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anonymous

  • 5 years ago

Find all critical points of R(t)=3t+77-15ln(t^2-3)

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  1. amistre64
    • 5 years ago
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    R' = 3 -(30t)/(t^2-3) i think

  2. amistre64
    • 5 years ago
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    3t^2 -30t -9 ------------- t^2- 3

  3. amistre64
    • 5 years ago
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    3(t^2 -10 -3) would be the top if I did it right....

  4. amistre64
    • 5 years ago
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    -10t....

  5. anonymous
    • 5 years ago
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    I Got R'=3-15/(t^2-3)(2t)

  6. amistre64
    • 5 years ago
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    yep, as long as thats just screen clutter and not a glaring error :)

  7. amistre64
    • 5 years ago
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    add the parts like fractions, 3(t^2 -3) -30t ------------ t^2 -3

  8. anonymous
    • 5 years ago
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    Sorry how did you get 3(t^2-3) on top?

  9. amistre64
    • 5 years ago
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    the top becomes a quadratic.... a fancy name for "highest power is a 2"

  10. amistre64
    • 5 years ago
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    ok... you pretty much have the form: 30 3 - ---- now how do we add fractions togther? b

  11. anonymous
    • 5 years ago
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    You have to have a common denominator?

  12. anonymous
    • 5 years ago
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    But if R'(t)=3- 15 (2t) ----- (t^2 -3)

  13. anonymous
    • 5 years ago
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    That would be 3 - 30t ------- (t^2 - 3)

  14. amistre64
    • 5 years ago
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    30t 3 - ------- is what we got right? (t^2 -3)

  15. anonymous
    • 5 years ago
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    That's what I got

  16. amistre64
    • 5 years ago
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    then to add the "fractions" mulitply the left part by: 3 t^2-3 -- x ----- right? 1 t^2-3

  17. anonymous
    • 5 years ago
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    Oh! Okay

  18. amistre64
    • 5 years ago
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    :) deriving...no problem.... fractions? ehhhh :)

  19. amistre64
    • 5 years ago
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    3t^2 -30t -9 ----------- t^2 -3

  20. amistre64
    • 5 years ago
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    quadratic formula for the top to find the critical points...

  21. amistre64
    • 5 years ago
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    5 +- [sqrt(900 - (4)(3)(-9))]/6

  22. anonymous
    • 5 years ago
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    Where is the 5 from?

  23. amistre64
    • 5 years ago
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    5 +- [sqrt(1008)]/6

  24. amistre64
    • 5 years ago
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    30/6 -b/2a

  25. amistre64
    • 5 years ago
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    you familiar with the quadratic formula?

  26. anonymous
    • 5 years ago
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    yes

  27. amistre64
    • 5 years ago
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    good heres what I get: t = 5 +- 2sqrt(63)/3

  28. anonymous
    • 5 years ago
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    Did you do quadratic formula of 3t^2-30t-9?

  29. amistre64
    • 5 years ago
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    yep..... as long as I kept track of myself

  30. amistre64
    • 5 years ago
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    30 sqrt(900-(4)(3)(-9)) --- +- -------------- 6 6

  31. anonymous
    • 5 years ago
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    So 30 +- sqrt(900+108) ---------------------- 6

  32. amistre64
    • 5 years ago
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    we can simplify that by factoring out a 3 first.... sneaky little bugger it is :)

  33. amistre64
    • 5 years ago
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    3(t^2 -10t -3) now quad the left side

  34. amistre64
    • 5 years ago
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    10 +- sqrt(100 -(4)(-3)(1)) ----------------------- 2

  35. amistre64
    • 5 years ago
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    10 +- sqrt(112) --- -------- same result, just easier to see 2 2

  36. amistre64
    • 5 years ago
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    112 = 16*7 4sqrt(7)

  37. anonymous
    • 5 years ago
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    So 5 +- 2sqrt(7)

  38. amistre64
    • 5 years ago
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    thats better :)

  39. amistre64
    • 5 years ago
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    you know how to test for min and max?

  40. anonymous
    • 5 years ago
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    So those are the critical points? The neg and pos value from that?

  41. anonymous
    • 5 years ago
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    No

  42. amistre64
    • 5 years ago
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    yep... those are the critical points. to see how they are behaving, you take a second derivative....derive the derivative.

  43. anonymous
    • 5 years ago
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    I don't think I have to do that.

  44. amistre64
    • 5 years ago
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    you sure? its fun :) if you have to determine if they are MIN MAX or inflections, then we would, but if you just gotta find criticals, thats it.... dont forget to include the end points of the graph as well, they might be important

  45. anonymous
    • 5 years ago
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    Wow, thanks! Can you help me with one more?

