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Math Help - Antiderivatives

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    Antiderivatives

    Hey,
    I have an idea of how to approach this problem, but I dont understand how to solve it. Can someone help me please?

    Problem:
    A charged particle moves along the x-axis under the influence of an electric field. The field strength varies with time, and as a result, the velocity of the particle is complicated. The position of the particle at time t is written as x=x(t) and the velocity of the particle at time t is written as v(t).
    Suppose we know that x(0)=0, and also that
    v(t)={2t-1, if 0<t<1
    {4t-3, if 1<t<2
    {6t-7, if 2<t<3

    What is x(1)? (...find x(2), x(3) and sketch x=x(t), v=v(t))

    So, obviously we can take the antiderivative of v(t) to find x(t) from which you get: t^2-t+C, 2*t^2-3*t+C, 3*t^2-7*t+C ..whats confusing me is the three different intervals for time t. how do i solve for x(1) if there are 2 different antiderivatives for it?? Im confused
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    Quote Originally Posted by coe236 View Post
    A charged particle moves along the x-axis under the influence of an electric field. The field strength varies with time, and as a result, the velocity of the particle is complicated. The position of the particle at time t is written as x=x(t) and the velocity of the particle at time t is written as v(t).
    Suppose we know that x(0)=0, and also that
    v(t)={2t-1, if 0<t<1
    {4t-3, if 1<t<2
    {6t-7, if 2<t<3

    What is x(1)? (...find x(2), x(3) and sketch x=x(t), v=v(t))

    So, obviously we can take the antiderivative of v(t) to find x(t) from which you get: t^2-t+C, 2*t^2-3*t+C, 3*t^2-7*t+C ..whats confusing me is the three different intervals for time t. how do i solve for x(1) if there are 2 different antiderivatives for it?? Im confused
    The function for x(t) is
    x = \int (2t - 1)~dt = t^2 - t + A for 0 \leq t < 1

    x = \int (4t - 3)~dt = 2t^2 - 3t + B for 1 \leq t < 2

    x = \int (6t - 7)~dt = 3t^2 - 7t + C for 2 \leq t < 3

    We know that x(0) = 0, so
    x(0) = 0^2 - 0 + A = 0 \implies A = 0

    And we know that x must be continuous at all times t. Thus
    \lim_{t \to 1}x(t) = 1^1 - 2 = 2 \cdot 1^2 - 3 \cdot 1 + B \implies B = 0

    and
    \lim_{t \to 2}x(t) = 2 \cdot 2^2 - 3 \cdot 2 = 3 \cdot 2^2 - 7 \cdot 2 + C \implies C = 4

    Thus
    x(t) = \left\{\begin{array}{rr} t^2 - t,\text{ if }0 \leq t < 1 \\<br />
2t^2 - 3t,\text{ if }1 \leq t < 2\\<br />
3t^2 - 7t + 4,\text{ if } 2 \leq t < 3\end{array}\right.

    -Dan
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