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anonymous

  • 5 years ago

The mechanics at Lincoln Automotive are reboring a 6-in deep cylinder to fit a new piston. The machine they are using increases the cylinders radius one-thousandth of an inch every 4min. How rapidly is the cylinder volume increasing when the bore (diameter) is 3.800in

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  1. anonymous
    • 5 years ago
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    okay, you can write it down like this in every 4 mins ---> radius increase = 1000in in every x mins? ----> radius increase R= D/2 =1.504 after finding the answer in mins, convert to seconds :) give it a try now :)

  2. anonymous
    • 5 years ago
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    cross multiply

  3. anonymous
    • 5 years ago
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    wait, I misread the question

  4. anonymous
    • 5 years ago
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    they want how much is the volume increasing right?

  5. anonymous
    • 5 years ago
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    Its okay :)

  6. anonymous
    • 5 years ago
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    at what rate is the volume increasing, supposed to use derivitives (this is calculus)

  7. anonymous
    • 5 years ago
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    then find the derivative of Volumeof Cylinder

  8. anonymous
    • 5 years ago
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    you have r' = 1000

  9. anonymous
    • 5 years ago
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    all you have to do is find h from the original equation which is Vol = 1/3 pi r^2 h using r = 1.504

  10. anonymous
    • 5 years ago
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    \[Vol = 1/3 \pi r^2h , Vol' = 2/3 \pi rr'h'\] I guess

  11. anonymous
    • 5 years ago
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    you have r' = 1000 r= 1.504 you'll have to find h'

  12. anonymous
    • 5 years ago
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    am I making any sense lol?

  13. anonymous
    • 5 years ago
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    Not really...

  14. anonymous
    • 5 years ago
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    How did you get r' ?

  15. anonymous
    • 5 years ago
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    jmcwhs2012, if you allow, may I say a few words

  16. anonymous
    • 5 years ago
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    it's given, it says that the radius is increasing 1000 in every 4 mins so r' = 1000in/min

  17. anonymous
    • 5 years ago
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    iam please go ahead :)

  18. anonymous
    • 5 years ago
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    sure:)

  19. anonymous
    • 5 years ago
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    "one-thousandth of an inch"

  20. anonymous
    • 5 years ago
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    Okay, as far as I understand, the rate at which the radius is increasing is equal to the radius bored in a second

  21. anonymous
    • 5 years ago
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    you have the original radius, you can computer r' I guess it's 1000(1.504)

  22. anonymous
    • 5 years ago
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    So we can write dr/dt= (2.54/100000)/(60*4)

  23. anonymous
    • 5 years ago
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    compute*

  24. anonymous
    • 5 years ago
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    Now all we need to find is dv/dt

  25. anonymous
    • 5 years ago
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    I dont understand where you got dr/dt?

  26. anonymous
    • 5 years ago
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    So we can use the result (dr/dt)x(dv/dr)

  27. anonymous
    • 5 years ago
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    (dr/dt)x(dv/dr)=dv/dt

  28. anonymous
    • 5 years ago
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    @jmcwhs2012 the thing is coming like this 2.54 cm is equal to one inch,

  29. anonymous
    • 5 years ago
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    Why do we have to use cm?

  30. anonymous
    • 5 years ago
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    I later convert it into m

  31. anonymous
    • 5 years ago
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    You see for that I divide it by 100000

  32. anonymous
    • 5 years ago
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    Ok let me explain the whole thing

  33. anonymous
    • 5 years ago
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    The machine is working at the rate of 1/1000 inch per 60*4 secs

  34. anonymous
    • 5 years ago
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    is that clear?

  35. anonymous
    • 5 years ago
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    wait @iam , he wants to find V'

  36. anonymous
    • 5 years ago
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    Thats what I am doing, if you all allow me

  37. anonymous
    • 5 years ago
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    lol, sorry, proceed :)

  38. anonymous
    • 5 years ago
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    The machine is working at the rate of 1/1000 inch per 60*4 secs, is that thing clear?

  39. anonymous
    • 5 years ago
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    yes, got you.

