91Ó°ÊÓ

Velocity is the rate at which an object changes its position. It's a vector quantity, meaning it has both magnitude and direction.
In differential calculus, to find the velocity from a position function, we take its derivative with respect to time, denoted as \( t \).
For example, in the given position function \( \mathbf{r}(t) = \cosh t \mathbf{i} + \sinh t \mathbf{j} + t \mathbf{k} \):

• \( \cosh t \) differentiates to \( \sinh t \)
• \( \sinh t \) differentiates to \( \cosh t \)
• \( t \) differentiates to \( 1 \)

Thus, the velocity function becomes \( \mathbf{v}(t) = \sinh t \mathbf{i} + \cosh t \mathbf{j} + \mathbf{k} \).
This function tells us how the position changes at any time \( t \). Each component of the vector describes the motion along the respective coordinate axes.

Acceleration is how quickly the velocity of an object is changing. Similar to velocity, it's a vector, so it includes both magnitude and direction.
Acceleration is the derivative of the velocity function.
Continuing from our velocity example, where \( \mathbf{v}(t) = \sinh t \mathbf{i} + \cosh t \mathbf{j} + \mathbf{k} \), we differentiate each part again:

• \( \sinh t \) differentiates to \( \cosh t \)
• \( \cosh t \) differentiates to \( \sinh t \)
• \( 1 \)'s derivative is \( 0 \)

The resulting acceleration function is \( \mathbf{a}(t) = \cosh t \mathbf{i} + \sinh t \mathbf{j} + 0 \mathbf{k} \).
This function indicates how the velocity vector changes over time, affecting the object's movement along different directions.

The magnitude of a vector is essentially the "length" of the vector in space. It gives us a scalar quantity that represents how far the vector extends, independent of its direction.
We find the magnitude of a velocity vector to calculate speed, which is the non-directional form of velocity.
To find the magnitude of a vector \( \mathbf{v}(t) \), such as \( \mathbf{v}(t) = \sinh t \mathbf{i} + \cosh t \mathbf{j} + \mathbf{k} \), use the formula:
\[|| \mathbf{v}(t) || = \sqrt{(\sinh t)^2 + (\cosh t)^2 + 1^2}\]Utilizing the identity \( \cosh^2 t - \sinh^2 t = 1 \), the calculation simplifies to:
\[|| \mathbf{v}(t) || = \sqrt{1 + 1} = \sqrt{2}\]The magnitude is vital as it provides a measurement of how fast an object is moving, regardless of the path it follows.