bmelyk 3 years ago I need someone to explain to me step by step the following problem:

1. bmelyk

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

2. bmelyk

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.

3. bmelyk

I figure I have to do this implicitly.

4. asnaseer

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)

5. Rezz5

6. bmelyk

7. Rezz5

its a plane....?

8. bmelyk

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')

9. bmelyk

a plane? huh?

10. bmelyk

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

11. bmelyk

$x^{2}(2y*y')+y^{2}(2x)=3(1+y')$

12. asnaseer

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

13. asnaseer

how would you have done this?

14. bmelyk

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

15. bmelyk

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

16. asnaseer

correct

17. bmelyk

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

18. asnaseer

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

19. bmelyk

so that will just equal 3y' ??

20. bmelyk

or 3+3y'

21. asnaseer

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

22. bmelyk

so it's 3(1+y')

23. asnaseer

yes - correct

24. bmelyk

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

25. bmelyk

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

26. asnaseer

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

27. bmelyk

haha maybe, maybe not...

28. bmelyk

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

29. bmelyk

and the right hand side over to the left?

30. asnaseer

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

31. asnaseer

first expand the expression on the right

32. bmelyk

do i leave my 3 on the right hand side?

33. asnaseer

snap! yes :)

34. bmelyk

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

35. bmelyk

that look right to you?

36. asnaseer

not quite :(

37. bmelyk

oh snappp

38. asnaseer

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:$x^{2}(2y*y')-3y'=3-y^{2}(2x)$

39. asnaseer

then factorise y' on the left

40. bmelyk

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

41. asnaseer

perfect!

42. bmelyk

woooo. so now that part is done, right?

43. asnaseer

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.

44. asnaseer

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)

45. bmelyk

okay thanks. : )

46. asnaseer

yw :) and good luck!

47. bmelyk

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

48. asnaseer

thats great! and you are more than welcome! :)