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mathcalculus

  • one year ago

Help:implicit differentiation with a fractions :(

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

  3. mathcalculus
    • one year ago
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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

  4. mathcalculus
    • one year ago
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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 ?

  5. mathcalculus
    • one year ago
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    isn't the point here to find dy/dx?

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

  7. mathcalculus
    • one year ago
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    huh? o.o

  8. mathcalculus
    • one year ago
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    howd you get this?? let F(x,y)=2x2+y2=36

  9. myko
    • one year ago
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    \(dF(x,y)\) is colled full differential it is: \(dF(x,y)=F_xdx+F_ydy\)

  10. mathcalculus
    • one year ago
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    i never seen that before

  11. myko
    • one year ago
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    \(F_x\) is partial derivative respect to x

  12. mathcalculus
    • one year ago
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    ok

  13. mathcalculus
    • one year ago
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    i know those, but where are the numbers coming from?

  14. mathcalculus
    • one year ago
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    in what order, where did you take them?

  15. mathcalculus
    • one year ago
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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

  16. myko
    • one year ago
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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'

  17. mathcalculus
    • one year ago
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    2x2+y(x)2=36?

  18. mathcalculus
    • one year ago
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    yes i know when to differentiate but how did you change the function to that

  19. myko
    • one year ago
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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|

  20. myko
    • one year ago
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    |dw:1381789088155:dw|

  21. mathcalculus
    • one year ago
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    i know I was going to tell you.

  22. mathcalculus
    • one year ago
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    you can't have 36 at the end.

  23. mathcalculus
    • one year ago
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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

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

  25. myko
    • one year ago
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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

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

  27. myko
    • one year ago
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    (2/36)=1/18

  28. mathcalculus
    • one year ago
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    how did it get simplified to y/18*dy/dx?

  29. myko
    • one year ago
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    your answer is same, just simplify 2 in numerator with 36 in denominator

  30. mathcalculus
    • one year ago
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    thanks! :) i dontkknow why i missed that silly stuff.

  31. myko
    • one year ago
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    :)

  32. mathcalculus
    • one year ago
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    then i plug in for x and y and find the common denominators rights?

  33. myko
    • one year ago
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    yes

  34. mathcalculus
    • one year ago
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    wait we're not done yet! :(

  35. mathcalculus
    • one year ago
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    i got -.005379

  36. mathcalculus
    • one year ago
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    the question is asking y'(1)

  37. myko
    • one year ago
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    y' = -(9x/4y) so: substitute x=1, y=5.80948 y'(1)=-9/4(5.80948)

  38. mathcalculus
    • one year ago
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    okay i got dy/dx=-.387298

  39. mathcalculus
    • one year ago
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    got it thanks @myko :))!

  40. myko
    • one year ago
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    yw

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