Results 1 to 5 of 5

Math Help - Integration of Sine, Pi, and Solving for t

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    29

    Integration of Sine, Pi, and Solving for t

    My class is doing a test review for homework, and I was doing well so far until I bumped into this:

    A particle moves along the x-axis in such a way that its acceleration at time t for t (greater than or equal to) 0 is given by a(t)=2 cos(t). At time t=0, the velocity of the particle is v(0)= -1 and its position is x(0)= 0.

    a. Write an equation for the velocity for the velocity v(t) of the particle.
    Already found this, it was: v(t)=2 sin t -1

    b. Write an equation for the position x(t) of the particle.
    Found this also, which was: x(t)= -2 cos t - t +2

    c. For what values of t, 0< t < \pi, is the particle at rest? (less thans are less than or equal to).
    The problem I have with this is isolating t using: -1= -2 cos t - t +2

    I already did a similar problem, but it was much simpler to isolate t to one side and solve for it. (2 \pi -2 \pi sin 2 \pit = 0 ----> t=0.25)


    Another little question while I'm on the subject, how do I \frac{d}{dx} tan (4x + 9x^2)? I'm too used to single terms.

    Some help would be very much appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by DarkestEvil View Post
    My class is doing a test review for homework, and I was doing well so far until I bumped into this:

    A particle moves along the x-axis in such a way that its acceleration at time t for t (greater than or equal to) 0 is given by a(t)=2 cos(t). At time t=0, the velocity of the particle is v(0)= -1 and its position is x(0)= 0.

    a. Write an equation for the velocity for the velocity v(t) of the particle.
    Already found this, it was: v(t)=2 sin t -1

    b. Write an equation for the position x(t) of the particle.
    Found this also, which was: x(t)= -2 cos t - t +2

    c. For what values of t, 0< t < \pi, is the particle at rest? (less thans are less than or equal to).
    The problem I have with this is isolating t using: -1= -2 cos t - t +2

    to find when the particle is at rest, set v(t) = 0, not x(t).

    I already did a similar problem, but it was much simpler to isolate t to one side and solve for it. (2 \pi -2 \pi sin 2 \pit = 0 ----> t=0.25)


    Another little question while I'm on the subject, how do I \frac{d}{dx} tan (4x + 9x^2)? I'm too used to single terms.

    chain rule ... \textcolor{red}{\frac{d}{dx}\tan{u} = \sec^2{u} \cdot \frac{du}{dx}}
    ...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2009
    Posts
    29
    Hmmm... I did what you said, and used v(t)=2 sin t - 1
    So: 2 sin t - 1 = 0 ---> sin t = 1/2 ---> t=30
    Ah, thanks. I just remembered \pi = 180 degrees, thanks.

    Do I do the same to this equation?
    v(t)=2 \pi - 2 \pi sin 2 \pit = 0
    But then I get stuck at sin 2 \pit = 1

    Sorry, I'm just a high school student. ^^"
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by DarkestEvil View Post

    So: 2 sin t - 1 = 0 ---> sin t = 1/2 ---> t=30 ... degrees are not the correct solution to these types of equations.

    note that \textcolor{red}{t = \frac{\pi}{6}} is not the only solution in the given interval


    v(t)=2 \pi - 2 \pi sin 2 \pit = 0
    But then I get stuck at sin 2 \pit = 1
    2\pi - 2\pi \sin(2\pi t) = 0

    2\pi[1 - \sin(2\pi t)] = 0

    \sin(2\pi t) = 1

    2\pi t = \frac{\pi}{2}

    t = \frac{1}{4}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2009
    Posts
    29
    Ah, I see, because 30 degrees is the same as \frac{\pi}{6}. Thanks a lot.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integration with Hyperbolic Sine
    Posted in the Calculus Forum
    Replies: 9
    Last Post: February 15th 2010, 04:55 PM
  2. solving sine function
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: April 23rd 2009, 02:03 AM
  3. Solving the Sine Equation
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: January 28th 2009, 05:33 PM
  4. Sine Integration
    Posted in the Calculus Forum
    Replies: 10
    Last Post: June 7th 2008, 02:00 PM
  5. solving sine
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: December 2nd 2007, 02:14 PM

Search Tags


/mathhelpforum @mathhelpforum