Thread: 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

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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.

That certainly should work! What, exactly did you try?

You are told that the acceleration vector is (which, by the way, because cosine is an "even" function and sine an "odd" function, is the same as ). What did you get when you integrated? How did you use the fact that V(0)= i+ k?

Originally Posted by HallsofIvy

That certainly should work! What, exactly did you try?

You are told that the acceleration vector is (which, by the way, because cosine is an "even" function and sine an "odd" function, is the same as ). 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

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

Originally Posted by topsquark

@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.

Originally Posted by operaphantom2003

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.