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anonymous

  • 5 years ago

When solving for the area bounded by curves, I often have trouble determing which function goes first. Can somebody help me with that?

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  1. anonymous
    • 5 years ago
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    It depends on which variable you're integrating, but generally you are solving the integral of the upper function minus the integral of the lower function.

  2. anonymous
    • 5 years ago
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    howabout x^2 + 2 AND x+8

  3. anonymous
    • 5 years ago
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    Drawing a sketch helps.

  4. anonymous
    • 5 years ago
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    I agree, draw a quick sketch and see which one is above. The only other thing you can do is solve for their intersections, then plug in values before and after they intersect to see which one is below and which is above.

  5. anonymous
    • 5 years ago
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    For example, we can see that they intersect at x = -2, and x = 3. So when you look in this interval at a few values (x=0 for example) you can see that x^2 +2 = 2, but x + 8 = 8 So the linear function is above over that interval.

  6. anonymous
    • 5 years ago
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    This is a nice way to figure it out also because you'll need the intersecting points in order to know your bounds of integration anyway.

  7. anonymous
    • 5 years ago
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    How did you determine those intervals?

  8. anonymous
    • 5 years ago
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    Set the two equations equal to eachother and solve for where they intersect. \[x^2 + 2 = x+8\] \[\implies x^2 -x - 6 = 0 \implies (x + 2)(x-3) = 0 \implies x \in {-2,3}\]

  9. anonymous
    • 5 years ago
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    err \(x \in \{-2,3\}\) rather. So the area you are looking for is in the interval \(x \in [-2,3]\). So I just picked an easy point in that interval to plug in for x and see which curve was higher than the other.

  10. anonymous
    • 5 years ago
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    Does that make sense?

  11. anonymous
    • 5 years ago
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    absolutely. Thank you.

  12. anonymous
    • 5 years ago
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    Can you assist me with a few problems? I have a thick calculus packet due tomorrow. Any assistance would be appreciated.

  13. anonymous
    • 5 years ago
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    Certainly.

  14. anonymous
    • 5 years ago
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    Given \[y^{2} = x^{2}(x+7)\] Find an equation for the tangent line at ( -3, 6) Find an equation for the normal line at ( -3, 6)

  15. anonymous
    • 5 years ago
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    Ok, so for the first part. What do you think would be a good way to find the equation of the tangent?

  16. anonymous
    • 5 years ago
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    take the derivative of the function to find the slope. then plug it into slope intercept form?

  17. anonymous
    • 5 years ago
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    Yes.

  18. anonymous
    • 5 years ago
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    Except I would use the point slope formula instead of slope intercept since you have a particular point, and the slope

  19. anonymous
    • 5 years ago
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    \[y-y_1 = m(x-x_1)\]

  20. anonymous
    • 5 years ago
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    Yes that one, sorry I was confused.

  21. anonymous
    • 5 years ago
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    However, when I took the derivative, it turned out complicated. Can you run through that with me?

  22. anonymous
    • 5 years ago
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    Sure. Since it's y^2 we'll have a \(\pm\), but the point they gave us is postive in y, so we'll rewrite the function as \[y = x\sqrt{x+7}\]

  23. anonymous
    • 5 years ago
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    Can you take the derivative of that?

  24. anonymous
    • 5 years ago
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    yes, by chain rule

  25. anonymous
    • 5 years ago
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    What do you get for the derivative

  26. anonymous
    • 5 years ago
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    Actually I'm thinking that we might want the derivative of \[-x\sqrt{x+7}\]

  27. anonymous
    • 5 years ago
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    why did it become negative?

  28. anonymous
    • 5 years ago
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    hey polpak the answer to my question was D right

  29. anonymous
    • 5 years ago
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    Well there are two options. \[y^2 = x^2(x+7) \implies y = \pm\sqrt{x^2(x+7)} = \pm x\sqrt{x+7}\]

  30. anonymous
    • 5 years ago
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    And I'm not sure how to tell which one we want. I think it's the negative one.

  31. anonymous
    • 5 years ago
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    yes! I'm sure it's the negative one

  32. anonymous
    • 5 years ago
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    Because y = 6, but x = -3. So that means that -x = 3

  33. anonymous
    • 5 years ago
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    And that's the only way you'll get a positive y value if x is negative is to have the negative in front.

  34. anonymous
    • 5 years ago
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    \[f'(x) = \sqrt{x + 7} + {x}\ {2 sqrt {x + 7}}]

  35. anonymous
    • 5 years ago
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    Its difficult for me to write out what Igot for the derivative, but if you please will show me what you got for the derivative then Ill tell you if I got the same thing?

