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mathcalculus Group Title

hello, help with related rates problem. ._. (attached below)

  • 11 months ago
  • 11 months ago

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  1. Paynesdad Group Title
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    ??

    • 11 months ago
  2. mathcalculus Group Title
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    • 11 months ago
  3. mathcalculus Group Title
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    @Paynesdad It's posted ^_^

    • 11 months ago
  4. mathcalculus Group Title
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    am i able to use the chain rule for this problem?

    • 11 months ago
  5. Paynesdad Group Title
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    Ok so take the derivative with respect of the equation using implicit differentiation. Since both x and y are variables that depend on t you will have to use the chain rule.

    • 11 months ago
  6. mathcalculus Group Title
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    okay so i began doing this: DQ/DT= (2X)(5Y)

    • 11 months ago
  7. mebs Group Title
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    \[\frac{dQ}{dt} = -2 = 2x\frac{dx}{dt}5y^{4}\frac{dy}{dt}\]

    • 11 months ago
  8. mathcalculus Group Title
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    got it.

    • 11 months ago
  9. mathcalculus Group Title
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    im getting there.

    • 11 months ago
  10. Paynesdad Group Title
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    no you can't do the derivative of x and y and the same time. \[\frac{ dQ }{ dt }= \frac{ d(x^2y^5) }{ dt }\]First take the derivative of the right side with respect to t of the x factor and treat y factor as a constant + the derivative of the right side with respect to t of the y factor treating the x factor as a constant.

    • 11 months ago
  11. mebs Group Title
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    we know that \[\frac{dx}{dt} = -2\] so when now find what \[\frac{dy}{dt} , at, x = 3, y = 1\]

    • 11 months ago
  12. mebs Group Title
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    So solve this \[-2 = 2(3)(-2)5(1)^{4}\frac{dy}{dt}\]

    • 11 months ago
  13. mebs Group Title
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    @Paynesdad I think this is implicit because they wanted it in terms of Q.

    • 11 months ago
  14. Paynesdad Group Title
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    Seems like you answered it mebs...not helped with it.

    • 11 months ago
  15. mebs Group Title
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    @Paynesdad I think your right....my apologies I get exited when I do math.

    • 11 months ago
  16. Paynesdad Group Title
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    LOL I know the feeling! I am glad I am not the only one.

    • 11 months ago
  17. mebs Group Title
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    Well @mathcalculus does this make sense to you.. I just subbed values that were given to me already from the question into the derivative of Q. and x and y were already defined.

    • 11 months ago
  18. mathcalculus Group Title
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    sorry just doing the math

    • 11 months ago
  19. mathcalculus Group Title
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    yes i see what you mean.

    • 11 months ago
  20. mathcalculus Group Title
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    im not up to the answer yet but close, i just need to check

    • 11 months ago
  21. mathcalculus Group Title
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    i got 48.96.... :? but it's not right..

    • 11 months ago
  22. mathcalculus Group Title
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    would using the chain rule be wrong?

    • 11 months ago
  23. mathcalculus Group Title
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    wait, which one's?

    • 11 months ago
  24. mebs Group Title
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    All of them

    • 11 months ago
  25. mathcalculus Group Title
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    • 11 months ago
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  26. mebs Group Title
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    You can't do what you just did in your 2nd and third step

    • 11 months ago
  27. mathcalculus Group Title
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    oh no, :? why not?

    • 11 months ago
  28. Paynesdad Group Title
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    Whoa whoa whoa.. Hang on here.

    • 11 months ago
  29. mathcalculus Group Title
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    ?

    • 11 months ago
  30. mebs Group Title
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    To use the product rule you do one at a time.\[\frac{dQ}{dt} = 2x\frac{dx}{dt}y^{5} + x^{2}5y^{4}\frac{dy}{dt}\]

    • 11 months ago
  31. Paynesdad Group Title
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    Right ^

    • 11 months ago
  32. mathcalculus Group Title
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    oh okay, i understand that.

    • 11 months ago
  33. mathcalculus Group Title
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    i used it twice before.

