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Math Help - Trigonometry

  1. #1
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    Trigonometry

    Can anyone please help
    An engineering process is tested over a period of 60secs. the equation can be given by y=e(to the power)0.06t.
    a) Use numerical integration with 10 second itervals to find the area.
    b) use integral calculus to find the area.
    c) Compare your answers to part (a) and (b), stating which is more accurate and giving your reasons.

    Thanks
    Headbazza
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  2. #2
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    Hello headbazza
    Quote Originally Posted by headbazza View Post
    Can anyone please help
    An engineering process is tested over a period of 60secs. the equation can be given by y=e(to the power)0.06t.
    a) Use numerical integration with 10 second itervals to find the area.
    b) use integral calculus to find the area.
    c) Compare your answers to part (a) and (b), stating which is more accurate and giving your reasons.

    Thanks
    Headbazza
    What attempt have you made? Have you worked out the values of y for t = 0, 10, ..., 60?

    Do you know how to integrate e^{0.06t}?

    Please give us some indication of where you're finding this question difficult.

    Grandad
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  3. #3
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    Hi Grandad
    to be totally honest with you i am completly stuck with this and could do with a complete answer, i will be doing an exam shortly with this question as an example of what i might encounter.

    Thanks
    headbazza
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  4. #4
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    Hello headbazza

    The simplest method of doing a numerical integration is to use the Trapezium Rule.


    Imagine the area under the curve y = f(x) divided vertically into n strips of equal width d, with the curve replaced by straight line segments. Then the area is approximately the same as the sum of the areas of the trapezia; which is:
    \Big(\frac{f(x_0)+f(x_n)}{2} + f(x_1) + f(x_2) + ... + f(x_{n-1})\Big)d, where x_0, x_1, ...,x_n are the n+1 equally spaced values of x.
    So, to use this formula in your question, you'll work out y(0), y(10), y(20), ... ,y(60), and then:
    Area = 10\Big(\frac{y(0)+y(60)}{2} + y(10) + y(20) + ... + y(50)\Big)
    Take care with the calculation, and it should be very straightforward.

    For (b):
    Area = \int_0^{60}e^{0.06t}dt
    = \frac{1}{0.06}\Big[e^{0.06t}\Big]_0^{60}

    = \frac{1}{0.06}(e^{3.6}-e^0)


    \approx 593
    For (c), the answer to (b) will be more accurate because in (a) the curve has been replaced by straight line segments joining the ordinates at x = 0, 10, ..., 60.

    Grandad
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