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anonymous

  • 5 years ago

Find the equation of the tangent line at the point (1,1) for the curve x^3+y^3-2xy=0. But I can't for the life of me figure out how to solve the equation for y.

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  1. amistre64
    • 5 years ago
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    a tangent line is vector + point

  2. amistre64
    • 5 years ago
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    the tangent vector to the curve matches the slope of the tangent line which is the derivative of the equation at the point right?

  3. amistre64
    • 5 years ago
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    D(x^3+y^3-2xy=0) 3x^2 x' +3y^3 y' -2xy' -2y x' = 0 right?

  4. amistre64
    • 5 years ago
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    now solve for y'

  5. amistre64
    • 5 years ago
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    x' = 1

  6. amistre64
    • 5 years ago
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    (y' 3y^2 - y' 2x)+(3x^2 -2y) = 0 y' (3y^2 - 2x) = -(3x^2 -2y) y' = -3x^2 +2y ---------- now fill in your x and y value from the point 3y^2 - 2x

  7. amistre64
    • 5 years ago
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    point (x=1,y=1) -3+2 -1 ----- = --- = -1 3-2 1

  8. amistre64
    • 5 years ago
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    teh tangent vector has the same slope as the tangent line here

  9. amistre64
    • 5 years ago
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    Tv <1,-1> would be suitable

  10. amistre64
    • 5 years ago
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    so lets define the line by define all magnitudes of this vector from that point x = 1 -t y = 1 +t z = 0

  11. amistre64
    • 5 years ago
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    but thats only if we want R^3; R^2 can be just the tangent lint thru the point now lol... i read to much into that

  12. amistre64
    • 5 years ago
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    y = mx +b 1 = (-1)1 + b 1 = -1 + b 2 = b Eq tan line: y = -x +2

  13. amistre64
    • 5 years ago
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    you implicitly differentiate the equation frist; then solve for dy/dx = y'

  14. amistre64
    • 5 years ago
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    then plug in your x and y amounts to determine the slope

  15. anonymous
    • 5 years ago
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    whoa - so I don't have to solve for y first. instead I find the derivative for each part of the product in terms of x; and for each part in terms of y ----no that is not what is going on sorry the vector thing is way over my head...trying also to get at the notation if everyone could agree to use newton or liebnitz my life would be easier am working on it thanks so far

  16. amistre64
    • 5 years ago
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    yeah, i had the wrong thought in me head at first :) No, you dont need to solve for y first; tell me: Does 3x +y = 4 and y = -3x +4 mean the same thing?

  17. anonymous
    • 5 years ago
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    hey amistre what happened to that derivative of integral question??

  18. amistre64
    • 5 years ago
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    lol... i left that to smarter people than me ;)

  19. anonymous
    • 5 years ago
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    bt i answered..nd no-one replied..lol

  20. amistre64
    • 5 years ago
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    kant; if they are equal, then we can derive them both and get the same results right?

  21. amistre64
    • 5 years ago
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    the thing is; deriving is deriving is deriving regardless of whether you can solve for y forst or not

  22. amistre64
    • 5 years ago
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    3x +y = 4 y = -3x +4 3 + y' = 0 y' = -3 y' = -3

  23. amistre64
    • 5 years ago
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    3xy^2 +y = 9 cant be solved for y; but it doesnt matter does it? just derive it like normal and solve for y' afterwards

  24. amistre64
    • 5 years ago
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    Dx(3xy^2) = 3x2y y' + xy^2 ; do you see why?

  25. anonymous
    • 5 years ago
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    well i get the 3x+y=4 == y=-3x+4 i can solve that one but was having trouble with the equation x^3-y^3-2xy could not figure out how to get the y's to one side. nothing factored out of each element ... let me fuss with the new item on paper

  26. amistre64
    • 5 years ago
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    just derive it as tho you could solve for y; becasue it aint the y that you want to find; its y'

  27. amistre64
    • 5 years ago
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    whats the derivative of 3x?

  28. amistre64
    • 5 years ago
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    well, with respect to x

  29. anonymous
    • 5 years ago
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    3

  30. amistre64
    • 5 years ago
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    ...... let me re ask that; i type it in wrong lol your right; but..

  31. amistre64
    • 5 years ago
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    what is the derivative with respect to x of: x^3 is what i meant :)

  32. anonymous
    • 5 years ago
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    3x^2

  33. amistre64
    • 5 years ago
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    good; what is the derivative with respect to x of: -y^2 ?

  34. amistre64
    • 5 years ago
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    .....of -y^3; i gotta learn to proof my work

  35. anonymous
    • 5 years ago
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    since there is no x...-y^3 (dx)?

