anonymous
  • anonymous
(x + y)^2 = Cx^2 I need to get the x's on one side and the y's on the other so i can take the derivative of them... how do i do that with the (x + y)^2???
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@zepdrix
anonymous
  • anonymous
Multiply it out.
anonymous
  • anonymous
did that, but how do you separate the xy term?

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AccessDenied
  • AccessDenied
If we're trying to avoid implicit differentiation, I wonder if taking the square root of both sides might help. We just have to be careful that we respect the properties of that (the square of either a positive or negative is positive, so we have two cases to deal with)
anonymous
  • anonymous
i like that idea... i dont remember what implicit differentiation is though, so maybe that might be better... I just need to be able to take the derivative of both sides, and I just thought the x's had to be on one side and y's on the other... b/c thats all the other examples are like that.
anonymous
  • anonymous
This has to be implicitly differentiated
dumbcow
  • dumbcow
http://www.wolframalpha.com/input/?i=%28x%2By%29^2+%3D+4x^2 you can use implicit differentiation or solve for y and get dy/dx directly either way works here
dumbcow
  • dumbcow
oh sorry , i plugged in 4 for C in wolfram example
AccessDenied
  • AccessDenied
Yea, it appears to work either way: http://puu.sh/1Xdb6 A little tricky dealing with the +/- cases, so I replaced it with an "A" for some of the work. :p \(A = \pm \sqrt{C} \)
zepdrix
  • zepdrix
You figure this one out yet zon? :D
anonymous
  • anonymous
I am putting it in that graphing calculator now to see if it worked.
anonymous
  • anonymous
well i guess i did something wrong, i ended up getting ln y - ln x = C
anonymous
  • anonymous
and it wont let me graph that.
zepdrix
  • zepdrix
\[\huge \ln\left(\frac{y}{x}\right)=C \qquad \rightarrow \qquad e^{\ln\left(\frac{y}{x}\right)}=e^C\] \[\large \frac{y}{x}=e^C \qquad \rightarrow \qquad \color{royalblue}{y=e^cx}\]
zepdrix
  • zepdrix
Try graphing that blue one maybe :o hmm
zepdrix
  • zepdrix
Where are these natural logs coming from? :D I don't understand what you did XD
dumbcow
  • dumbcow
?? i have graph in wolfram link
anonymous
  • anonymous
i did the first suggestion and took the square root. (x+y)^2 = Cx^2 .... that is the problem. C = (x+y)^2/x^2 but back to the original, i took the square root. x + y = C^.5 x , then i moved all around, took the derivative... and then substituted C back in and I ended up with 1/y = 1/x.... then took the integral of that.
zepdrix
  • zepdrix
oh the next step is to take the integral? Hmm yah i guess I'm coming up with that answer also then :o
anonymous
  • anonymous
when i graphed that blue equation it was a straight line.. but didnt look to be orthogonal to the originals
zepdrix
  • zepdrix
yah it sure didn't :( hmm
dumbcow
  • dumbcow
??? integral ... i thought the point was to find dy/dx
anonymous
  • anonymous
ohhh my... okay i know what i did wrong... i took the y' and not the 1/y' to make it orthogonal... but there are two parts, the first is to get the derivative of the orthogonal trajectories, and then to take the integral to find the general solution.
anonymous
  • anonymous
back where it was y' = y/x, it needed to be -1/y' = y/x
anonymous
  • anonymous
so now i get 1/2x^2 + 1/2y^2 = C
anonymous
  • anonymous
sweet, it worked...
zepdrix
  • zepdrix
Multiply both sides by 2, get rid of the halves. Just absorb the 2 into the C. Ooo you have a circle :O interesting.
zepdrix
  • zepdrix
oh cool! c:
anonymous
  • anonymous
im trying to get the link of the graph in here
zepdrix
  • zepdrix
Upper right corner, theres a blue button, it's a like...a square with an arrow in it
anonymous
  • anonymous
https://s3.amazonaws.com/grapher/exports/vg28uobv4e.png
anonymous
  • anonymous
anyways, there is the picture... so it worked out... that was a tough one... thanks for all the help.

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