1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Calc problem (area of parametric curves)
  1. Parametric curve (Replies: 3)

  2. Parametrized curve (Replies: 1)

Loading...