91Ó°ÊÓ

In the realm of multivariate calculus, a velocity vector provides insights into the direction and rate at which a particle moves through space. It's a derivative of the position vector, indicating how the position changes over time. This is crucial in understanding motion dynamics.
When given a position vector \[ \mathbf{r}(t) = 12t \mathbf{i} + (5\sin(2t)) \mathbf{j} - (5\cos(2t)) \mathbf{k} \]one can find the velocity vector by differentiating each component with respect to time:

• The derivative of a linear term like \(12t\) is simply the coefficient, \(12\).
• For \(5\sin(2t)\), apply the chain rule to obtain \(10\cos(2t)\).
• For \(-5\cos(2t)\), using the chain rule gives you \(10\sin(2t)\).

Thus, the velocity vector is
\[ \mathbf{v}(t) = 12 \mathbf{i} + 10\cos(2t) \mathbf{j} + 10\sin(2t) \mathbf{k} \].
The velocity vector's significance extends beyond just speed; it provides the direction of motion, showing how the orientation of movement changes over time.

The acceleration vector tells how the velocity of a particle changes over time. It’s the derivative of the velocity vector with respect to time, embedding the concept of dynamic change in motion.
Given the velocity vector:\[ \mathbf{v}(t) = 12 \mathbf{i} + 10\cos(2t) \mathbf{j} + 10\sin(2t) \mathbf{k} \]the next step is differentiation:

• The derivative of \(12\) is zero, as it's constant in time.
• Differentiating \(10\cos(2t)\) produces \(-20\sin(2t)\).
• Differentiating \(10\sin(2t)\) results in \(20\cos(2t)\).

The resulting acceleration vector is
\[ \mathbf{a}(t) = 0 \mathbf{i} - 20\sin(2t) \mathbf{j} + 20\cos(2t) \mathbf{k} \].
This vector provides insight into how swiftly the particle's path curves or changes speed, capturing the essence of dynamic forces in action.

Speed, unlike velocity, is a scalar quantity and represents how fast a particle is moving regardless of its direction. To find the speed, or the magnitude of the velocity vector, involves calculating its length.
Using the velocity vector \[ \mathbf{v}(t) = 12 \mathbf{i} + 10\cos(2t) \mathbf{j} + 10\sin(2t) \mathbf{k} \]we find the magnitude:

• Compute the squared values of each component: \(12^2 = 144\), \((10\cos(2t))^2\), and \((10\sin(2t))^2\).
• Recognizing that \(\cos^2(2t) + \sin^2(2t) = 1\), simplifies the sum \((10\cos(2t))^2 + (10\sin(2t))^2 = 100\).
• Hence, the magnitude \(\|\mathbf{v}(t)\| = \sqrt{144 + 100} = \sqrt{244}\).

The resulting speed describes how quickly the particle's position changes per unit of time, providing a straightforward measure of motion intensity.