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anonymous

  • one year ago

what is the slope of the tangent line to the curve (x^3 + y)^2 -2a^2xy = b^2 at the point P(0,1) when a=(2)^(1/2) and b = 1

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  1. anonymous
    • one year ago
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    @ganeshie8 @Robert136 @nincompoop

  2. anonymous
    • one year ago
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    I'm getting DNE but its wrong

  3. Jhannybean
    • one year ago
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    For this problem can you input the values for a and b, then find the derivative? Or does the inputting of values happen after differentiating..

  4. Jhannybean
    • one year ago
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    It looks really ugly if you differentiate it along with a and b :\

  5. Jhannybean
    • one year ago
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    I would use implicit differentiation to find the derivative.

  6. anonymous
    • one year ago
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    i really have no idea , iv imputed the values for a and be and then solved for y and derived that , but that seems to be wrong

  7. Jhannybean
    • one year ago
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    Yeah it is hahahahaha

  8. Jhannybean
    • one year ago
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    Because that's basically stating the tangent line is found at a point before even being found.

  9. anonymous
    • one year ago
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    you only need to find f'(x,y)

  10. anonymous
    • one year ago
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    but before u start it says when a=(2)^(1/2) and b = 1 , a and b seems like small values and you can sub them why it mention them in ur question ? seems like there is a formula in ur book or something

  11. IrishBoy123
    • one year ago
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    \[\vec n = <2(x^3+y)3x^2 - 2a^2y, 2(x^3+y)-2a^2x>\] \[= <6x^2(x^3+y) - 2a^2y, 2(x^3+y)-2a^2x>\] x = 0 \[= <-2a^2y,2>\] \[y = 1, a = \sqrt{2}\] \[\vec n = <-4,2,>\]

  12. freckles
    • one year ago
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    don't solve for y just find the derivative of both sides w.r.t. x a and b are constants you can input those values in later or not once you have differentiated both sides you can input the values for a and b if you haven't already and then also input 0 for x and 1 for y then solve for y' to find the slope.

  13. anonymous
    • one year ago
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    see thats what i'm talking about :D like @irishboy123

  14. IrishBoy123
    • one year ago
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    |dw:1441778557798:dw|

  15. anonymous
    • one year ago
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    just noticed the name of the chapter is implicit differention

  16. freckles
    • one year ago
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    why is there y'=that mess=0?

  17. freckles
    • one year ago
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    wait I just had a problem with y'= part

  18. anonymous
    • one year ago
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    well the question said (tangent line) so it's line equation u can do it in both ways with irish normal vector or with slope and a point

  19. freckles
    • one year ago
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    \[2(x^3+y)(3x^2+y')-(2a^2xy'+2a^2y)=0 \text{ was right }\]

  20. Jhannybean
    • one year ago
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    Thanks.

  21. freckles
    • one year ago
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    I just didn't agree with the y'=that stuff=0

  22. freckles
    • one year ago
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    because saying y'=0 is saying you already know the slope to be 0

  23. freckles
    • one year ago
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    \[2(x^3+y)(3x^2+y')-(2a^2xy'+2a^2y)=0 \text{ was right }\] so copying this @Jdosio just enter in 0 for x and 1 for y simplify and solve for y'

  24. anonymous
    • one year ago
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    okay

  25. anonymous
    • one year ago
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    a^2 ?

  26. freckles
    • one year ago
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    yes and a was?

  27. anonymous
    • one year ago
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    a = 2^(1/2) so final answer is 2

  28. freckles
    • one year ago
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    right but hey listen do you understand how she got that equation?

  29. anonymous
    • one year ago
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    not a single clue ^_^

  30. freckles
    • one year ago
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    she used chain rule and product rule (there are some other small potato rules she used)

  31. freckles
    • one year ago
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    \[\frac{d}{dx}(x^3+y)^2 \] for this you should use chain rule

  32. Jhannybean
    • one year ago
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    \[ (x^3 + y)^2 -2a^2xy = b^2\]\[2(x^3+y)(3x^2+y')-(2a^2xy'+2a^2y)=0 \]\[6x^5+2x^3y'+6x^3y+2yy'-2a^2xy'-2a^2y=0\]\[y'(2x^3+2y-2a^2x)=-6x^5 -6x^3y+2a^2y\]\[y'=...\]

  33. freckles
    • one year ago
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    \[\frac{d}{dx}(x^3+y)^2 =2(x^3+y) \cdot (x^3+y)'\] derivative of outside times the derivative of inside

  34. freckles
    • one year ago
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    \[\frac{d}{dx}(x^3+y)=3x^2+y' \\ \text{ so } \frac{d}{dx}(x^3+y)^2=2(x^3+y) \cdot (3x^2+y')\]

  35. Jhannybean
    • one year ago
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    chain rule: \(f(g(x)) = f'(x) \cdot g'(x)\)

  36. anonymous
    • one year ago
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    oh okay i see i see

  37. anonymous
    • one year ago
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    i was just rather lost on the second part of the subtraction where you used the product rule , im familiar with it i just wasn't entirely sure how to go about the a but i see what you did

  38. freckles
    • one year ago
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    the other term you can use constant multiple rule and product rule \[\frac{d}{dx}-2a^2xy \\ -2a^2 \frac{d}{dx}(xy) \text{ by constant multiple rule } \\ \text{ do you think you can use product rule \to find } \frac{d}{dx}(xy)\]

  39. anonymous
    • one year ago
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    yea x' y + x y'

  40. Jhannybean
    • one year ago
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    You're trying to keep in mind that taking the derivative of \(y\) will give you \(\dfrac{dy}{dx}\) and that you are taking the derivative of \(x\) as well.

  41. freckles
    • one year ago
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    that is correct but x'=?

  42. freckles
    • one year ago
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    or dx/dx=?

  43. anonymous
    • one year ago
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    wait , are you asking about the example or the actual question?

  44. freckles
    • one year ago
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    well you said \[\frac{d}{dx}(xy)=\frac{dx}{dx} y+x \frac{dy}{dx}\] and I was asking you to simplify dx/dx

  45. anonymous
    • one year ago
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    well 1 right?

  46. freckles
    • one year ago
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    yep

  47. freckles
    • one year ago
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    \frac{d}{dx}-2a^2xy \\ -2a^2 \frac{d}{dx}(xy) \text{ by constant multiple rule } \\ \text{ do you think you can use product rule \to find } \frac{d}{dx}(xy) right \[\frac{d}{dx}(xy)=y+xy' \\ \text{ so } \frac{d}{dx}-2a^2xy=-2a^2(y+xy') =-2a^2y-2a^2xy'\]

  48. freckles
    • one year ago
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    and of course the other side of the equation is b^2 so when differentiating a constant you get 0

  49. freckles
    • one year ago
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    @Jdosio do you have any questions on this?

  50. anonymous
    • one year ago
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    not on this question specifically , if i come up with anything else i missed ill ask you guys if your still around. thanks for all your help once again.

  51. Jhannybean
    • one year ago
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    No problem.

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