Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

bmelyk Group Title

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

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  2. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • 2 years ago
  3. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I figure I have to do this implicitly.

    • 2 years ago
  4. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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)

    • 2 years ago
  5. Rezz5 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    is the answer y=4-5x?

    • 2 years ago
  6. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  7. Rezz5 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    its a plane....?

    • 2 years ago
  8. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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')

    • 2 years ago
  9. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    a plane? huh?

    • 2 years ago
  10. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  11. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  12. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  13. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    how would you have done this?

    • 2 years ago
  14. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  15. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  16. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    correct

    • 2 years ago
  17. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  18. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  19. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    so that will just equal 3y' ??

    • 2 years ago
  20. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    or 3+3y'

    • 2 years ago
  21. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  22. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    so it's 3(1+y')

    • 2 years ago
  23. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yes - correct

    • 2 years ago
  24. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  25. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  26. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  27. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    haha maybe, maybe not...

    • 2 years ago
  28. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  29. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    and the right hand side over to the left?

    • 2 years ago
  30. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • 2 years ago
  31. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    first expand the expression on the right

    • 2 years ago
  32. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    do i leave my 3 on the right hand side?

    • 2 years ago
  33. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    snap! yes :)

    • 2 years ago
  34. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  35. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    that look right to you?

    • 2 years ago
  36. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    not quite :(

    • 2 years ago
  37. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    oh snappp

    • 2 years ago
  38. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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:\[x^{2}(2y*y')-3y'=3-y^{2}(2x)\]

    • 2 years ago
  39. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    then factorise y' on the left

    • 2 years ago
  40. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  41. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    perfect!

    • 2 years ago
  42. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    woooo. so now that part is done, right?

    • 2 years ago
  43. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • 2 years ago
  44. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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)

    • 2 years ago
  45. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    okay thanks. : )

    • 2 years ago
  46. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yw :) and good luck!

    • 2 years ago
  47. bmelyk Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  48. asnaseer Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.