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anonymous

  • 5 years ago

Areas between graphs of 2 functions

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  1. anonymous
    • 5 years ago
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    y=x^2, x+y=2

  2. anonymous
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    Alright, I'll lead you through the steps, how about that? :)

  3. anonymous
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    1) Sketch a graph of both functions

  4. anonymous
    • 5 years ago
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    sounds fantastic! thank you! lol

  5. anonymous
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    np ^_^

  6. anonymous
    • 5 years ago
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    graph was given lol (way easier)

  7. anonymous
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    lol, okay!

  8. anonymous
    • 5 years ago
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    2) find the intersection points when y1= y2

  9. anonymous
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    set equal to zero?

  10. anonymous
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    yep, find the x's :)

  11. anonymous
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    got em...-2 and 1

  12. anonymous
    • 5 years ago
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    alright, good :)

  13. anonymous
    • 5 years ago
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    now, we must find out which function is bigger :) in other words, which graph is on the top of the other?

  14. anonymous
    • 5 years ago
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    I think y= x^2 is on top of the other one, right? :)

  15. anonymous
    • 5 years ago
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    yeah i remember that from class but it sort of looks like an upside down "V" so neither is on top. Do i use the right one because its a graph of y not x?

  16. anonymous
    • 5 years ago
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    exactly! you can see the intersection in the graph right? which curve is on the top?

  17. anonymous
    • 5 years ago
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    i think for y he said right-left and for x top-bottom

  18. anonymous
    • 5 years ago
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    let me sketch the graph lol just to make sure :)

  19. anonymous
    • 5 years ago
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    yep i see the intersection lol thanks. i would upload the pic but that would take longer than a sketch

  20. anonymous
    • 5 years ago
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    hold on :)

  21. anonymous
    • 5 years ago
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    okay got it, look at your sketch, color the region bounded (where the intersection lies/ closed region lol)

  22. anonymous
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    which line is on the top? :)

  23. anonymous
    • 5 years ago
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    in my sketch, I see y = 2-x on top :)

  24. anonymous
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    1 Attachment
  25. anonymous
    • 5 years ago
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    hmm, I got a different sketch , I've sovled for y for the second equation (y=2-x)

  26. anonymous
    • 5 years ago
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    try to sketch the graph using y = x^2 and y = 2-x :)

  27. anonymous
    • 5 years ago
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    A plot related to this problem is attached.

    1 Attachment
  28. anonymous
    • 5 years ago
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    yeah, that's what I got robto ^_^, birdsell are you following?

  29. anonymous
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    its the same graph its just the given one is a snippet

  30. anonymous
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    LOL, then draw your own to make your life easier :) , look at robto's sketch , which line is on the top ? ^_^

  31. anonymous
    • 5 years ago
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    intersection point of the 2 lines in Q1 is at 1,1

  32. anonymous
    • 5 years ago
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    are you sure? I got x = -2 and x = 1

  33. anonymous
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    so did i :-X but the picture shows it at 1,1

  34. anonymous
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    birdsell, dear :) , let's take it step by step 1) sketch both graphs, on your own , ignore the given :)

  35. anonymous
    • 5 years ago
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    you have y =x^2 and y = 2-x

  36. anonymous
    • 5 years ago
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    sketch them

  37. anonymous
    • 5 years ago
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    you'll have to get the same sketch as robto's ^_^

  38. anonymous
    • 5 years ago
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    im so confused :( most of them are clear that one is on top of the other

  39. anonymous
    • 5 years ago
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    calm down, did you draw the graph?

  40. anonymous
    • 5 years ago
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    and im attempting to use my calculator to graph the points but cannnt

  41. anonymous
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    don't! you can try out values, alright for example, sketch y = x^2 first, take values for x and find y, then plot them :) same with the other equation, do that

  42. anonymous
    • 5 years ago
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    after that, you'll get a graph similar to robto's

  43. anonymous
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    try :)

  44. anonymous
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    The second plot version uses the same formulas except the vertical and horizontal scaling is the same.

