Question

find the tangential and normal components \(\left(a_{T}\right.\) and \(\left.a_{N}\right)\) of the acceleration vector at \(t .\) Then evaluate at \(t=t_{1} .\) $$ \mathbf{r}(t)=(t-2)^{2} \mathbf{i}-t^{2} \mathbf{j}+t \mathbf{k} ; t_{1}=2 $$

Short answer

At \( t = 2 \), \( a_T = \frac{8}{\sqrt{17}} \) and \( a_N = \sqrt{\frac{72}{17}} \).

Step-by-step solution

Find the Velocity Vector

The velocity vector, \( \mathbf{v}(t) \), is the first derivative of the position vector \( \mathbf{r}(t) \) with respect to \( t \). Compute this by differentiating each component of \( \mathbf{r}(t) \).\[\mathbf{v}(t) = \frac{d}{dt}[(t-2)^2 \mathbf{i} - t^2 \mathbf{j} + t \mathbf{k}] = 2(t-2)\mathbf{i} - 2t\mathbf{j} + \mathbf{k}\]

Find the Acceleration Vector

The acceleration vector, \( \mathbf{a}(t) \), is the derivative of the velocity vector \( \mathbf{v}(t) \). Differentiate each component of \( \mathbf{v}(t) \) with respect to \( t \).\[\mathbf{a}(t) = \frac{d}{dt}[2(t-2)\mathbf{i} - 2t\mathbf{j} + \mathbf{k}] = 2\mathbf{i} - 2\mathbf{j}\]

Calculate the Magnitude of the Velocity Vector

To find the tangential component, we need the magnitude of the velocity vector \( \mathbf{v}(t) \).The magnitude is given by:\[|\mathbf{v}(t)| = \sqrt{(2(t-2))^2 + (-2t)^2 + 1^2} = \sqrt{4(t-2)^2 + 4t^2 + 1}\]

Tangential Component of Acceleration

The tangential component \( a_T \) is given by the dot product of the acceleration vector and the unit vector of velocity:\[a_T = \frac{\mathbf{a}(t) \cdot \mathbf{v}(t)}{|\mathbf{v}(t)|}\]First, calculate the dot product:\[\mathbf{a}(t) \cdot \mathbf{v}(t) = (2\mathbf{i} - 2\mathbf{j}) \cdot (2(t-2)\mathbf{i} - 2t\mathbf{j} + \mathbf{k}) = 4(t-2) + 4t = 8t - 8\]Thus, \(a_T = \frac{8t - 8}{|\mathbf{v}(t)|}\). Substitute \( |\mathbf{v}(t)| \) from Step 3.

Normal Component of Acceleration

The normal component of acceleration \( a_N \) can be calculated using:\[a_N = \sqrt{|\mathbf{a}(t)|^2 - a_T^2}\]First, find \(|\mathbf{a}(t)|\):\[|\mathbf{a}(t)| = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}\]Then substitute \( a_T \) from Step 4.

Evaluate at \( t = t_1 = 2 \)

First, identify \( t_1 = 2 \) and calculate \( |\mathbf{v}(t_1)|, a_T(t_1), \) and \( a_N(t_1) \).Calculate \[|\mathbf{v}(2)| = \sqrt{4(0)^2 + 4(2)^2 + 1} = \sqrt{17}\]Substitute into \(a_T\): \[a_T = \frac{8(2) - 8}{\sqrt{17}} = \frac{8}{\sqrt{17}}\]Calculate \(a_N\): \[a_N = \sqrt{(2\sqrt{2})^2 - \left(\frac{8}{\sqrt{17}}\right)^2} = \sqrt{8 - \frac{64}{17}} = \sqrt{\frac{72}{17}}\]

Key concepts

Tangential Component of Acceleration

The tangential component of acceleration, often denoted as \( a_T \), measures how much the velocity vector is increasing or decreasing in magnitude along the curve of motion. It is concerned only with changes in speed, not the direction of the velocity.

In the given exercise, you're first tasked with finding the velocity vector \( \mathbf{v}(t) \) by differentiating the position vector \( \mathbf{r}(t) \). Once you have the velocity vector, calculate the magnitude \( |\mathbf{v}(t)| \), which is crucial for determining \( a_T \).

Next, the acceleration vector \( \mathbf{a}(t) \) is obtained by differentiating \( \mathbf{v}(t) \). With both \( \mathbf{a}(t) \) and \( \mathbf{v}(t) \) at hand, the tangential component is calculated using the dot product formula:

• \( a_T = \frac{\mathbf{a}(t) \cdot \mathbf{v}(t)}{|\mathbf{v}(t)|} \)

This formula showcases how \( \mathbf{a}(t) \) aligns or opposes \( \mathbf{v}(t) \), impacting how the speed changes along the path.

Evaluation at \( t = t_1 = 2 \) involves substituting into the expression for \( a_T \), providing specific insights at that point on the curve.

Normal Component of Acceleration

The normal component of acceleration, denoted as \( a_N \), captures how the acceleration vector causes a change in direction, keeping track of how sharply or gently the curve is turning.

Unlike \( a_T \), the normal component is independent of speed changes and purely represents altering direction. With \( \mathbf{a}(t) \) and \( a_T \) known, you calculate \( a_N \) using:

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

This formula essentially decomposes the length (magnitude) of the total acceleration vector into tangential and normal parts.

By substituting the computed \( a_T \), along with using the magnitude of \( \mathbf{a}(t) \), you zero in on how the curve twists or arcs at any moment, such as at \( t = t_1 = 2 \), offering critical insight into the motion's geometry.

Velocity Vector

The velocity vector, \( \mathbf{v}(t) \), acts as a vital descriptor of the motion defined by its speed and direction at any given time. It points in the direction of movement along a curve and describes how fast the position changes.

In the problem, deriving \( \mathbf{v}(t) \) involves differentiating each component of the position vector \( \mathbf{r}(t) \). The result represents how the system's location alters over time, componet-wise:

• \( \mathbf{v}(t) = 2(t-2)\mathbf{i} - 2t\mathbf{j} + \mathbf{k} \)

This computed vector captures both direction and speed changes in the whole 3D space.

Understanding \( \mathbf{v}(t) \) is foundational because it paves the way to assess other aspects, like acceleration or curvature, by directly impacting the tangential and normal components of motion.

Acceleration Vector

The acceleration vector, \( \mathbf{a}(t) \), is all about how the velocity vector changes over time. It's not just about getting faster or slower; it's also about changing direction.

You find \( \mathbf{a}(t) \) by differentiating the velocity vector \( \mathbf{v}(t) \). This second derivative captures the nuances of motion dynamics and includes all effects impacting an object's speed and path:

• \( \mathbf{a}(t) = 2\mathbf{i} - 2\mathbf{j} \)

The acceleration vector stands perpendicular to the velocity vector when the motion is purely curvilinear.

Clearly understanding \( \mathbf{a}(t) \) allows one to distill it into tangential and normal components, each shedding light on different facets of dynamic behavior at any point in time, such as at \( t = t_1 = 2 \). It leads to deeper interpretations of how external forces or constraints interact with motion dynamics.