Question

My question is too long to copy and paste here so here's the link if any of you also use peeranswer.com: http://www.peeranswer.com/question/562804c9991909e7216bb6d1

Answer 1

Why dont u tag someone on peeranswer?

Answer 2

I've shot a friend a message on there to see if he could help but he's too busy.

Answer 3

Well peeranswer.com questions or anything related to peeranswer.com is not allowed in OpenStudy.

Answer 4

I was just wondering if anyone from here could help, that's all.

Answer 5

Well copy and paste the question as a OpenStudy question instead of making people go to another website!

Answer 6

Look at what I said in my original post!: "My question is too long to copy and paste here..."

Answer 7

Well... were having the burden to help you do YOUR hw, how about you do the burden of pasting YOUR hw. After all this is supposed to be YOUR own work!

Answer 8

Doing my homework?! I'm stuck half way through figuring out a problem! YOU don't have to be rude!

Answer 9

Im not being rude, im just saying you should cooperate. Im trying to help you do your hw, and you dont want to cooperate

Answer 10

No, I am cooperating, you are making it difficult to cooperate with.

Answer 11

Well how about you copy and paste the question and ill help you with your hw. How about that?

Answer 12

Have you not look what I said twice? IT IS TOO LONG TO COPY AND PASTE!

Answer 13

Well if you dont want to paste the question, then im out -_-

Answer 14

Seriously? I have stated that it's too long to be copied and pasted.

Answer 15

Not my problem... im trying to help you, and you don't want to help me

Answer 16

You telling me to copy and paste when I have stated that it's too long to do so three times is not helping me at all.

Answer 17

Well maybe im not helping you

Answer 18

The question is: Find the points of the lemniscate where the tangent is horizontal. 8(x^2+y^2)^2=49(x^2-y^2).

This is what I have done for work so up right up to where I got stuck:
d/dx(8(x^2+y^2)^2)=d/dx49(x^2-y^2) (2)8(x^2+y^2)^1(x^2+y^2)^1=49(2x-2y(dy/dx)) 16(x^2+y^2)(2x+2y(dy/dx))=98x-98ydy/dx 16(2x^3+2xy^2+2x^2ydy/dx+2y^3dy/dx)=98x-98ydy/dx 32x^2ydy/dx+32y^3dy/dx+98ydy/dx=98x-32x^2-32xy^2 dy/dx(32x^2y+32y^3+98y)=2x(49-8x^2-8y^2)
dy/dx=2x(49-8(x^2+y^2))/2y(8x^2+8y^2+49)
2x(49-8(x^2+y^2))=0 (insert arrow under the 2x)=> x=0 or x^2+y^2=49/8
Going back to original => 8(x^2+y^2)^2=49(x^2-y^2)
8y^4=-49y^2
8y^4+49y^2=0
y^2(8y^2+49)=0=>y=0 (insert arrow under 8y^2+49)=> =/=0
x^2+y^2=49/8
8(49/8)^2=49(x^2-y^2)
8/49(49^2/64)=x^2-y^2
The 49 under the 8 get's crossed out, the 8 also get's crossed out because it goes into 64 8 times so the 64 under the 49 turns into an 8. I'm confused as to what happens with ^2.
So I am stuck here now:
x^2-y^2=49/8
+x^2+y^2=49/8