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Math Help - Calc III...position and acceleration vectors

  1. #1
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    Calc III...position and acceleration vectors

    This is probably easy but I am completely stumped.

    Given that the acceleration vector is a(t)=(-9cos(-3t))I+(-9sin(-3t))J+(-5t)K, the initial velocity is v(0)=I+K, and the initial position vector is X(0)=I+J+K, compute:
    A. The velocity vector v(t)=_____I+_____J+_____K
    B. The position vector X(t)=_____I+_____J+_____K

    *note the coefficients in your answer must be entered in the form of the expressions in the variable t

    ----------------
    If the acceleration vector is the derivative of the velocity vector then would you just integrate to work backwards? I tried that but was wrong.
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  2. #2
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    That certainly should work! What, exactly did you try?

    You are told that the acceleration vector is (-9cos(-3t))i+ (-9sin(3t))j-5tk (which, by the way, because cosine is an "even" function and sine an "odd" function, is the same as -9 cos(3t)i- 9sin(3t)j- 5tk). What did you get when you integrated? How did you use the fact that V(0)= i+ k?
    Last edited by HallsofIvy; February 1st 2011 at 12:20 PM.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    That certainly should work! What, exactly did you try?

    You are told that the acceleration vector is (-9cos(-3t)i+ (-9sin(3t)j-5tk (which, by the way, because cosine is an "even" function and sine an "odd" function, is the same as 9 cos(3t)i- 9sin(3t)j- 5tk). What did you get when you integrated? How did you use the fact that V(0)= i+ k?
    @HallsofIvy: You have a typo. Since sine is an odd function we have that -9sin(-3t) = +9sin(3t). Since cosine is an even function we have that -9cos(-3t) = -9cos(3t). So the overall acceleration function is:
    -9cos(3t)*i + 9sin(3t)*j - 5t*k

    -Dan
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    That's where I must be getting messed up because I know I need to find a constant, but where does it go? When I integrate I get (-3sin(3t))I+(-3cos(3t))J+(-5/2 t^2)K if I plug in 0 for t then I get -3 but I am getting confused....ugg
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  5. #5
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    Quote Originally Posted by topsquark View Post
    @HallsofIvy: You have a typo. Since sine is an odd function we have that -9sin(-3t) = +9sin(3t). Since cosine is an even function we have that -9cos(-3t) = -9cos(3t). So the overall acceleration function is:
    -9cos(3t)*i + 9sin(3t)*j - 5t*k

    -Dan
    Thanks, I have fixed that.
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  6. #6
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    Quote Originally Posted by operaphantom2003 View Post
    That's where I must be getting messed up because I know I need to find a constant, but where does it go? When I integrate I get (-3sin(3t))I+(-3cos(3t))J+(-5/2 t^2)K if I plug in 0 for t then I get -3 but I am getting confused....ugg
    First, "if I plug in 0 for t then I get -3" indicates a serious misunderstanding! This is a vector function. If you plug in any number for t, you should get a vector, NOT a number! When you put t= 0 in what you have, you should get 0i- 3j+ 0k= -3j, not -3.

    You had to do three integrations, didn't you, one for each component? You need a constant for each integration.
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