How do you find the velocity and position vectors if you are given that the acceleration vector is $a (t) = (- 4 cos (- 2 t)) i + (- 4 sin (- 2 t)) j + (- 2 t) k$ $a(t)= (-4 cos (-2t))i + (-4sin (-2t))j + (-2t)k$ and the initial velocity is $v (0) = i + k$ $v(0)=i+k$ and the initial position vector is $r (0) = i + j + k$ $r(0)= i+j+k$ ?

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How do you find the velocity and position vectors if you are given that the acceleration vector is $a (t) = (- 4 cos (- 2 t)) i + (- 4 sin (- 2 t)) j + (- 2 t) k$ $a(t)= (-4 cos (-2t))i + (-4sin (-2t))j + (-2t)k$ and the initial velocity is $v (0) = i + k$ $v(0)=i+k$ and the initial position vector is $r (0) = i + j + k$ $r(0)= i+j+k$ ?

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Answer 1 · 2015-03-30T01:55:01

First, note that $cos (- 2 t) = cos (2 t)$ $cos(-2t)=cos(2t)$ and $sin (- 2 t) = - sin (2 t)$ $sin(-2t)=-sin(2t)$ since cosine is an even function and sine is an odd function.

Now integrate to get $v (t) = - 2 sin (2 t) i - 2 cos (2 t) j - t^{2} k + \to c$ $v(t)=-2sin(2t)i-2cos(2t)j-t^2k+\vec{c}$ , where $\to c$ $\vec{c}$ is a constant of integration. Since $v (0) = i + k$ $v(0)=i+k$ and since $sin (0) = 0$ $sin(0)=0$ and $cos (0) = 1$ $cos(0)=1$ , it follows that $\to c = i + 2 j + k$ $\vec{c}=i+2j+k$ so that $v (t) = (- 2 sin (2 t) + 1) i + (- 2 cos (2 t) + 2) j + (1 - t^{2}) k$ $v(t)=(-2sin(2t)+1)i+(-2cos(2t)+2)j+(1-t^2)k$ .

Now integrate again to get $r (t) = (cos (2 t) + t) i + (- sin (2 t) + 2 t) j + (t - \frac{1}{3} t^{3}) k + \to c$ $r(t)=(cos(2t)+t)i+(-sin(2t)+2t)j+(t-\frac{1}{3}t^{3})k+\vec{c}$ . Since $r (0) = i + j + k$ $r(0)=i+j+k$ , it follows that $\to c = j + k$ $\vec{c}=j+k$ so that $r (t) = (cos (2 t) + t) i + (- sin (2 t) + 2 t + 1) j + (t - \frac{1}{3} t^{3} + 1) k$ $r(t)=(cos(2t)+t)i+(-sin(2t)+2t+1)j+(t-\frac{1}{3}t^{3}+1)k$

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