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Arc length formula help

  1. Nov 26, 2006 #1
    1 Find the area bounded by the curve [tex] x = t - \frac{1}{t} [/tex], [tex] y = t + \frac{1}{t} [/tex] and the line [tex] y = 2.5 [/tex].

    I know that [tex] A = \int_{\alpha}^{\beta} g(t)f'(t) \; dt [/tex]


    I ended up with [tex] \int_{1}^{2} 2.5-(t+\frac{1}{t})(1+\frac{1}{t^{2}}) [/tex]


    2 Find the length of the curve: [tex] x = a(\cos \theta + \theta \sin \theta) [/tex], [tex] y = a(\sin \theta-\theta \cos \theta) [/tex], [tex] 0\leq \theta\leq \pi [/tex]

    I obtained [tex] \frac{a\pi^{2}}{2} [/tex]. Does this look correct? I used the arc length formula for parametric equations.


    Is this correct?


    3 Find the surface area obtained by rotating the given curve about the x-axis: [tex] x = 3t-t^{3} [/tex] [tex] y = 3t^{2} [/tex], [tex] 0\leq t\leq 1 [/tex].

    So [tex] S = \int_{a}^{b} 2\pi y \sqrt{(\frac{dx}{dt}^{2})+(\frac{dy}{dt}^{2})} \; dt [/tex]

    So would I do the following: [tex] \int_{0}^{1} 2\pi(3t^{2})\sqrt{(3-3t^{2})+36t^{2}} \; dt [/tex]?
     
    Last edited: Nov 26, 2006
  2. jcsd
  3. Nov 27, 2006 #2

    HallsofIvy

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    Now quite. if x= f(t), y= g(t), then dx= f'(t)dt and ydx= g(t)f'(t)dt but you want (2.5- y)dx. You should have (2.5- t+ 1/t)(1+ 1/t2)dt. Also, recheck your limits of integration. When t= 1, y= 2, not 2.5.


    Yes, it is. Don't you just love it when things cancel out?


    No. You forgot a square: it should be
    [tex] \int_{0}^{1} 2\pi(3t^{2})\sqrt{(3-3t^{2})^2+36t^{2}} \; dt [/tex]
    (Once again, that square root simplifies nicely. Your teacher is being nice to you!)
     
  4. Nov 27, 2006 #3
    Instead if the question read as:

    Find the surface area generated by rotating the given curve about the y-axis:

    [tex] x = 3t^{2}, y = 2t^{3}, 0\leq t \leq 5 [/tex] would it be:

    [tex] \int_{0}^{5} 2\pi(3t^{2})\sqrt{72t^{2}} \; dt [/tex]
     
  5. Nov 27, 2006 #4

    HallsofIvy

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    No, it wouldn't. You want [itex]\sqrt{(dx/dt)^2+ (dy/dt)^2}[/itex]. Here, dx/dt= 6t and dy/dt= 6t2. You need [itex]\sqrt{36t^2+ 36t^4}= 6t\sqrt{1+ t^2}[/itex].
     
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