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mathcalculus Group Title

Help:implicit differentiation with a fractions :(

  • 10 months ago
  • 10 months ago

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  1. mathcalculus Group Title
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    • 10 months ago
  2. myko Group Title
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    multiplying everything by 36 you get; 2x^2+y^2=36 which maybe more clear for you, :)

    • 10 months ago
  3. mathcalculus Group Title
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    hey! :) i tried this: x²/16 + y²/36 = 1 (1/16)x² + (1/36)y² = 1 2(1/16)x + 2(1/36)yy' = 0

    • 10 months ago
  4. mathcalculus Group Title
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    but so far im not sure what to do after because i know i get x/8 + ?? but then i can't really multiply (1/36) 2y*dy/dx ?

    • 10 months ago
  5. mathcalculus Group Title
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    isn't the point here to find dy/dx?

    • 10 months ago
  6. myko Group Title
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    let \(F(x,y)=2x^2+y^2=36\) then: \(dF(x,y)=4xdx+2ydy=0\) from here: \(y'=dy/dx=-4x/2y=-2x/y\)

    • 10 months ago
  7. mathcalculus Group Title
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    huh? o.o

    • 10 months ago
  8. mathcalculus Group Title
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    howd you get this?? let F(x,y)=2x2+y2=36

    • 10 months ago
  9. myko Group Title
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    \(dF(x,y)\) is colled full differential it is: \(dF(x,y)=F_xdx+F_ydy\)

    • 10 months ago
  10. mathcalculus Group Title
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    i never seen that before

    • 10 months ago
  11. myko Group Title
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    \(F_x\) is partial derivative respect to x

    • 10 months ago
  12. mathcalculus Group Title
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    ok

    • 10 months ago
  13. mathcalculus Group Title
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    i know those, but where are the numbers coming from?

    • 10 months ago
  14. mathcalculus Group Title
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    in what order, where did you take them?

    • 10 months ago
  15. mathcalculus Group Title
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    i know they're given but how did you get 36 on the other side. and 2x^2+y^2let F(x,y)=2x2+y2=36

    • 10 months ago
  16. myko Group Title
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    ok, other way: think that y is actualy a function of x, so it is y=y(x). Then: \(2x^2+y(x)^2=36\) now we differentiate keeping this in mind 4x+2yy'=0 now just solve for y'

    • 10 months ago
  17. mathcalculus Group Title
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    2x2+y(x)2=36?

    • 10 months ago
  18. mathcalculus Group Title
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    yes i know when to differentiate but how did you change the function to that

    • 10 months ago
  19. myko Group Title
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    Ya you right I made a mistake. I was treating 36 like 32, :). Shold be. Rest of steps same. dw:1381788857534:dw|

    • 10 months ago
  20. myko Group Title
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    |dw:1381789088155:dw|

    • 10 months ago
  21. mathcalculus Group Title
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    i know I was going to tell you.

    • 10 months ago
  22. mathcalculus Group Title
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    you can't have 36 at the end.

    • 10 months ago
  23. mathcalculus Group Title
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    i saw a different way on yahoo, would be be able to help me with this one x²/16 + y²/36 = 1 (1/16)x² + (1/36)y² = 1 2(1/16)x + 2(1/36)yy' = 0 (x/8) + (y/18)y' = 0 (y/18)y' = -(x/8) y' = -(x/8)/(y/18) y' = -(x/8)(18/y) y' = -(9x/4y) y'(2) = -(9(2))/(4(5.2)) y'(2) = -18/20.8 y'(2) = -9/10.4 y'(2) = -0.865

    • 10 months ago
  24. mathcalculus Group Title
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    I'm not really sure how they got the 4th step :/.... this part----> + (y/18)*dy/dx ??

    • 10 months ago
  25. myko Group Title
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    2(1/16)x + 2(1/36)yy' = 0 (2/16)x+(2/36)yy'=0 (x/8) + (y/18)y' = 0

    • 10 months ago
  26. mathcalculus Group Title
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    yes my question is exactly what you wrote up there .... |dw:1381789520628:dw|(x/8) + (y/18)y' = 0

    • 10 months ago
  27. myko Group Title
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    (2/36)=1/18

    • 10 months ago
  28. mathcalculus Group Title
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    how did it get simplified to y/18*dy/dx?

    • 10 months ago
  29. myko Group Title
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    your answer is same, just simplify 2 in numerator with 36 in denominator

    • 10 months ago
  30. mathcalculus Group Title
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    thanks! :) i dontkknow why i missed that silly stuff.

    • 10 months ago
  31. myko Group Title
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    :)

    • 10 months ago
  32. mathcalculus Group Title
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    then i plug in for x and y and find the common denominators rights?

    • 10 months ago
  33. myko Group Title
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    yes

    • 10 months ago
  34. mathcalculus Group Title
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    wait we're not done yet! :(

    • 10 months ago
  35. mathcalculus Group Title
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    i got -.005379

    • 10 months ago
  36. mathcalculus Group Title
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    the question is asking y'(1)

    • 10 months ago
  37. myko Group Title
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    y' = -(9x/4y) so: substitute x=1, y=5.80948 y'(1)=-9/4(5.80948)

    • 10 months ago
  38. mathcalculus Group Title
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    okay i got dy/dx=-.387298

    • 10 months ago
  39. mathcalculus Group Title
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    got it thanks @myko :))!

    • 10 months ago
  40. myko Group Title
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    yw

    • 10 months ago
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