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bmelyk

  • 3 years ago

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

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  1. bmelyk
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    I figure I have to do this implicitly.

  4. asnaseer
    • 3 years ago
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    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
    • 3 years ago
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    is the answer y=4-5x?

  6. bmelyk
    • 3 years ago
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    that's not the answer.... the answer is: y=-1/5x+4/5

  7. Rezz5
    • 3 years ago
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    its a plane....?

  8. bmelyk
    • 3 years ago
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    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
    • 3 years ago
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    a plane? huh?

  10. bmelyk
    • 3 years ago
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    that's not how i would start the implicit differentiation.

  11. bmelyk
    • 3 years ago
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    \[x^{2}(2y*y')+y^{2}(2x)=3(1+y')\]

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

  13. asnaseer
    • 3 years ago
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    how would you have done this?

  14. bmelyk
    • 3 years ago
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    haha no idea... i suck at math : (

  15. bmelyk
    • 3 years ago
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    so i use the product rule for x^2*y^2 ??

  16. asnaseer
    • 3 years ago
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    correct

  17. bmelyk
    • 3 years ago
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    and for the 3*(x+y) ??

  18. asnaseer
    • 3 years ago
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    that is just straight differentiation since it can be written as: 3x + 3y

  19. bmelyk
    • 3 years ago
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    so that will just equal 3y' ??

  20. bmelyk
    • 3 years ago
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    or 3+3y'

  21. asnaseer
    • 3 years ago
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    so you can effectively just ignore the 3 outside the braces and just differentiate what is inside the braces

  22. bmelyk
    • 3 years ago
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    so it's 3(1+y')

  23. asnaseer
    • 3 years ago
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    yes - correct

  24. bmelyk
    • 3 years ago
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    okay so now i have the same first line as he does haha.

  25. bmelyk
    • 3 years ago
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    i ignored the 3 outside the brackets because it's a constant right?

  26. asnaseer
    • 3 years ago
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    yes. thats good - you are making progress - maybe you are a lot better at maths than you think. :)

  27. bmelyk
    • 3 years ago
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    haha maybe, maybe not...

  28. bmelyk
    • 3 years ago
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    so now i can take the y^2(2x) over to the other side?

  29. bmelyk
    • 3 years ago
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    and the right hand side over to the left?

  30. asnaseer
    • 3 years ago
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    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
    • 3 years ago
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    first expand the expression on the right

  32. bmelyk
    • 3 years ago
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    do i leave my 3 on the right hand side?

  33. asnaseer
    • 3 years ago
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    snap! yes :)

  34. bmelyk
    • 3 years ago
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    okay so now i have: (x^2)(2y*y')-(1-y')=3(-2xy^2)

  35. bmelyk
    • 3 years ago
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    that look right to you?

  36. asnaseer
    • 3 years ago
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    not quite :(

  37. bmelyk
    • 3 years ago
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    oh snappp

  38. asnaseer
    • 3 years ago
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    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
    • 3 years ago
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    then factorise y' on the left

  40. bmelyk
    • 3 years ago
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    \[y'=\frac{ 3-2xy^{2} }{ 2x^{2}y-3 }\]

  41. asnaseer
    • 3 years ago
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    perfect!

  42. bmelyk
    • 3 years ago
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    woooo. so now that part is done, right?

  43. asnaseer
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    okay thanks. : )

  46. asnaseer
    • 3 years ago
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    yw :) and good luck!

  47. bmelyk
    • 3 years ago
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    i got the rest of it now ; ) thanks a bunch!

  48. asnaseer
    • 3 years ago
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    thats great! and you are more than welcome! :)

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