponding curve.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss various attempts at implicit differentiation and express confusion regarding the implications of their results. Some question how to demonstrate that no points satisfy the equation, while others explore the simplification of the equation to show contradictions.

Discussion Status

There is ongoing exploration of the equation's implications, with some participants suggesting methods to simplify the equation and check for points. Multiple interpretations of the problem are being considered, particularly regarding the relationship between the derivative and the existence of points on the curve.

Contextual Notes

Participants note the challenge of proving that no real number pairs \((x, y)\) satisfy the equation, with discussions about the implications of the derivative being undefined or meaningless in this context.

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Homework Statement



Use implicit differentiation to find y' if (x-y)/(x+y)=(x/y)+1. Now show that there are, in fact, no points on that curve, so the derivative you calculated is meaningless.

Homework Equations





The Attempt at a Solution



I managed to get it into the form:
dy/dx = [(x+y)2-2y3]/[(x)(-2y2+(x+y)2]
This doesn't seem like it proves anything though. Someone help please!
 
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kylem1994 said:

Homework Statement



Use implicit differentiation to find y' if (x-y)/(x+y)=(x/y)+1. Now show that there are, in fact, no points on that curve, so the derivative you calculated is meaningless.

Homework Equations



The Attempt at a Solution



I managed to get it into the form:
dy/dx = [(x+y)2-2y3]/[(x)(-2y2+(x+y)2]
This doesn't seem like it proves anything though. Someone help please!
Find an equation equivalent to \displaystyle \frac{x-y}{x+y}=\frac{x}{y}+1 which shows that this equation cannot be satisfied by any combination of x & y.

Alternatively, solve this equation for x or y.
 
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SammyS said:
Find an equation equivalent to \displaystyle \frac{x-y}{x+y}=\frac{x}{y}+1 which shows that this equation cannot be satisfied by any combination of x & y.

Alternatively, solve this equation for x or y.

I understand this, but how would I go about doing so? I already derived it and what I got wasn't all that nice.



Also the picture is very helpful! How did you go about simplifying it down so far?
 
Nvm, I simplified it down lol. My brain froze for a minute there.
 
hedipaldi said:
Attached

So where I substitute in (2) do I have to get (2) in terms of one of the variables and then substitute that in for the variable?
 
Substitute -1/4 for y/x and see that (2) is not satisfied.
 
hedipaldi said:
Substitute -1/4 for y/x and see that (2) is not satisfied.

I just realized that right before I seen your post (I'm not having a good night lol) So last question, I promise, how did you come up with the -1/4 ?
 
the denominator of y' is 0 where x+4y=0,that is,y/x=-1/4
 
  • #10
Ok, I see that. So when you plug that into the initial equation (2), it shows that both sides are not equal. Wouldn't that show that there are in fact points on that curve? I'm just totally missing this whole question in general. I really appreciate all you help for this by the way!
 
  • #11
kylem1994 said:
Ok, I see that. So when you plug that into the initial equation (2), it shows that both sides are not equal. Wouldn't that show that there are in fact points on that curve? I'm just totally missing this whole question in general. I really appreciate all you help for this by the way!
The question as to whether or not there are points which fit \displaystyle \frac{x-y}{x+y}=\frac{x}{y}+1 has nothing to do with finding y'.

The thing is that there is no pair or real numbers, (x, y), which can make the left side of the equation equal to the right side. The problem asks you to show this.
 
  • #12
Ok, thanks Sammy, making some sense now (the question asks to find y' I guess just to show you can and whatnot) So now, how would I begin to show this?
 
  • #13
kylem1994 said:
Ok, thanks Sammy, making some sense now (the question asks to find y' I guess just to show you can and whatnot) So now, how would I begin to show this?

Start by trying to simplify it. Put everything to one side of the equation and put everything over a common denominator.
 
  • #14
I think the point in this exercise is to practice implicit differentiation.It seems that the choice of the equation is not so good.
 
  • #15
Ok Dick, I have that so far, everything to one side simplified. I'm so fed up with this question.
 
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  • #16
I've spent hours on this, many, many hours, can someone please help me. I've been looking at it so long my brain is mush.
 
  • #17
I know that no matter which number I select for x or y, I can't solve for the other one, but I need a way to prove this for ALL values. I'm really stumped.
 
  • #18
kylem1994 said:
I've spent hours on this, many, many hours, can someone please help me. I've been looking at it so long my brain is mush.

What did you get? The idea will be to write the numerator as a sum of squares.
 
  • #19
See answer 3.The simplification is by common denominator and canceling.Then perform implicit differentiation and find y'.y' is valid when its denominator is not zero,i.e when x is not equal -4y.Finally substitute x=4y in the given equation and show that it is not satisfied.Conclude that the derivative exists everywhere on the given curve.
 
  • #20
I got : x2+xy+2y2=0
 
  • #21
hedipaldi said:
See answer 3.The simplification is by common denominator and canceling.Then perform implicit differentiation and find y'.y' is valid when its denominator is not zero,i.e when x is not equal -4y.Finally substitute x=4y in the given equation and show that it is not satisfied.Conclude that the derivative exists everywhere on the given curve.

I see what you're saying, but the question really doesn't have anything to do with the derivative, its just to show the dangers of calculating something if you don't know whether it exists or not. The question is to prove that there are no points on the curve (x-y)/(x+y)=(x/y)+1 I know I have to prove that for any x value, there is no y value to satisfy the equation but I don't know how to prove for ALL.
 
  • #22
kylem1994 said:
I got : x2+xy+2y2=0

Complete the square on the x^2+xy part. You can write it as (x+y/2)^2 minus something. What's the something?
 
  • #23
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  • #24
Dick said:
Complete the square on the x^2+xy part. You can write it as (x+y/2)^2 minus something. What's the something?

Thank you Dick, my brain is mush at this point. (Been at this for the past 4 1/2 hours)
 
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  • #25
hedipaldi said:
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Thanks, this is great. You and Dick are life savers! I understand this completely now.
 
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  • #26
it suffices to indicate that the discriminant is negative for non zero y.
 
  • #27
Ok great! Thank you so much! I think I need an ibuprofen and a good sleep now!
 
  • #28
Actually the point (0,0) is also not on the curve since it gives 0 in the denominator,so the "curve" contains no points at all.
 
  • #29
kylem1994 said:
Thank you Dick, my brain is mush at this point. (Been at this for the past 4 1/2 hours)

I was trying to get you to write x^2+xy+2y^2=(x+y/2)^2+7*y^2/4. That gives the sum of two nonnegative things equal to zero. So both must be zero. Same conclusion, really.
 

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