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zonazoo

(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???

  • one year ago
  • one year ago

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  1. zonazoo
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    @zepdrix

    • one year ago
  2. azolotor
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    Multiply it out.

    • one year ago
  3. zonazoo
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    did that, but how do you separate the xy term?

    • one year ago
  4. AccessDenied
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    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)

    • one year ago
  5. zonazoo
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    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.

    • one year ago
  6. azolotor
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    This has to be implicitly differentiated

    • one year ago
  7. dumbcow
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    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

    • one year ago
  8. dumbcow
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    oh sorry , i plugged in 4 for C in wolfram example

    • one year ago
  9. AccessDenied
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    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 \(A = \pm \sqrt{C} \)

    • one year ago
  10. zepdrix
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    You figure this one out yet zon? :D

    • one year ago
  11. zonazoo
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    I am putting it in that graphing calculator now to see if it worked.

    • one year ago
  12. zonazoo
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    well i guess i did something wrong, i ended up getting ln y - ln x = C

    • one year ago
  13. zonazoo
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    and it wont let me graph that.

    • one year ago
  14. zepdrix
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    \[\huge \ln\left(\frac{y}{x}\right)=C \qquad \rightarrow \qquad e^{\ln\left(\frac{y}{x}\right)}=e^C\] \[\large \frac{y}{x}=e^C \qquad \rightarrow \qquad \color{royalblue}{y=e^cx}\]

    • one year ago
  15. zepdrix
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    Try graphing that blue one maybe :o hmm

    • one year ago
  16. zepdrix
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    Where are these natural logs coming from? :D I don't understand what you did XD

    • one year ago
  17. dumbcow
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    ?? i have graph in wolfram link

    • one year ago
  18. zonazoo
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    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.

    • one year ago
  19. zepdrix
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    oh the next step is to take the integral? Hmm yah i guess I'm coming up with that answer also then :o

    • one year ago
  20. zonazoo
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    when i graphed that blue equation it was a straight line.. but didnt look to be orthogonal to the originals

    • one year ago
  21. zepdrix
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    yah it sure didn't :( hmm

    • one year ago
  22. dumbcow
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    ??? integral ... i thought the point was to find dy/dx

    • one year ago
  23. zonazoo
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    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.

    • one year ago
  24. zonazoo
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    back where it was y' = y/x, it needed to be -1/y' = y/x

    • one year ago
  25. zonazoo
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    so now i get 1/2x^2 + 1/2y^2 = C

    • one year ago
  26. zonazoo
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    sweet, it worked...

    • one year ago
  27. zepdrix
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    Multiply both sides by 2, get rid of the halves. Just absorb the 2 into the C. Ooo you have a circle :O interesting.

    • one year ago
  28. zepdrix
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    oh cool! c:

    • one year ago
  29. zonazoo
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    im trying to get the link of the graph in here

    • one year ago
  30. zepdrix
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    Upper right corner, theres a blue button, it's a like...a square with an arrow in it

    • one year ago
  31. zonazoo
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    https://s3.amazonaws.com/grapher/exports/vg28uobv4e.png

    • one year ago
  32. zonazoo
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    anyways, there is the picture... so it worked out... that was a tough one... thanks for all the help.

    • one year ago
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