# Thread: Calc III...position and acceleration vectors

1. ## 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.

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

You are told that the acceleration vector is $\displaystyle (-9cos(-3t))i+ (-9sin(3t))j-5tk$ (which, by the way, because cosine is an "even" function and sine an "odd" function, is the same as $\displaystyle -9 cos(3t)i- 9sin(3t)j- 5tk$). What did you get when you integrated? How did you use the fact that V(0)= i+ k?

3. Originally Posted by HallsofIvy
That certainly should work! What, exactly did you try?

You are told that the acceleration vector is $\displaystyle (-9cos(-3t)i+ (-9sin(3t)j-5tk$ (which, by the way, because cosine is an "even" function and sine an "odd" function, is the same as $\displaystyle 9 cos(3t)i- 9sin(3t)j- 5tk$). 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

4. 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

5. 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.

6. 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.