Results 1 to 7 of 7

Thread: Antiderivative/Integral Problem

  1. #1
    Member
    Joined
    Jan 2010
    Posts
    79

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    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
    $\displaystyle a(t) = 10\sin{t} + 3\cos{t}$


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

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


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

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


    Now using the intital conditions:

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

    $\displaystyle 9 = -3 + D$

    $\displaystyle D = 12$.


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

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

    $\displaystyle -17 = 2\pi C$

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


    Therefore $\displaystyle s(t) = -10\sin{t} - 3\cos{t} - \frac{17t}{2\pi} + 12$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2010
    Posts
    79
    Quote Originally Posted by Prove It View Post
    $\displaystyle a(t) = 10\sin{t} + 3\cos{t}$


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

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


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

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


    Now using the intital conditions:

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

    $\displaystyle 9 = -3 + D$

    $\displaystyle D = 12$.


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

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

    $\displaystyle 17 = 2\pi C$

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


    Therefore $\displaystyle s(t) = -10\sin{t} - 3\cos{t} + \frac{17t}{2\pi} + 12$.
    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..
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    3
    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

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

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

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

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

    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 $\displaystyle a(t)$ ...

    Tonio
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    3
    Quote Originally Posted by Prove It View Post
    $\displaystyle a(t) = 10\sin{t} + 3\cos{t}$


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

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


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

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


    Now using the intital conditions:

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

    $\displaystyle 9 = -3 + D$

    $\displaystyle D = 12$.


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

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

    $\displaystyle 17 = 2\pi C$

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


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

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

    Tonio
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by tonio View Post
    I think a minus sign is missing in $\displaystyle C = \frac{17}{2\pi}$

    Tonio
    Yes, edited. Thanks.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jan 2010
    Posts
    79
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Antiderivative problem.
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: Dec 2nd 2009, 08:13 PM
  2. Antiderivative Problem
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Nov 10th 2009, 06:29 PM
  3. Antiderivative problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Jun 3rd 2009, 04:54 AM
  4. Improper Integral Antiderivative
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Jan 11th 2009, 08:39 PM
  5. Antiderivative problem
    Posted in the Calculus Forum
    Replies: 10
    Last Post: May 31st 2008, 08:42 PM

Search Tags


/mathhelpforum @mathhelpforum