  36. amistre64
    • 5 years ago
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    dont worry about the "no x" part; go ahead and 'imply' that there is an x. It is the same derivative as tho you were doing: x^3 with respect to x but we got a y in there..

  37. anonymous
    • 5 years ago
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    so I do take the derivative of y: 3y^2?

  38. amistre64
    • 5 years ago
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    exactly :).... but the thing is, when they first teach you derivatives they have you throw out an important part.... and we actually need to keep that part in to see what happens

  39. amistre64
    • 5 years ago
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    lets do x^3 again but keep all the parts and see if you can figure out the secret that they are keeping from you ok?

  40. anonymous
    • 5 years ago
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    those b*st*rds.... then i deal with the mixed item the same way 2x and 2y ( the one im deriving drops off)

  41. amistre64
    • 5 years ago
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    d(x^3) dx ------ = --- 3x^2 ; you see the dx/dx part? that equals 1 dx dx so they simply tell you to ignore it

  42. anonymous
    • 5 years ago
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    right at first i thought the y' notation would be easier but then we started canceling the dy/dx and then that seemed easier

  43. amistre64
    • 5 years ago
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    now lets derive the y^3 and see what happens: d (y^3) dy ------ = --- 3y^2 ; you recognize that dy/dx? dx dx

  44. amistre64
    • 5 years ago
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    that dy/dx doesnt equal 1 so we cant just ignore it; in fact, its what youve always done. watch d (y) d (x^3) ---- = ------- dx dx dy dx -- 1 = -- 3x^2 dx dx dy/dx = 1 * 3x^2 dy/dx = 3x^2

  45. anonymous
    • 5 years ago
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    looking at it on paper that is easier for me

  46. amistre64
    • 5 years ago
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    when you do implicits.... KEEP ALL YOUR DERIVED bits in place till the end

  47. amistre64
    • 5 years ago
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    derive (-2x)(y) for me then; and remember to use the product rule for it

  48. anonymous
    • 5 years ago
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    f'g+g'f= so -2y+(1)-2x = -2y+1-2x

  49. amistre64
    • 5 years ago
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    you threw out your derived bits.... but you got the idea down, now watch this: f = -2x ; f' = -2 x' g = y ; g' = 1 y' f'g + fg': -2 x' y + -2x y' ; since x' = dx/dx = 1 we get -2y - 2x y' rigth?

  50. anonymous
    • 5 years ago
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    akkk not sure where that +1 came from typing math = harder math but the extra y' is just 1 ya?

  51. amistre64
    • 5 years ago
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    lets put this into your equation and see if we can peice together the mystery :) x^3 +y^3 -2xy = 0 3x^2 x' + 3y^2 y' -2y x' -2x y' = 0 now we can take out the derived bits that equal 1: x' = dx/dx = 1

  52. amistre64
    • 5 years ago
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    teh extra y' is just ... y' cant determine that yet becasue we have to solve for it :) it is actually what we are looking for

  53. anonymous
    • 5 years ago
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    ok...i wont jump the gun

  54. anonymous
    • 5 years ago
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    3x^2+3y^2(dy)-2y-2x(dy)

  55. amistre64
    • 5 years ago
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    exactly :) now 3x^2 + 3y^2 dy -2y -2x dy = 0 get all the y's to one side and the rest to the other: 3y^2 dy -2x dy = -3x^2 +2y dy (3y^2 -2x) = -3x^2 +2y dy = -3x^2 +2y --------- now plug in your point (x=1, y=1) to find 3y^2 -2x the slope

  56. amistre64
    • 5 years ago
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    I get dy = slope = -1 now we put that into the equation for the line: y = mx + b and use the point to calibrate it: 1 = -1(1) + b 1 = -1 + b 1+1 = b the equation of the tangent line at (1,1) is: y = -x + 2

  57. anonymous
    • 5 years ago
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    once again let me transfer to paper to see it

  58. anonymous
    • 5 years ago
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    alright i can see that ....one second while i recompute the derivative part im not clear on who got the (dy) part and who did not

  59. amistre64
    • 5 years ago
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    if theres a y involved; it will end up with a dy part with it

  60. anonymous
    • 5 years ago
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    there i like hard fast rule like that

  61. anonymous
    • 5 years ago
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    I think I have it --- find the derivatives for x just like normal; derivative of y just like normal but whenever there is a derivative of y multiply by (dy)....whoa like when using u substitution...thanks for the help

  62. amistre64
    • 5 years ago
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    you got it :)

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