  40. anonymous
    • 5 years ago
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    Which means 2.54/100000 m per 240 secs

  41. anonymous
    • 5 years ago
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    So now we can find dr/dt from there

  42. anonymous
    • 5 years ago
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    okay

  43. anonymous
    • 5 years ago
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    But we need to find dv/dt

  44. anonymous
    • 5 years ago
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    so, let's find it

  45. anonymous
    • 5 years ago
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    So we can use the formula dv/dt=(dr/dt)x(dv/dr)

  46. anonymous
    • 5 years ago
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    Is this step clear

  47. anonymous
    • 5 years ago
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    honestly, no, why are you multiplying them?

  48. anonymous
    • 5 years ago
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    wait! chain rule?

  49. anonymous
    • 5 years ago
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    Write that thing on a piece of paper, and try to cancel the commong terms, you will get why I am doing that

  50. anonymous
    • 5 years ago
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    Yes chain rule

  51. anonymous
    • 5 years ago
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    lol, got you! okay then what?

  52. anonymous
    • 5 years ago
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    So we have already found dr/dt which is 2.54/100000/240

  53. anonymous
    • 5 years ago
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    Now we need to find dv/dr

  54. anonymous
    • 5 years ago
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    alright

  55. anonymous
    • 5 years ago
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    For that we will have to differenciate 6 x pi x r^2 with respect to r

  56. anonymous
    • 5 years ago
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    why 6 pi r^2?

  57. anonymous
    • 5 years ago
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    Oops I am sorry, it would be 6*2.54*pi*r^2

  58. anonymous
    • 5 years ago
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    6 is the length of the cylinder given on the problem

  59. anonymous
    • 5 years ago
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    I don't get where you got 6?

  60. anonymous
    • 5 years ago
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    oh, the height

  61. anonymous
    • 5 years ago
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    yes

  62. anonymous
    • 5 years ago
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    So now just we will have to multiply both the things

  63. anonymous
    • 5 years ago
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    but isn't the vol = 1/3 pi r^2 h?

  64. anonymous
    • 5 years ago
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    Thats for a cone

  65. anonymous
    • 5 years ago
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    which will be 2 pi r^2?

  66. anonymous
    • 5 years ago
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    lol...sorry ^^"

  67. anonymous
    • 5 years ago
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    So we just need to multiply both the things

  68. anonymous
    • 5 years ago
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    then differentiate with respect to r :)

  69. anonymous
    • 5 years ago
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    But unfortunately the person who gave this problem is not much interested

  70. anonymous
    • 5 years ago
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    No no no need to differenciate again

  71. anonymous
    • 5 years ago
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    lol, he'll come back later :)

  72. anonymous
    • 5 years ago
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    Its over

  73. anonymous
    • 5 years ago
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    so V = 15.34 pi r^2?

  74. anonymous
    • 5 years ago
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    yes

  75. anonymous
    • 5 years ago
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    what is r? 1.504?

  76. anonymous
    • 5 years ago
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    I didn't get you

  77. anonymous
    • 5 years ago
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    r is a variable

  78. anonymous
    • 5 years ago
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    wouldn't you derive to V' then substitute r = 1.504 and r' in the equation and then u are done?

  79. anonymous
    • 5 years ago
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    No not 1.504, we will have to substitute 1.900

  80. anonymous
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    then by then you have found V' which is the volume rate

  81. anonymous
    • 5 years ago
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    I didn't get you

  82. anonymous
    • 5 years ago
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    alright, 1.900 and r' the one you have found which is dr/dt, then you're done

  83. anonymous
    • 5 years ago
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    right? :)

  84. anonymous
    • 5 years ago
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    Actually I am not getting you at all

  85. anonymous
    • 5 years ago
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    {(2.54/100000)/240}*2.54*6*pi*2*1.900=dv/dt Thats it and thats all

  86. anonymous
    • 5 years ago
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    so the given is: r = 1.900 r' = dr/dt = 2.54 so V' = 15.34(2) pi rr' = 30.68 pi (1.900)(2.54) = 465.15 m / secs So the volume is increasing = 465.15m/sec right?

  87. anonymous
    • 5 years ago
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    simply ^_^

  88. anonymous
    • 5 years ago
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    dr/dt = 2.54 this is not right

  89. anonymous
    • 5 years ago
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    so then it's 2.54 /1000/240?

  90. anonymous
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    = 1.05 x10^-5

  91. anonymous
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    dr/dt ={(2.54/100000)/240} dv/dr= 2.54*6*pi*2*1.900

  92. anonymous
    • 5 years ago
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    so: V' = dv/dr = 2.54*12 pi *1.900 = 181.94 m/secs

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