91Ó°ÊÓ

In calculus, the velocity vector of a moving object is crucial for understanding its dynamics. It tells us the rate and direction at which an object is shifting its position. To find the velocity vector, \(\mathbf{v}(t)\), we take the derivative of the position vector \(\mathbf{r}(t)\) with respect to time \(t\). This derivative represents the instantaneous rate of change of the object's position vector. For the given problem, the position vector was \(\mathbf{r}(t)=\left\langle e^{t} \cos t, e^{t} \sin t, e^{t}\right\rangle\). Upon differentiation, we apply the product and chain rules of calculus because the position components involve products of functions.

• The x-component: \(\frac{d}{dt}(e^t \cos t) = e^t \cos t - e^t \sin t\).
• The y-component: \(\frac{d}{dt}(e^t \sin t) = e^t \sin t + e^t \cos t\).
• The z-component: \(\frac{d}{dt}(e^t) = e^t\).

Thus, the velocity vector is \(< e^t\cos t - e^t\sin t, e^t\sin t + e^t\cos t, e^t>\). It gives a clear insight into how each component of the position changes over time.

Acceleration describes how the velocity of an object changes over time, and it's depicted by the acceleration vector \(\mathbf{a}(t)\). To find it, we differentiate the velocity vector \(\mathbf{v}(t)\) obtained earlier. This provides a detailed view of the acceleration in each direction as the object moves through its trajectory.The given velocity vector was \(\mathbf{v}(t)=\left\langle e^t\cos t - e^t\sin t, e^t\sin t + e^t\cos t, e^t \right\rangle\). By further applying the chain rule:

• The x-component accelerates as \(\frac{d}{dt}(e^t\cos t - e^t\sin t) = -2e^t\sin t\).
• The y-component accelerates as \(\frac{d}{dt}(e^t\sin t + e^t\cos t) = 2e^t\cos t\).
• The z-component is consistent with \(\frac{d}{dt}(e^t) = e^t\).

Therefore, the acceleration vector is \(< -2e^t\sin t, 2e^t\cos t, e^t >\). It fundamentally tells us the rates at which the components of velocity change – indicating how the motion's nature in terms of speed and direction is varying.

Breaking down the acceleration into tangential and normal components is a method to analyze an object's motion along a path. The tangential component, \(a_T\), reflects the rate of change of speed along the trajectory, while the normal component, \(a_N\), represents the rate of directional change, or how the path is curving.To compute these:The tangential component is found by projecting the acceleration vector onto the tangent vector \(\mathbf{T}(t)\) – which is the normalized velocity vector. For this problem:

• Calculate \(\mathbf{T}(t) = \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|} = \frac{1}{\sqrt{2}}\langle\cos t - \sin t, \sin t + \cos t, 1 \rangle\).
• The tangential component is then \(a_T = \mathbf{a}(t) \cdot \mathbf{T}(t) = e^t\).

The normal component gives insight into how sharply the path is curving and is calculated by:

• Using \(a_N = \sqrt{\|\mathbf{a}(t)\|^2 - a_T^2} = e^t\).

Thus, for this specific trajectory, both the tangential and normal components have identical magnitudes \(e^t\), indicating uniform aspects of both speed and curvature along its motion.