91Ó°ÊÓ

An acceleration vector describes the rate at which an object's velocity changes over time. It includes both direction and magnitude, providing a complete description of how quickly and in what direction an object is accelerating. In vector calculus, the acceleration vector is often specified as a function of time.

For example, in our problem, the acceleration vector is given by \( \mathbf{a}(t) = \langle 1, t, 4t \rangle \). Each component represents the acceleration in the x, y, and z directions, respectively.

• The first component, \(1\), is constant, indicating a steady increase in velocity along the x-axis.
• The second component, \(t\), shows a linearly increasing acceleration along the y-axis.
• The third component, \(4t\), indicates a more rapid increase along the z-axis, as the acceleration is proportional to time.
  

The velocity vector \( \mathbf{v}(t) \) is derived by integrating the acceleration vector. This vector tells us the rate of change of position with respect to time and combines both speed and direction.

After integrating the given acceleration vector \( \mathbf{a}(t) = \langle 1, t, 4t \rangle \), we obtain:

• \( \mathbf{v}(t) = \langle t + c_1, \frac{1}{2} t^2 + c_2, 2t^2 + c_3 \rangle \)

To find the specific constants of integration \( c_1, c_2, \) and \( c_3 \), initial velocity conditions are utilized. Substituting \( \langle 20, 0, 0 \rangle \), the velocity at \( t = 0 \), leads to:

• \( c_1 = 20,\)
• \( c_2 = 0,\)
• \( c_3 = 0. \)

Thus, the velocity vector simplifies as:

• \( \mathbf{v}(t) = \langle t + 20, \frac{1}{2} t^2, 2t^2 \rangle \)

The position vector \( \mathbf{r}(t) \) indicates where an object is located relative to a given starting point. This vector is obtained by integrating the velocity vector, extending our insight from how fast an object is moving to where it actually is.

Starting from \( \mathbf{v}(t) = \langle t + 20, \frac{1}{2} t^2, 2t^2 \rangle \), we integrate to find:

• \( \mathbf{r}(t) = \langle \frac{1}{2}t^2 + 20t + c_4, \frac{1}{6} t^3 + c_5, \frac{2}{3}t^3 + c_6 \rangle \)

Again, we apply initial position conditions \( \langle 0, 0, 0 \rangle \) to solve for the constants \( c_4, c_5, \) and \( c_6 \):

• \( c_4 = 0,\)
• \( c_5 = 0,\)
• \( c_6 = 0.\)

This results in the position vector:

• \( \mathbf{r}(t) = \langle \frac{1}{2}t^2 + 20t, \frac{1}{6} t^3, \frac{2}{3}t^3 \rangle \)

Integration is a fundamental process in calculus, used to find quantities such as velocity from acceleration or position from velocity. In essence, integration reverses differentiation, accumulating values over time to understand how a variable changes.

• To find the velocity vector, integrate each component of the acceleration vector with respect to time:

\[ \int \mathbf{a}(t) dt = \int \langle 1, t, 4t \rangle dt = \langle t + c_1, \frac{1}{2} t^2 + c_2, 2t^2 + c_3 \rangle \]

• Similarly, to find the position vector from the velocity vector, integrate each component with respect to time:

\[ \int \mathbf{v}(t) dt = \int \langle t + 20, \frac{1}{2} t^2, 2t^2 \rangle dt = \langle \frac{1}{2}t^2 + 20t + c_4, \frac{1}{6} t^3 + c_5, \frac{2}{3}t^3 + c_6 \rangle\]
Integration leads to solutions containing constants which are resolved using initial conditions, ensuring solutions are tailored to specific situations.