  46. amistre64
    • 5 years ago
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    i dunno....... i feel a bout of stupidy brewing amongst me brain cells :) ........sure why not :)

  47. anonymous
    • 5 years ago
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    Find all critical points of P(t)=-cos(4t)+39-2t

  48. anonymous
    • 5 years ago
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    Holy cow sorry my computer spazzed out

  49. amistre64
    • 5 years ago
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    thats just the cosine function moved about.... should be rather simple to determine

  50. anonymous
    • 5 years ago
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    I have found P'(t) if that helps

  51. amistre64
    • 5 years ago
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    P' = 4sin(4t) - 2 right?

  52. anonymous
    • 5 years ago
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    Yes that's what I got

  53. amistre64
    • 5 years ago
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    good, then we are either both genuises or idiots :) Ill go with genuisesss

  54. anonymous
    • 5 years ago
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    (=

  55. amistre64
    • 5 years ago
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    0 = 4sins(4t) -2 solve for sin(4t)

  56. anonymous
    • 5 years ago
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    sin(4t)=1/2

  57. amistre64
    • 5 years ago
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    what does 4t have to equal for sin to be 1/2?

  58. amistre64
    • 5 years ago
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    there are 2 angles, that will satisfy this

  59. anonymous
    • 5 years ago
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    I have no idea.

  60. amistre64
    • 5 years ago
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    what? you gotta have SOME idea.... its just a circle :)

  61. amistre64
    • 5 years ago
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    sin(30) = 1/2 and sin(150) = 1/2

  62. anonymous
    • 5 years ago
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    How do I figure the two angles that will satisfy that?

  63. amistre64
    • 5 years ago
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    you can either remember back to trig that a 30-60-90 triangle has sides of 1-2-sqrt(3) and interpolate from that, OR use the sin inverse function on a calculator

  64. anonymous
    • 5 years ago
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    so sin^-1(1/2)=4t

  65. amistre64
    • 5 years ago
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    yep

  66. anonymous
    • 5 years ago
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    what from that tells me that 30 and 150 are what I'm looking for?

  67. anonymous
    • 5 years ago
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    Sorry I want to understand

  68. amistre64
    • 5 years ago
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    sin^-1 will only give you one angle tho...30. You have to remember to cross over the yaxis to get to the other one...150

  69. amistre64
    • 5 years ago
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    4t = 30 and 4t = 150 solve for t :)

  70. amistre64
    • 5 years ago
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    might be better to use radians instead of degrees, half dozen of one or 6 of the other

  71. anonymous
    • 5 years ago
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    Oh! Okay thanks. Moving on

  72. amistre64
    • 5 years ago
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    4t = pi/6 4t = 5pi/6

  73. anonymous
    • 5 years ago
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    whoa where did that come from?

  74. amistre64
    • 5 years ago
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    what!!! you see a bee??

  75. anonymous
    • 5 years ago
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    No I mean the 4t=pi/6 and 4t=5pi/6

  76. anonymous
    • 5 years ago
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    Wait is that 30 degrees and 150 degrees in radians?

  77. amistre64
    • 5 years ago
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    :) there are a few ways to express angles. you can use degrees: 30 and 150 are degrees. OR you can use "radians" which measures the angle using the radius itself as a measuring stick. a circle is 360 degrees around OR 2pi radians pi/6 = 30 degrees 5pi/6 = 150 degrees

  78. anonymous
    • 5 years ago
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    Okay

  79. amistre64
    • 5 years ago
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    To get a number for "t" we should use radians :)

  80. anonymous
    • 5 years ago
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    Okay

  81. amistre64
    • 5 years ago
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    t = pi/24 t = 5pi/24 is what I come up with

  82. amistre64
    • 5 years ago
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    30/4 or 150/4 depending on what your answers are looking for

  83. anonymous
    • 5 years ago
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    so t=.1309 + pin/2

  84. amistre64
    • 5 years ago
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    t = pi/24 = .1309 or t = 5pi/24 = .6545

  85. amistre64
    • 5 years ago
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    for degrees... t = 7.5 degrees or t = 37.5 degrees

  86. anonymous
    • 5 years ago
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    Okay. I have a question.

  87. amistre64
    • 5 years ago
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    go ahead and ask :)

  88. anonymous
    • 5 years ago
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    usually before solving for t my professor asks "which quadrants is sin negative in" why did we not have to look for that in this problem?

  89. amistre64
    • 5 years ago
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    sin(angle) = 1/2 sin is only positive in Q1 and Q2 .... no need to look any where else, unless you can think of a reason to..

  90. anonymous
    • 5 years ago
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    Okay. That makes sense. Thank you sooo much!

  91. amistre64
    • 5 years ago
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    when sin = -1/2, it puts a different value into the equation that is not a critical point

  92. amistre64
    • 5 years ago
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    you are soooo welcome :)

  93. amistre64
    • 5 years ago
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    if you have any more questions, just start a new posting :)

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