## zonazoo 2 years 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???

1. zonazoo

@zepdrix

2. azolotor

Multiply it out.

3. zonazoo

did that, but how do you separate the xy term?

4. 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)

5. zonazoo

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

This has to be implicitly differentiated

7. 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

8. dumbcow

oh sorry , i plugged in 4 for C in wolfram example

9. 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 $$A = \pm \sqrt{C}$$

10. zepdrix

You figure this one out yet zon? :D

11. zonazoo

I am putting it in that graphing calculator now to see if it worked.

12. zonazoo

well i guess i did something wrong, i ended up getting ln y - ln x = C

13. zonazoo

and it wont let me graph that.

14. zepdrix

$\huge \ln\left(\frac{y}{x}\right)=C \qquad \rightarrow \qquad e^{\ln\left(\frac{y}{x}\right)}=e^C$ $\large \frac{y}{x}=e^C \qquad \rightarrow \qquad \color{royalblue}{y=e^cx}$

15. zepdrix

Try graphing that blue one maybe :o hmm

16. zepdrix

Where are these natural logs coming from? :D I don't understand what you did XD

17. dumbcow

?? i have graph in wolfram link

18. zonazoo

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

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

when i graphed that blue equation it was a straight line.. but didnt look to be orthogonal to the originals

21. zepdrix

yah it sure didn't :( hmm

22. dumbcow

??? integral ... i thought the point was to find dy/dx

23. zonazoo

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

back where it was y' = y/x, it needed to be -1/y' = y/x

25. zonazoo

so now i get 1/2x^2 + 1/2y^2 = C

26. zonazoo

sweet, it worked...

27. 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.

28. zepdrix

oh cool! c:

29. zonazoo

im trying to get the link of the graph in here

30. zepdrix

Upper right corner, theres a blue button, it's a like...a square with an arrow in it

31. zonazoo
32. zonazoo

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