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Homework Help: Calc problem (area of parametric curves)

  1. Jul 5, 2005 #1
    calc 2 problem (area bound by parametric eq.)

    I'm having a problem with this question:
    Find the area bounded by the curve [tex] x=cos{t}\ y= e^t[/tex],
    [tex] 0\geq t\leq\pi/2\ [/tex],
    and the lines [tex] y=1\ x=0 [/tex]

    ... I came up with [tex] \int e^t(-sin{t})dt [/tex] from [tex] 0\to\pi/2[/tex]

    But apparently I'm missing steps... which i'm not aware of! Please help!
     
    Last edited: Jul 5, 2005
  2. jcsd
  3. Jul 5, 2005 #2
    try to write t in terms of x and then replace t with what you have found in the other equation...
     
  4. Jul 5, 2005 #3
    What i did was i used the formula [tex] Area= \int f(t)g'(t)dt [/tex] and plugged in [tex] f(t) = e^t\ g'(t) = -sint [/tex],
    which is how i came to the this: [tex] \int e^t(-sin{t})dt [/tex]
    with a bound of [tex] 0\to\pi/2[/tex]

    Although I'm suppose to start of by
    1. letting the Area [tex] = \int (y-1)dx [/tex]
    with a bound of [tex] 0\to1 [/tex],
    2. And then that would equal to [tex] \int (e^t - 1)(-sint)dt\ [/tex]
    with a bound of [tex] \pi/2\to 0 [/tex]
    3. Which would result in [tex] \int e^t sint - sint dt [/tex]
    with a bound [tex] 0\to\pi/2 [/tex]

    I don't get how the integral in 1. was reached :confused: or why bounds were changed in each step.... any help would be appreciated
     
  5. Jul 5, 2005 #4

    lurflurf

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    Homework Helper

    I think the parametric equations are throwing you of think back to when you first did area problems. Think of y(x)=exp(Arccos(x))
    to find the area bounded by x=0 y=1 and y=exp(Arccos(x)) do as you would do then
    [tex] =\int_0^1 (y-1)dx [/tex]
    or
    [tex] =\int_0^1 (e^{{Arccos}(x)}-1)dx [/tex]
    now do a change of variable
    x=cos(t)
    dx=-sin(t)dt
    0<x<1
    becomes
    0<t<pi/2
    this is how the bounds change
    so we have
    [tex] = \int_0^\frac{pi}{2} (e^t-1)(-sin(t))dt [/tex]
     
  6. Jul 5, 2005 #5
    I'm so sorry... but I still don't understand why the area wouldn't be given by:
    [tex] \int_0^\frac{pi}{2} e^t(-sin{t})dt [/tex]

    How did arrive to [tex] = \int_0^1 (y-1)dx [/tex]
     
  7. Jul 5, 2005 #6

    lurflurf

    User Avatar
    Homework Helper

    It is because you want the area between y and 1. Recall that to find the area between f(x) and g(x) where f(x)>g(x) and x=a, x=b you find

    [tex] =\int_a^b (f(x)-g(x))dx [/tex]
    This is the same, but in this example f(x)=y g(x)=1 a=0 b=1 so
    [tex] =\int_0^1 (y-1)dx [/tex]
     
  8. Jul 6, 2005 #7
    Thanks a bunch! That explanation really helped =) I can solve it from here on.
     
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