    • 11 months ago
  34. Paynesdad Group Title
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    Remember d(xy)/dt = y (dx/dt) + x (dy/dt). Also mathcalculus when you took the derivative you left off the dx/dt and dy/dt out of your equations.

    • 11 months ago
  35. mebs Group Title
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    You already know that \[\frac{dQ}{dt}= -2 \] You know that \[\frac{dx}{dt} = -2\] and you know that \[x = 3 , y = 1\] just substitute it into the derivative of dQ/dt and solve.

    • 11 months ago
  36. mathcalculus Group Title
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    yes

    • 11 months ago
  37. Paynesdad Group Title
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    So using mebs equation you should get...a fractional answer.

    • 11 months ago
  38. Paynesdad Group Title
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    What did you get?

    • 11 months ago
  39. Paynesdad Group Title
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    @mathcalculus

    • 11 months ago
  40. mathcalculus Group Title
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    gotta start over, i'll keep you posted.

    • 11 months ago
  41. Paynesdad Group Title
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    Ok... let us know.

    • 11 months ago
  42. mathcalculus Group Title
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    okay, jus wondering. so am i able to substitute all of them right away after i found the derivative?

    • 11 months ago
  43. mathcalculus Group Title
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    damn it. wrong again. i got -2/33 this time

    • 11 months ago
  44. mathcalculus Group Title
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    @mebs

    • 11 months ago
  45. mathcalculus Group Title
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    @Paynesdad

    • 11 months ago
  46. Paynesdad Group Title
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    Show us what you did again if you can.

    • 11 months ago
  47. mathcalculus Group Title
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    ok. 1 sec

    • 11 months ago
  48. mathcalculus Group Title
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    • 11 months ago
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  49. Paynesdad Group Title
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    Got it...Order of operations!

    • 11 months ago
  50. mathcalculus Group Title
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    ?

    • 11 months ago
  51. mathcalculus Group Title
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    where did i miss?

    • 11 months ago
  52. mebs Group Title
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    Let me show you what must be done. \[-2 = 2\times 3 \times (-2) \times (1)^{5} + (3)^{2} \times 5 \times 1^{4} \frac{dy}{dx}\]

    • 11 months ago
  53. mathcalculus Group Title
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    yes that i got.

    • 11 months ago
  54. mathcalculus Group Title
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    I used PEMDAS

    • 11 months ago
  55. mebs Group Title
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    You can't add them they both don't have dy/dx dude.

    • 11 months ago
  56. Paynesdad Group Title
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    you have -12 + 45*dy/dt=-2 ...You can't add -12 to 45 until you multiply 45 times dy/dt...It's like 4=-3+7x....you don't simplify that as 4 = 4x ... right

    • 11 months ago
  57. mebs Group Title
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    you get \[-2 = -12 + 45\frac{dy}{dx}\]

    • 11 months ago
  58. mathcalculus Group Title
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    ohhhhh 45 is part of dy/dx

    • 11 months ago
  59. mebs Group Title
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    yea bro, step It up man dude.

    • 11 months ago
  60. mathcalculus Group Title
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    oops.

    • 11 months ago
  61. Paynesdad Group Title
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    lol yep... it is a coefficient...come on bro!

    • 11 months ago
  62. mathcalculus Group Title
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    i wasn't looking closely.

    • 11 months ago
  63. mathcalculus Group Title
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    p.s not a bro lol but thanks! ^_^

    • 11 months ago
  64. mebs Group Title
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    Its fine I was jking haha.hhaa

    • 11 months ago
  65. mebs Group Title
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    I can tell that by your handwriting haha.haha

    • 11 months ago
  66. mathcalculus Group Title
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    kay, well thanks guys!!! appreciate it tons tons.

    • 11 months ago
  67. mebs Group Title
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    Good question @mathcalculus keep up the hard work.

    • 11 months ago
  68. Paynesdad Group Title
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    @mebs you crack me up

    • 11 months ago
  69. Paynesdad Group Title
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    Keep it up @mathcalculus

    • 11 months ago
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