    1 Attachment
  45. anonymous
    • 5 years ago
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    so x^2 is on top?

  46. anonymous
    • 5 years ago
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    nope, look at the sketch again :)

  47. anonymous
    • 5 years ago
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    The curve passing through (0,0) is x^2.

  48. anonymous
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    look at the closed region,=)

  49. anonymous
    • 5 years ago
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    lmfao im dumb :)... got it

  50. anonymous
    • 5 years ago
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    no you're not, you panicked :D, alright so which line is on the top?

  51. anonymous
    • 5 years ago
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    y=2-x

  52. anonymous
    • 5 years ago
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    excellent, so this wil be you g(x) :) and the other curve is f(x) okay?

  53. anonymous
    • 5 years ago
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    got it

  54. anonymous
    • 5 years ago
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    I actually think I got it! lol wanna help with another?

  55. anonymous
    • 5 years ago
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    now to find the Area, the general formula is :\[A =\int\limits_{a}^{b}[g(x) - f(x)] dx\] where g(x) is the curve that's in the top and f(x) in the bottom, replace the equations and solve : \[A = \int\limits_{-2}^{1} (2-x) -(x^2) dx\] \[A = \int\limits_{-2}^{1}( 2-x-x^2)dx\] \[A = [2x-x^2/2 + x^3/3]\] from -2 to 1 so you'll get : \[A = [2(1)-(1)^2/2-(1)^3/3] - [2(-2) -(-2)^2/2 - (-2)^3/3]\] compute it and you'll get A lol :)

  56. anonymous
    • 5 years ago
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    sure , I'll be glad to help ^_^

  57. anonymous
    • 5 years ago
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    hanks! new thread or current?

  58. anonymous
    • 5 years ago
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    thanks**

  59. anonymous
    • 5 years ago
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    lol whatever you want :), np ^_^

  60. anonymous
    • 5 years ago
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    not sure if you are familiar with slicing?

  61. anonymous
    • 5 years ago
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    yes I am , it has to do with finding the volume right =)?

  62. anonymous
    • 5 years ago
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    yep! :):)

  63. anonymous
    • 5 years ago
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    I had a midterm on it last sunday lol!

  64. anonymous
    • 5 years ago
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    alright, which method does it want you to use? :)

  65. anonymous
    • 5 years ago
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    ughhh college... calc is killin me... up until now i have been great

  66. anonymous
    • 5 years ago
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    lol, calculus II right? practice, work hard, you'll perfect everything. Work hard now to relax in the end, believe me I sucked in this part =-= but now it's just SIMPLE :)

  67. anonymous
    • 5 years ago
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    Sorry to intterrupt, your question ask area of between x^2 , y+x=2 and x axis according to photo but the solution answers area of two function.I am confused

  68. anonymous
    • 5 years ago
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    you're finding the area of the 2 functions between the 2 curves Hakan, to do that you have to find the enclosed region :)

  69. anonymous
    • 5 years ago
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    actually calc 1. I go to a tough school and this is calc 1 carriculum

  70. anonymous
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    serious? I'm taking this in calculus II in college lol

  71. anonymous
    • 5 years ago
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    Ill start a new thread sstarcia

  72. anonymous
    • 5 years ago
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    well, I took the area b/w the curves in calculus I but the volume in calculus II

  73. anonymous
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    alright :)

  74. anonymous
    • 5 years ago
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    ahhh it isnt working lol

  75. anonymous
    • 5 years ago
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    how come?

