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Math Help - Find location using Taylor poly

  1. #1
    Junior Member
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    Find location using Taylor poly

    A car is at location f(0) = 10 miles with a velocity of f'(0) = 10 mi/min, acceleration f''(0) = 2 mi/min^2 and f'''(0) = -1 mi/min^3. Predict the location of the car at time t = 2 min.

    Here is what I think needs to be done. A Taylor polynomial needs to be made using the above info. Once that is done plug in the time value and I should get the location.

    Now how to interpet the info? All the functions are given with 0 as the input. So does that mean the start time of 0 or is this a Maclaurin Polynomial? I know that the dirivatives of f(x) form the coeffients of the powers of x in a Taylor/Maclaurin Polynomial.

    The second dirivative of the location function is acceration but i do not know what the third dirivative is. the question says f'''(0) = -1 mi/min^3, i just don't see how this relates to finding location.

    So I think the answer is something like this

    [edit]
    P(x) = 10 + 10x + (2/2!)x + (2/2!)x^2 -(1/3!)x^3
    P(2) = 32.67

    i did some adding wrong but fixed it.
    Last edited by diroga; April 23rd 2009 at 11:09 AM.
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  2. #2
    MHF Contributor
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    Hi

    f(x) = f(0+x) = f(0) + f'(0) \:x + \frac{f"(0)}{2}\:x^2 + \frac{f^{(3)}(0)}{6}\:x^3

    Spoiler:

    f(2) = 10 + 10 \times 2 + \frac{2}{2}\times 2^2 - \frac{1}{6}\times 2^3 = 32.67 m
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  3. #3
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    ok that looks like what I did. why did you choose to pick that formula?
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  4. #4
    MHF Contributor
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    This is Taylor-Maclaurin formula
    It is the most appropriate one since you know all the derivatives values at t=0
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