Question

find all points ob the curve x^2y^2+xy=2
where the slope of the tangent line is -1. Ok so i know you find y prime first, but then what? Thanks in advance...

Answer 1

Did you try taking a derivative yet? :o
What'd you come up with?
is it really ugly?

Answer 2

y prime is: -y-2xy^2/2yx^2+x

Answer 3

Hmm you're missing a 2 in your numerator I think.

Answer 4

so the two doesnt go to zero when you take the derivative?

Answer 5

Oooo good call.. XD

Answer 6

\[\Large\bf\sf y'\quad=\quad \frac{-y-2xy^2}{2yx^2+x}\]Bahhh ya caught me :D

Answer 7

:)

Answer 8

Gotta watch my math there, heh.

Answer 9

so i know how to find the equation if given the points but i dont know how to find them with only a slope..any ideas?

Answer 10

Ok so instead of giving a point, they gave a slope.
The slope is the value for y'.\[\Large\bf\sf -1\quad=\quad \frac{-y-2xy^2}{2yx^2+x}\]

Answer 11

And then we have to try and uhhhh.. do something with this.

Answer 12

ok that seems right...i think lol

Answer 13

ooo this works out really nicely.
I won't spoil the fun.
I bet you'll figure it out! :O

Answer 14

so i assume that you times the denominator by the -1 to get -2yx^2-x=-y-2xy^2, then the 2yx^2 cancel

Answer 15

x=y+2xy^2

Answer 16

and y= x-2xy^2

Answer 17

what? I'm confused.
If you cancel the 2xy^2, aren't you just left with -x=-y ?

Answer 18

oh my gosh yes oops lol

Answer 19

so then do you plug in the slope to find points?

Answer 20

so we've found that our curve has slope -1 along y=x.
Let's see where our curve `intersects` the line y=x to find the actual points.

Answer 21

So we'll plug x in for all of our y's.

Answer 22

\[\Large\bf\sf x^2y^2+xy=2\qquad\quad\to\quad\qquad x^2(x^2)+x(x)=2\]

Answer 23

to find x's?

Answer 24

so X^4+X^2=2

Answer 25

ok good.
Understand how to solve for x?
It's a little tricky.

Answer 26

so i assume you bring the 2 over then equal it to 0?

Answer 27

yes you subtract 2 from each side, leaving zero on the right.

Answer 28

ok then it looks like X^4+x^2-2=0

Answer 29

i dont think i know how to factor this one:(

Answer 30

Let, \(\large\bf\sf u=x^2\)

Answer 31

\[\Large\bf\sf u^2+u-2=0\]

Answer 32

can you factor it now? :o

Answer 33

No.
(u-1)(u+2)=0

Answer 34

but what about the x^4?

Answer 35

You can't factor out an x^2 from each term.
Is that what you were trying to do?

Just do the substitution I mentioned.

Answer 36

so are we not factoring X^4+x^2-2=0? cause there is an x^4

Answer 37

4th degree polynomials are not easy to factor unless you can immediately understand what is going on.

So I was making a substitution to reduce the function to a QUADRATIC.
We can factor it easily once it's a quadratic.

We will undo the substitution later.

Answer 38

u=x^2

Answer 39

u^2 = x^4

Answer 40

x^4+x^2-2=0

--> u^2+u-2=0

Answer 41

oh i see what ya did there