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Math Help - Antiderivative/Integral Problem

  1. #1
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    Antiderivative/Integral Problem

    Let s(t) denote the position of a particle at time t, and let v and a be the velocity and acceleration respectively. The particle is moving according to the data


    Find a function describing the position of the particle.

    If I could get help on this I would greatly appreciate it. Basically, I found the v(t) equation and then the s(t) equation through integrals/antiderivatives, but my solution isn't right when I try to submit it. s(t) = -(10sint)-(3cost)-((17/2pi)t)+12
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  2. #2
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    Quote Originally Posted by cdlegendary View Post
    Let s(t) denote the position of a particle at time t, and let v and a be the velocity and acceleration respectively. The particle is moving according to the data


    Find a function describing the position of the particle.

    If I could get help on this I would greatly appreciate it. Basically, I found the v(t) equation and then the s(t) equation through integrals/antiderivatives, but my solution isn't right when I try to submit it. s(t) = -(10sint)-(3cost)-((17/2pi)t)+12
    a(t) = 10\sin{t} + 3\cos{t}


    v(t) = \int{10\sin{t} + 3\cos{t}\,dt}

     = -10\cos{t} + 3\sin{t} + C.


    s(t) = \int{-10\cos{t} + 3\sin{t} + C\,dt}

     = -10\sin{t} - 3\cos{t} + Ct + D.


    Now using the intital conditions:

    9 = -10\sin{0} - 3\cos{0} + c(0) + D

    9 = -3 + D

    D = 12.


    -8 = -10\sin{2\pi} - 3\cos{2\pi} + 2\pi C + D

    -8 = -3 + 2\pi C + 12

    -17 = 2\pi C

    C = -\frac{17}{2\pi}.


    Therefore s(t) = -10\sin{t} - 3\cos{t} - \frac{17t}{2\pi} + 12.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    a(t) = 10\sin{t} + 3\cos{t}


    v(t) = \int{10\sin{t} + 3\cos{t}\,dt}

     = -10\cos{t} + 3\sin{t} + C.


    s(t) = \int{-10\cos{t} + 3\sin{t} + C\,dt}

     = -10\sin{t} - 3\cos{t} + Ct + D.


    Now using the intital conditions:

    9 = -10\sin{0} - 3\cos{0} + c(0) + D

    9 = -3 + D

    D = 12.


    -8 = -10\sin{2\pi} - 3\cos{2\pi} + 2\pi C + D

    -8 = -3 + 2\pi C + 12

    17 = 2\pi C

    C = \frac{17}{2\pi}.


    Therefore s(t) = -10\sin{t} - 3\cos{t} + \frac{17t}{2\pi} + 12.
    Thanks! That's the same answer I came up with, but my Math homework is entered on an online site, and it seems to not like my notation I suppose..
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    Quote Originally Posted by cdlegendary View Post
    Let s(t) denote the position of a particle at time t, and let v and a be the velocity and acceleration respectively. The particle is moving according to the data


    Find a function describing the position of the particle.

    If I could get help on this I would greatly appreciate it. Basically, I found the v(t) equation and then the s(t) equation through integrals/antiderivatives, but my solution isn't right when I try to submit it. s(t) = -(10sint)-(3cost)-((17/2pi)t)+12

    v(t)=\int a(t)\,dt=\int\left(10\sin t+3\cos t\right)\,dt=-10\cos t+3\sin t+C_1\,,\,\,s(t)= \int v(t)\,dt\int\left(-10\cos t+3\sin t+C_1\right)\,dt =-10\sin t-3\cos t+C_1t+C_2 ,

     C_1\,,\,C_2= constants wrt the variable t. Now:

    9=s(0)=-3+C_2\,\Longrightarrow,C_2=12

    -8=s(2\pi)=-3+2\pi C_1+12\,\Longrightarrow\,C_1=-\frac{17}{2\pi}

    And, as far as I can see, your solution is correct...check there isn't a minus sign or something like that in the definition of a(t) ...

    Tonio
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    Quote Originally Posted by Prove It View Post
    a(t) = 10\sin{t} + 3\cos{t}


    v(t) = \int{10\sin{t} + 3\cos{t}\,dt}

     = -10\cos{t} + 3\sin{t} + C.


    s(t) = \int{-10\cos{t} + 3\sin{t} + C\,dt}

     = -10\sin{t} - 3\cos{t} + Ct + D.


    Now using the intital conditions:

    9 = -10\sin{0} - 3\cos{0} + c(0) + D

    9 = -3 + D

    D = 12.


    -8 = -10\sin{2\pi} - 3\cos{2\pi} + 2\pi C + D

    -8 = -3 + 2\pi C + 12

    17 = 2\pi C

    C = \frac{17}{2\pi}.


    Therefore s(t) = -10\sin{t} - 3\cos{t} + \frac{17t}{2\pi} + 12.

    I think a minus sign is missing in  C = \frac{17}{2\pi}

    Tonio
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  6. #6
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    Quote Originally Posted by tonio View Post
    I think a minus sign is missing in  C = \frac{17}{2\pi}

    Tonio
    Yes, edited. Thanks.
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  7. #7
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    Thanks again you two. I figured it out, the application accepted the answer when I entered it as -(10sin(t))-(3cos(t))-(((17t)/(2pi)))+12
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