Question

I need someone to explain to me step by step the following problem:

Answer 1

Find the equation of the tangent line to the curve
\[x^{2}y^{2}-1=3(x+y)\] at the points (1,-1)

Answer 2

I have the problem worked out in front of me (prof gave us the answers) but, his work is disorganized and I don't understand what it is he's doing.

Answer 3

I figure I have to do this implicitly.

Answer 4

yes - you are correct - use implicit differentiation to get an expression for slope of the tangent line at any point (x,y).
then substitute x=1, y=-1 to find the value of the slope of the tangent line at (1, -1)
then use the equation of a straight line: y = mx + c to find c (since you now know m) by using the fact that it passes through the point (1, -1)

Answer 5

is the answer y=4-5x?

Answer 6

that's not the answer.... the answer is: y=-1/5x+4/5

Answer 7

its a plane....?

Answer 8

okay and the way my prof did his differentiation was really weird.
his first step looks like: x^{2}2y*y'+y^{2}(2x)=3(1+y')

Answer 9

a plane? huh?

Answer 10

that's not how i would start the implicit differentiation.

Answer 11

\[x^{2}(2y*y')+y^{2}(2x)=3(1+y')\]

Answer 12

what your prof did is correct - he is using the product rule, see here: http://www.1728.org/chainrul.htm

Answer 13

how would you have done this?

Answer 14

haha no idea... i suck at math : (

Answer 15

so i use the product rule for x^2*y^2 ??

Answer 16

correct

Answer 17

and for the 3*(x+y) ??

Answer 18

that is just straight differentiation since it can be written as: 3x + 3y

Answer 19

so that will just equal 3y' ??

Answer 20

or 3+3y'

Answer 21

so you can effectively just ignore the 3 outside the braces and just differentiate what is inside the braces

Answer 22

so it's 3(1+y')

Answer 23

yes - correct

Answer 24

okay so now i have the same first line as he does haha.

Answer 25

i ignored the 3 outside the brackets because it's a constant right?

Answer 26

yes.

thats good - you are making progress - maybe you are a lot better at maths than you think. :)

Answer 27

haha maybe, maybe not...

Answer 28

so now i can take the y^2(2x) over to the other side?

Answer 29

and the right hand side over to the left?

Answer 30

yes - what you need to do is to get an expression for y'.
so move all terms involving y' to the left and the rest to the right

Answer 31

first expand the expression on the right

Answer 32

do i leave my 3 on the right hand side?

Answer 33

snap! yes :)

Answer 34

okay so now i have: (x^2)(2y*y')-(1-y')=3(-2xy^2)

Answer 35

that look right to you?

Answer 36

not quite :(

Answer 37

oh snappp

Answer 38

you started with this:\[x^{2}(2y*y')+y^{2}(2x)=3(1+y')\]so first expand the right to get:\[x^{2}(2y*y')+y^{2}(2x)=3+3y'\]then move terms involving y' to the left and the rest to the right to get:\[x^{2}(2y*y')-3y'=3-y^{2}(2x)\]

Answer 39

then factorise y' on the left

Answer 40

\[y'=\frac{ 3-2xy^{2} }{ 2x^{2}y-3 }\]

Answer 41

perfect!

Answer 42

woooo. so now that part is done, right?

Answer 43

yup - the rest should be straight forward. I need to go for dinner now. I'll check back afterwards to see if you had any further problems.

Answer 44

the steps left are:

then substitute x=1, y=-1 to find the value of the slope of the tangent line at (1, -1)
then use the equation of a straight line: y = mx + c to find c (since you now know m) by using the fact that it passes through the point (1, -1)

Answer 45

okay thanks. : )

Answer 46

yw :) and good luck!

Answer 47

i got the rest of it now ; ) thanks a bunch!

Answer 48

thats great!

and you are more than welcome! :)