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mathcalculus

  • one year ago

Find the slope of the tangent line to the curve (a lemniscate) 2(x^2+y^2)^2=25(x^2-y^2) at the point (-3,1)?

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  1. hartnn
    • one year ago
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    could you implicitly differentiate that ?

  2. hartnn
    • one year ago
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    you will need chain rule here whats d/dx x^2 =...? whats d/dx y^2=...?

  3. hartnn
    • one year ago
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    what ? where does 2*2 come from ?

  4. hartnn
    • one year ago
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    d/dx[2(x^2+y^2)^2] = 2*(2 (x^2+y^2)) times d/dx(x^2+y^2) this is because of chain rule, got this step ?

  5. hartnn
    • one year ago
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    d/dx (x^2) = 2x thats why and the other 2 was already there

  6. hartnn
    • one year ago
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    ok, so now what about d/dx (x^2+y^2) =...?

  7. mathcalculus
    • one year ago
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    2x+2y?

  8. hartnn
    • one year ago
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    here, chain rule again comes into picture d/dx f(y) = f'(y) dy/dx so d/dx (y^2) = 2y dy/dx got this ?

  9. hartnn
    • one year ago
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    if you got that we are almost done :)

  10. hartnn
    • one year ago
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    d/dx[2(x^2+y^2)^2] = 2*(2 (x^2+y^2)) times d/dx(x^2+y^2) this part ?

  11. hartnn
    • one year ago
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    whats d/dx x^2 =.... ?

  12. hartnn
    • one year ago
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    we don't rewrite anything

  13. mathcalculus
    • one year ago
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    ohman :/

  14. hartnn
    • one year ago
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    derivative of outer times derivative of inner like for (sin x)^3 we have outer = x^3, derivative = 3x^2 inner = sin x derivative = cos x so, 3 (sin x)^2 cos x got this ?

  15. hartnn
    • one year ago
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    in same way d/dx [(x^2+y^2)^2] = 2 (x^2+y^2) times d/dx (x^2+y^2)

  16. mathcalculus
    • one year ago
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    i get that, but in this problem im confused with why the exponent goes away

  17. hartnn
    • one year ago
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    ok, you said d/dx (x^2) = 2x why did the exponent go away here ?

  18. mathcalculus
    • one year ago
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    2(x^2+y^2)^2=25(x^2-y^2)

  19. mathcalculus
    • one year ago
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    so you're finding the derivative of the inside?

  20. hartnn
    • one year ago
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    we do find derivative of inside in chain rule, right ?

  21. mathcalculus
    • one year ago
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    yup

  22. mathcalculus
    • one year ago
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    my professor also taught us a way of outer first then rewrite inner and then multiple by the derivative of the inner.

  23. hartnn
    • one year ago
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    he must have meant "derivative of outer first ....."

  24. mathcalculus
    • one year ago
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    d/dx[f(g(x))]= f' (g(x))*g'(x)

  25. hartnn
    • one year ago
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    yes, here f(x) was x2

  26. mathcalculus
    • one year ago
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    well isn't it 4?

  27. mathcalculus
    • one year ago
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    2(x^2+y^2)^2 |dw:1381774415672:dw|

  28. hartnn
    • one year ago
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    yes, it is 4.

  29. mathcalculus
    • one year ago
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    oh so then it is. where are we getting x2?

  30. hartnn
    • one year ago
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    d/dx[2(x^2+y^2)^2] = 2*(2 (x^2+y^2)) times d/dx(x^2+y^2) = 4(x^2+y^2)times d/dx(x^2+y^2)

  31. hartnn
    • one year ago
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    sorry i didn't get your question?

  32. mathcalculus
    • one year ago
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    got that =]

  33. mathcalculus
    • one year ago
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    so can it be written like this? : d/dx 4(x^2+y^2) (2x+2y) ?

  34. hartnn
    • one year ago
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    no, why would d/dx come in front again ?

  35. mathcalculus
    • one year ago
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    so i get rid of it?

  36. mathcalculus
    • one year ago
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    since i have the derivative already?

  37. hartnn
    • one year ago
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    d/dx [(x^2+y^2)^2] = 2 (x^2+y^2) times d/dx (x^2+y^2) = 4 (x^2+y^2) times (2x+2y dy/dx) see whether you get this?

  38. mathcalculus
    • one year ago
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    yeah i got it, but why don't we do dy/dx here: 4 (x^2+y^2) times (2x+2y dy/dx)|dw:1381774790286:dw|

  39. mathcalculus
    • one year ago
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    we can't bc we have to leave it as it is right? according to the chain rule

  40. hartnn
    • one year ago
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    because there, you have not differentiated x^2+y^2 yrt

  41. hartnn
    • one year ago
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    *yet

  42. hartnn
    • one year ago
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    and yes, we need to leave it as it is, that too

  43. mathcalculus
    • one year ago
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    ok

  44. mathcalculus
    • one year ago
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    now what do we do

  45. mathcalculus
    • one year ago
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    same for the other side?

  46. hartnn
    • one year ago
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    yes, other side is too simple. try ?

  47. hartnn
    • one year ago
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    right side is just 25(2x-2y dy/dx) isn't it ?

  48. mathcalculus
    • one year ago
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    25(x^2-y^2)*(2x-2y*dy/dx)

  49. hartnn
    • one year ago
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    what why ? we need chain rule for just y^2 here, isn't it ?

  50. mathcalculus
    • one year ago
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    huh i did?

  51. hartnn
    • one year ago
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    its like 25x^2 - 25y^2 derivative = 25(2x) - 25 (2y dy/dx)

  52. mathcalculus
    • one year ago
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    where did i go wrong?

  53. hartnn
    • one year ago
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    the x^2-y^2 part was unnecessary, you though it was chain rule again ?

  54. mathcalculus
    • one year ago
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    wait it wasn't because it had no exponent?

  55. hartnn
    • one year ago
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    yes, correct.

  56. mathcalculus
    • one year ago
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    25*(2x-2y*dy/dx)

  57. mathcalculus
    • one year ago
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    so its this?

  58. mathcalculus
    • one year ago
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    got cha.

  59. mathcalculus
    • one year ago
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    now do we plug in x and y?

  60. hartnn
    • one year ago
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    correct! put x=-3, y=1 in that...

  61. hartnn
    • one year ago
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    i assume you are simplifying...

  62. mathcalculus
    • one year ago
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    i don't know i keep getting the answer wrong whenever i try to submit it

  63. hartnn
    • one year ago
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    what did u get for dy/dx ?

  64. mathcalculus
    • one year ago
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    i keep getting incorrects :/

  65. mathcalculus
    • one year ago
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    i got 3/119

  66. hartnn
    • one year ago
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    4 (x^2+y^2) times (2x+2y dy/dx) = 25(2x-2y dy/dx) x=-3, y=1 4(9+1) (-6 +2 dy/dx) = 25 (-6 -2 dy/dx)

  67. mathcalculus
    • one year ago
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    ohi wrote one of the signs wrong.

  68. mathcalculus
    • one year ago
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    got it.

  69. mathcalculus
    • one year ago
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    thank you so much!

  70. mathcalculus
    • one year ago
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    :)

  71. hartnn
    • one year ago
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    welcome ^_^

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