# Math Help - I need help finding a position vector

1. ## I need help finding a position vector

A particle starts at the initial position r(0)=<0,0,1> with the initial velocity v(0)=<1,0,0> and the accelerlation a(t<sin(t),e^(t),12t>. Find its position function r(t).

2. ## Re: I need help finding a position vector

Hello, allstar2!

$\text{A particle starts at the initial position: }\:r(0)\:=\:\langle 0,0,1\rangle$
$\text{with initial velocity: }\:v(0)\:=\:\langle1,0,0\rangle$
$\text{and accelerlation: }\:a(t) \:=\:\langle \sin t,\,e^t,\,12t\rangle.$

$\text{Find its position function, }r(t).$

The velocity function is the integral of the acceleration function.

$v(t) \;=\;\int\langle \sin t,\;e^t,\;12t\rangle\,dt \;=\;\langle \text{-}\cos t\!+\!c_1,\; e^t\!+\!c_2,\; 6t^2\!+\!c_3\rangle$

We are told that: . $v(0) \:=\:\langle 1,\,0,\,0\rangle$

So we have: . $\langle \text{-}\cos0\!+\!c_1,\:e^0\!+\!c_2,\:0\!+\!c_3\rangle \;=\;\langle 1,\,0,\,0\rangle$

. . . . . . . . . . . . . . $\langle \text{-}1\!+\!c_1,\;1\!+\!c_2,\;c_3\rangle \;=\;\langle 1,\,0,\,0\rangle$

Then we have: . $\begin{Bmatrix}\text{-}1+c_1 \:=\:1 & \Rightarrow & c_1 \:=\:2 \\ 1+c_2 \:=\:0 & \Rightarrow & c_2 \:=\:\text{-}1 \\ c_3 \:=\:0\end{Bmatrix}$

Hence: . $v(t) \;=\;\langle \text{-}\cos t\!+\!2,\;e^t\!-\!1,\; 6t^2\rangle$

The position function is the integral of the velocity function.

$r(t) \;=\;\int \langle \text{-}\cos t\!+\!2,\;e^t\!-\!1,\;6t^2\rangle\,dt \;=\;\langle -\sin t\!+\!2t\!+\!k_1,\;e^t\!-\!t\!+\!k_2,\;2t^3\!+\!k_3\rangle$

We are told: . $r(0) \:=\:\langle 0,0,1\rangle$

So we have: . $\langle \text{-}\sin0\!+\!0\!+\!k_1,\;e^0\!-\!0\!+\!k_2,\;0\!+\!k_3\rangle \;=\;\langle 0,0,1\rangle$

. . . . . . . . . . . . . . $\langle k_1,\;1\!+\!k_2,\;k_3\rangle \;=\;\langle 0,0,1\rangle$

Then we have: . $\begin{Bmatrix}k_1 \:=\:0 \\ 1+k_2 \:=\:0 & \Rightarrow & k_2 \:=\:\text{-}1 \\ k_3\:=\:1 \end{Bmatrix}$

Therefore: . $r(t) \;=\;\langle \text{-}\sin t\!+\!2t,\;e^t\!-\!t\!-\!1,\;2t^3\!+\!1 \rangle$