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Math Help - I'm not sure how to solve this question?

  1. #1
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    I'm not sure how to solve this question?

    Question: At t = 55 seconds, an object moving along a straight path is 47 meters from its origin. If the velocity at that time is v(55) = −3 meters per second, use a tangent line approximation to predict the objects position at one minute (t = 60 seconds).

    So I know that the equation for a tangent line is:
    T(x) = f(x) + f'(x) (x-c)

    But I am not sure if that is how I solve this question as I'm trying to predict the object's position.

    Any help is appreciated.
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  2. #2
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    Let p(t) be the position in meters of the object at time t seconds. Then, p(t) = -4.9t^2 + v_0t + s_0 where v_0 is the initial velocity in meters per second of the object and s_0 is the initial position in meters of the object.

    Let L(t) be the tangent line approximation to p(t) at t = 55. Then, L(t) = p(55) + p'(55)(t - 55). Because p(55) = 47 and p'(55) = v(55) = -3, L(t) = 47 - 3(t - 55).

    Evaluate L(60).
    Last edited by NOX Andrew; October 18th 2010 at 10:11 AM.
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  3. #3
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    Thank you for answering!
    So to evaluate L(60)..
    L(60) = p(60) + p'(60)(x - 60)
    Is that correct?
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  4. #4
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    Sorry, I had let L(t) = p(55) + p'(55)(x - 55) in my original post. The variable x shouldn't be on the left-hand side of the equation.

    I meant to let L(t) = p(55) + p'(55)(t - 55).

    Then, L(60) = p(55) + p'(55)(60 - 55) = 47 - 3(5) = 47 - 15 = 32.
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