  76. anonymous
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    anyway... The solid lies between planes perpendicular to the y axis at y=0 and y=2 the cross-sections perpendicular to the y-axis are circular discs with diameters running from the y-axis to the parabola \[x=\sqrt{5y^2}\]

  77. anonymous
    • 5 years ago
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    But the photo of question doesn't ask that as i see.Anyway.Thanx :)

  78. anonymous
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    lol we re-sketched it, look at robto's sketch :) np :)

  79. anonymous
    • 5 years ago
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    alright, to make your life easier bird, find y of this equation

  80. anonymous
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    \[y = x/\sqrt(5)\] I guess

  81. anonymous
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    i got y=x/5 lol

  82. anonymous
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    lol, take y^2 out and you'll get y, then divide by sqrt of 5 :)

  83. anonymous
    • 5 years ago
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    nvm i see now lol gotta take Sqrt of both top and bottom

  84. anonymous
    • 5 years ago
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    gotcha

  85. anonymous
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    okay :)

  86. anonymous
    • 5 years ago
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    now: 1) sketch the given in a graph, y=0, y=2 and y=x/sqrt(5)

  87. anonymous
    • 5 years ago
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    use your calculator for y = x/sqrt(5) you'll get something like a slight curve on the right, right?

  88. anonymous
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    it wont give me anything... when i type y=x/sqrt[5]

  89. anonymous
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    lol, forget your graphying calculator and do it by hand. Take values for x and calculate y using the calculator :)

  90. anonymous
    • 5 years ago
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    graphing*

  91. anonymous
    • 5 years ago
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    okay

  92. anonymous
    • 5 years ago
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    got it

  93. anonymous
    • 5 years ago
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    I think you'll get something like this , I think

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  94. anonymous
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    did you get something similar to it?

  95. anonymous
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    yep

  96. anonymous
    • 5 years ago
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    alright, now what did the question say, how does the curve rotate? about the y-axis, x-axis?

  97. anonymous
    • 5 years ago
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    find vol of disc that runs perpendicular top y axis w diameter running from x to the parabola

  98. anonymous
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    so you'll have to use the disk method :). So that means it's rotating about the y-axis right?

  99. anonymous
    • 5 years ago
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    so cut the curve perpendiculary to the y-axis ( draw a horizontal line in the bounded region I showed you )

  100. anonymous
    • 5 years ago
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    ughh im so sorry but i have to go to class now. thank you for all of your help :)

  101. anonymous
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    lol, sure :) good luck, and try to finish it ^_^

  102. anonymous
    • 5 years ago
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    thanks!

  103. anonymous
    • 5 years ago
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    :)

  104. anonymous
    • 5 years ago
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    :):) hiya ppl!!!!

  105. anonymous
    • 5 years ago
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    need any more help @birdsell ???

  106. anonymous
    • 5 years ago
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    lol, he's in class :)

  107. anonymous
    • 5 years ago
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    hey sstarica, couldn't type ther in grp chat ..lol...its stuck again !

  108. anonymous
    • 5 years ago
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    I don't like it either ways :) it's alright.

  109. anonymous
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    which grade are you? ^_^

  110. anonymous
    • 5 years ago
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    im gonna be engineer too :)

  111. anonymous
    • 5 years ago
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    which field? :)

  112. anonymous
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    software

  113. anonymous
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    nah, I'm in the hardware field =P

  114. anonymous
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    never mind..we can help each other:)

  115. anonymous
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    lol, yep. I'll deal with the hardware, you deal with the software, then we'll combine them together in the end

  116. anonymous
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    sure :):)

  117. anonymous
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    I find it rather weird in someway. I thought software = more girls work in that field and hardware = men, but now it's the opposite LOL

  118. anonymous
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    half of the guys I know are all wanting software and girls working in hardware.

  119. anonymous
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    lol..girls work on both like us:)

  120. anonymous
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    the majority of the boys in here are all hardware. It is very rare to see a girl working in the hardware field :)

  121. anonymous
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    they say it's a guy thing, but I disagree.

  122. anonymous
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    good...we can work on both :) right :)

  123. anonymous
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    yes ^_^, anyways, I wish you the best of luck :)

  124. anonymous
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    thank you !!! best of luck to u too...:) lets see wen we meet here again :)

  125. anonymous
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    np :) and thank you ^_^, good night ~

  126. anonymous
    • 5 years ago
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    good night star..:):)

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