  36. anonymous
    • 5 years ago
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    Close. It should be \[\sqrt{x+7} + \frac{x}{2\sqrt{x+7}}\]

  37. anonymous
    • 5 years ago
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    Oh, except negative.

  38. anonymous
    • 5 years ago
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    -1 times that whole thing.

  39. anonymous
    • 5 years ago
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    That exacty what I got.

  40. anonymous
    • 5 years ago
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    We can simplify a bit also. \[-\sqrt{x+7} - \frac{x}{2\sqrt{x+7}} = \frac{-2x -14}{2\sqrt{x+7}} - \frac{x}{2\sqrt{x+7}} \] \[= \frac{-3x-14}{2\sqrt{x+7}}\]

  41. anonymous
    • 5 years ago
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    So now plug in -3 for x.

  42. anonymous
    • 5 years ago
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    how did you get -2x - 14 for your numerator?

  43. anonymous
    • 5 years ago
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    Multiply the top and bottom of the first term by \(2\sqrt{x+7}\) \[\frac{-\sqrt{x+7}}{1} *\frac{2\sqrt{x+7}}{2\sqrt{x+7}}\]

  44. anonymous
    • 5 years ago
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    For getting everything over the same denominator.

  45. anonymous
    • 5 years ago
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    Gives \[\frac{-2(x+7)}{2\sqrt{x+7}}\]

  46. anonymous
    • 5 years ago
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    allright now what?

  47. anonymous
    • 5 years ago
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    Now plug in -3 for x and find your slope at that point.

  48. anonymous
    • 5 years ago
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    Then plug that slope, and the point (-3,6) into the point slope formula to find the equation of the tangent line

  49. anonymous
    • 5 years ago
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    I got y = -5/4x - 9/2

  50. anonymous
    • 5 years ago
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    y-6 = -5/4(x+3) y = (-5/4)x -15/4 + 24/4 y = (-5/4)x - 9/4

  51. anonymous
    • 5 years ago
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    Not -9/2

  52. anonymous
    • 5 years ago
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    Did you use point slope?

  53. anonymous
    • 5 years ago
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    Or did you use slope intercept?

  54. anonymous
    • 5 years ago
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    slope intercept. my points were (-3, 6) and I plugges that into y2-y1= m (x2-x1)

  55. anonymous
    • 5 years ago
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    Ok, 2 things. First, you don't use two points for slope intercept. Second, that's not slope intercept. =)

  56. anonymous
    • 5 years ago
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    Just use point slope.

  57. anonymous
    • 5 years ago
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    \[y - y_1 = m(x-x_1)\] Plug in the slope for m, and \(x_1,y_1\) is the x, y value of one point on the line.

  58. anonymous
    • 5 years ago
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    I know where I went wrong, i reversed my x and y values

  59. anonymous
    • 5 years ago
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    How do I find the equation of a normal line now that I found the equation of a tangent line?

  60. anonymous
    • 5 years ago
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    Well the normal line is at that same point, but the slope is perpendicular.

  61. anonymous
    • 5 years ago
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    Perpendicular to the tangent line that is

  62. anonymous
    • 5 years ago
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    4/5 is my slope?

  63. anonymous
    • 5 years ago
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    Yes, the slope of the normal line is 4/5

  64. anonymous
    • 5 years ago
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    okayI know how to find the slope of that line. How do I find points at where the tangent line is vertical?

  65. anonymous
    • 5 years ago
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    You don't need to. The normal at that point still touches that point (-3,6)

  66. anonymous
    • 5 years ago
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    You just need one point and the slope to find the equation of the line.

  67. anonymous
    • 5 years ago
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    for where the tangent line is vertical?

  68. anonymous
    • 5 years ago
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    Oh, sorry I misunderstood. The tangent line will be vertical when the slope is something of the form \[\frac{a}{0}\] So you are looking for the point where you will have a 0 in the denominator.

  69. anonymous
    • 5 years ago
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    0 in the denominator of the derivative that is.

  70. anonymous
    • 5 years ago
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    Help? :/

  71. anonymous
    • 5 years ago
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    What is the denominator of the derivative?

  72. anonymous
    • 5 years ago
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    oh, set the denominator to zero?>

  73. anonymous
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    oh, set the denominator to zero?>

  74. anonymous
    • 5 years ago
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    x is greater than or equal to -7

  75. anonymous
    • 5 years ago
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    x cannot be less than -7 or the radical becomes imaginary, but if x is greater than -7, the denominator is non-zero (positive). Only when x = -7 do you have a 0 in the denominator. Therefore at x= -7, you have a vertical tangent line.

  76. anonymous
    • 5 years ago
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    Can I bother you with one more problem?

  77. anonymous
    • 5 years ago
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    yes, but go ahead and make a new question for it. this one is getting long ;)

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