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zonazoo

  • one year ago

(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???

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  1. zonazoo
    • one year ago
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    @zepdrix

  2. azolotor
    • one year ago
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    Multiply it out.

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

  4. AccessDenied
    • one year ago
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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)

  5. zonazoo
    • one year ago
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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.

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

  7. dumbcow
    • one year ago
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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

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

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

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

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

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

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

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

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

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

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

  18. zonazoo
    • one year ago
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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.

  19. zepdrix
    • one year ago
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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

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

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

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

  23. zonazoo
    • one year ago
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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.

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

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

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

  27. zepdrix
    • one year ago
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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.

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

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

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

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

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

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