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Math Help - trapezoid rule

  1. #1
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    trapezoid rule

    Ocean currents (miles/hr) at a certain location are given below

    Time of day: Speed mph:
    8:00 am 17
    8:10 am 20
    8:20 am 22
    8:30 am 21
    8:40 am 17

    How can I estimate the average river current speed from 8:00 am to 8:40 am using the trapezoidal rule? I tried but I got a huge number.
    Last edited by chukie; June 29th 2008 at 08:46 PM.
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    Quote Originally Posted by chukie View Post
    Ocean currents (miles/hr) at a certain location are given below

    Time of day: Speed mph:
    8:00 am 17
    8:10 am 20
    8:20 am 22
    8:30 am 21
    8:40 am 17

    How can I estimate the average river current speed from 8:00 am to 8:40 am using the trapezoidal rule? I tried but I got a huge number.
    The average current speed is:

    \bar{v}=\left(\int_{t=8:00}^{t=8:40}v(t) \; dt\right)/T,

    where T is the total duration, and in this case is 40 minutes. Now use the trapezoid rule to estimate the integral

    RonL
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    Quote Originally Posted by CaptainBlack View Post
    The average current speed is:

    \bar{v}=\left(\int_{t=8:00}^{t=8:40}v(t) \; dt\right)/T,

    where T is the total duration, and in this case is 40 minutes. Now use the trapezoid rule to estimate the integral

    RonL
    But the speed is miles per hour. If I just divide by 40, I get an answer of 4, which is too small?
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    Quote Originally Posted by chukie View Post
    But the speed is miles per hour. If I just divide by 40, I get an answer of 4, which is too small?
    So do some unit conversion. What's the trouble?

    40\text{ min} = 40\text{ min}\left(\frac{1\text{ hr}}{60\text{ min}}\right) = \frac23\text{ hr}

    Just make sure you use this value when applying the trapezoidal rule (for b - a) or else your units won't line up correctly.
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    Quote Originally Posted by Reckoner View Post
    So do some unit conversion. What's the trouble?

    40\text{ min} = 40\text{ min}\left(\frac{1\text{ hr}}{60\text{ min}}\right) = \frac23\text{ hr}

    Just make sure you use this value when applying the trapezoidal rule (for b - a) or else your units won't line up correctly.
    I tried that too, but then the answer becomes 240, which is too big?
    (1/(2/3))(17+40+44+42+17)=240
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    Quote Originally Posted by chukie View Post
    I tried that too, but then the answer becomes 240, which is too big?
    (1/(2/3))(17+40+44+42+17)=240
    What are you doing?

    The trapezoidal rule is

    \int_a^b f(x)\,dx\approx\frac{b-a}{2n}\left[f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n)\right]

    Substituting the appropriate values, this gives us

    \frac{\left(\frac23\right)}{\left(2\cdot4\right)}\  left(17+40+44+42+17\right) = \frac{40}3

    and dividing by \frac23 we get an average velocity of 20\text{ mi/hr}, which seems about right to me.
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  7. #7
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    Quote Originally Posted by Reckoner View Post
    What are you doing?

    The trapezoidal rule is

    \int_a^b f(x)\,dx\approx\frac{b-a}{2n}\left[f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n)\right]

    Substituting the appropriate values, this gives us

    \frac{\left(\frac23\right)}{\left(2\cdot4\right)}\  left(17+40+44+42+17\right) = \frac{40}3

    and dividing by \frac23 we get an average velocity of 20\text{ mi/hr}, which seems about right to me.
    oops! sorry, i completely left out the (b-a)/2n!!
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    Quote Originally Posted by chukie View Post
    But the speed is miles per hour. If I just divide by 40, I get an answer of 4, which is too small?
    The units are unimportant, the integrand is of the order of 20 mph, so the integral over 40 mins is ~ 20 \times 40, and dividing by 40 gives a mean ~ 20 mph, there is no problem of units.

    Applying the trapesiaum rule gives for the integral:

    I \approx \frac{40}{2 \times 4}(17+2 \times 20+2 \times 22+2 \times 21 +17) = 800

    So to get the average speed we divide the integral by 40 to get 20 mph

    RonL
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