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mathcalculus

  • one year ago

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

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  1. Paynesdad
    • one year ago
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    ??

  2. mathcalculus
    • one year ago
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  3. mathcalculus
    • one year ago
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    @Paynesdad It's posted ^_^

  4. mathcalculus
    • one year ago
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    am i able to use the chain rule for this problem?

  5. Paynesdad
    • one year ago
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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.

  6. mathcalculus
    • one year ago
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    okay so i began doing this: DQ/DT= (2X)(5Y)

  7. mebs
    • one year ago
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    \[\frac{dQ}{dt} = -2 = 2x\frac{dx}{dt}5y^{4}\frac{dy}{dt}\]

  8. mathcalculus
    • one year ago
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    got it.

  9. mathcalculus
    • one year ago
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    im getting there.

  10. Paynesdad
    • one year ago
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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. mebs
    • one year ago
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    we know that \[\frac{dx}{dt} = -2\] so when now find what \[\frac{dy}{dt} , at, x = 3, y = 1\]

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

  13. mebs
    • one year ago
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    @Paynesdad I think this is implicit because they wanted it in terms of Q.

  14. Paynesdad
    • one year ago
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    Seems like you answered it mebs...not helped with it.

  15. mebs
    • one year ago
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    @Paynesdad I think your right....my apologies I get exited when I do math.

  16. Paynesdad
    • one year ago
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    LOL I know the feeling! I am glad I am not the only one.

  17. mebs
    • one year ago
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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.

  18. mathcalculus
    • one year ago
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    sorry just doing the math

  19. mathcalculus
    • one year ago
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    yes i see what you mean.

  20. mathcalculus
    • one year ago
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    im not up to the answer yet but close, i just need to check

  21. mathcalculus
    • one year ago
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    i got 48.96.... :? but it's not right..

  22. mathcalculus
    • one year ago
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    would using the chain rule be wrong?

  23. mathcalculus
    • one year ago
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    wait, which one's?

  24. mebs
    • one year ago
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    All of them

  25. mathcalculus
    • one year ago
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  26. mebs
    • one year ago
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    You can't do what you just did in your 2nd and third step

  27. mathcalculus
    • one year ago
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    oh no, :? why not?

  28. Paynesdad
    • one year ago
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    Whoa whoa whoa.. Hang on here.

  29. mathcalculus
    • one year ago
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    ?

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

  31. Paynesdad
    • one year ago
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    Right ^

  32. mathcalculus
    • one year ago
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    oh okay, i understand that.

  33. mathcalculus
    • one year ago
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    i used it twice before.

  34. Paynesdad
    • one year ago
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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.

  35. mebs
    • one year ago
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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.

  36. mathcalculus
    • one year ago
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    yes

  37. Paynesdad
    • one year ago
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    So using mebs equation you should get...a fractional answer.

  38. Paynesdad
    • one year ago
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    What did you get?

  39. Paynesdad
    • one year ago
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    @mathcalculus

  40. mathcalculus
    • one year ago
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    gotta start over, i'll keep you posted.

  41. Paynesdad
    • one year ago
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    Ok... let us know.

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

  43. mathcalculus
    • one year ago
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    damn it. wrong again. i got -2/33 this time

  44. mathcalculus
    • one year ago
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    @mebs

  45. mathcalculus
    • one year ago
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    @Paynesdad

  46. Paynesdad
    • one year ago
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    Show us what you did again if you can.

  47. mathcalculus
    • one year ago
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    ok. 1 sec

  48. mathcalculus
    • one year ago
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  49. Paynesdad
    • one year ago
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    Got it...Order of operations!

  50. mathcalculus
    • one year ago
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    ?

  51. mathcalculus
    • one year ago
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    where did i miss?

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

  53. mathcalculus
    • one year ago
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    yes that i got.

  54. mathcalculus
    • one year ago
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    I used PEMDAS

  55. mebs
    • one year ago
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    You can't add them they both don't have dy/dx dude.

  56. Paynesdad
    • one year ago
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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

  57. mebs
    • one year ago
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    you get \[-2 = -12 + 45\frac{dy}{dx}\]

  58. mathcalculus
    • one year ago
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    ohhhhh 45 is part of dy/dx

  59. mebs
    • one year ago
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    yea bro, step It up man dude.

  60. mathcalculus
    • one year ago
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    oops.

  61. Paynesdad
    • one year ago
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    lol yep... it is a coefficient...come on bro!

  62. mathcalculus
    • one year ago
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    i wasn't looking closely.

  63. mathcalculus
    • one year ago
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    p.s not a bro lol but thanks! ^_^

  64. mebs
    • one year ago
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    Its fine I was jking haha.hhaa

  65. mebs
    • one year ago
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    I can tell that by your handwriting haha.haha

  66. mathcalculus
    • one year ago
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    kay, well thanks guys!!! appreciate it tons tons.

  67. mebs
    • one year ago
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    Good question @mathcalculus keep up the hard work.

  68. Paynesdad
    • one year ago
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    @mebs you crack me up

  69. Paynesdad
    • one year ago
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    Keep it up @mathcalculus

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