Question

(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???

Answer 1

@zepdrix

Answer 2

Multiply it out.

Answer 3

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

Answer 4

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)

Answer 5

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.

Answer 6

This has to be implicitly differentiated

Answer 7

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

Answer 8

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

Answer 9

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 \(A = \pm \sqrt{C} \)

Answer 10

You figure this one out yet zon? :D

Answer 11

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

Answer 12

well i guess i did something wrong,

i ended up getting ln y - ln x = C

Answer 13

and it wont let me graph that.

Answer 14

\[\huge \ln\left(\frac{y}{x}\right)=C \qquad \rightarrow \qquad e^{\ln\left(\frac{y}{x}\right)}=e^C\]

\[\large \frac{y}{x}=e^C \qquad \rightarrow \qquad \color{royalblue}{y=e^cx}\]

Answer 15

Try graphing that blue one maybe :o hmm

Answer 16

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

Answer 17

??
i have graph in wolfram link

Answer 18

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.

Answer 19

oh the next step is to take the integral? Hmm yah i guess I'm coming up with that answer also then :o

Answer 20

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

Answer 21

yah it sure didn't :( hmm

Answer 22

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

Answer 23

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.

Answer 24

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

Answer 25

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

Answer 26

sweet, it worked...

Answer 27

Multiply both sides by 2, get rid of the halves.
Just absorb the 2 into the C.

Ooo you have a circle :O interesting.

Answer 28

oh cool! c:

Answer 29

im trying to get the link of the graph in here

Answer 30

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

Answer 31

https://s3.amazonaws.com/grapher/exports/vg28uobv4e.png

Answer 32

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