Problem 10 - Thomas Calculus

Question

\({r}(t)\) is the position of a particle in space at time \(t .\) Find the particle's velocity and acceleration vectors. Then find the particle's speed and direction of motion at the given value of \(t\) . Write the particle's velocity at that time as the product of its speed and direction. \begin{equation} \mathbf{r}(t)=(1+t) \mathbf{i}+\frac{t^{2}}{\sqrt{2}} \mathbf{j}+\frac{t^{3}}{3} \mathbf{k}, \quad t=1 \end{equation}

Step-by-Step Solution

Answer

Velocity at \( t=1 \) is \( \mathbf{v}(1) = \mathbf{i} + \sqrt{2} \mathbf{j} + \mathbf{k} \), speed is 2, direction is \( \frac{1}{2}(\mathbf{i} + \sqrt{2} \mathbf{j} + \mathbf{k}) \).

Step 1: Find the Velocity Vector

To find the velocity vector of the particle, differentiate the position vector \( \mathbf{r}(t) \) with respect to time \( t \). The velocity vector is given by \( \mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t) \). Differentiating each component, we get: \( \mathbf{v}(t) = \frac{d}{dt}((1+t) \mathbf{i}) + \frac{d}{dt}\left(\frac{t^{2}}{\sqrt{2}} \mathbf{j}\right) + \frac{d}{dt}\left(\frac{t^{3}}{3} \mathbf{k}\right) \). This simplifies to \( \mathbf{v}(t) = \mathbf{i} + \frac{2t}{\sqrt{2}} \mathbf{j} + t^2 \mathbf{k} \).

Step 2: Calculate the Acceleration Vector

The acceleration vector is found by differentiating the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \). So \( \mathbf{a}(t) = \frac{d}{dt}(\mathbf{v}(t)) \). Differentiating, we get: \( \mathbf{a}(t) = \frac{d}{dt}(\mathbf{i}) + \frac{d}{dt}\left(\frac{2t}{\sqrt{2}} \mathbf{j}\right) + \frac{d}{dt}(t^2 \mathbf{k}) \). Therefore, \( \mathbf{a}(t) = \frac{2}{\sqrt{2}} \mathbf{j} + 2t \mathbf{k} \).

Step 3: Evaluate Velocity and Acceleration at t = 1

Substitute \( t = 1 \) into the expressions for \( \mathbf{v}(t) \) and \( \mathbf{a}(t) \). For velocity, we have \( \mathbf{v}(1) = \mathbf{i} + \frac{2 \times 1}{\sqrt{2}} \mathbf{j} + 1^2 \mathbf{k} = \mathbf{i} + \sqrt{2} \mathbf{j} + \mathbf{k} \). For acceleration, \( \mathbf{a}(1) = \frac{2}{\sqrt{2}} \mathbf{j} + 2 \times 1 \mathbf{k} = \sqrt{2} \mathbf{j} + 2 \mathbf{k} \).

Step 4: Find the Speed and Direction at t = 1

The speed is the magnitude of the velocity vector at \( t = 1 \). Compute: \|\mathbf{v}(1)\| = \sqrt{ (1)^2 + (\sqrt{2})^2 + (1)^2 } = \sqrt{1 + 2 + 1} = \sqrt{4} = 2.\ The direction of the motion is the unit vector in the direction of \( \mathbf{v}(1) \), which is \( \frac{1}{2}(\mathbf{i} + \sqrt{2} \mathbf{j} + \mathbf{k}) \).

Step 5: Express Velocity as the Product of Speed and Direction

The velocity \( \mathbf{v}(1) \) can be expressed as the product of its speed and direction. We have: \( \mathbf{v}(1) = 2 \times \left( \frac{1}{2}(\mathbf{i} + \sqrt{2} \mathbf{j} + \mathbf{k}) \right) = \mathbf{i} + \sqrt{2} \mathbf{j} + \mathbf{k} \), which matches \( \mathbf{v}(1) \) from earlier.

Key Concepts

Velocity Vector

In particle motion, the velocity vector is crucial because it reveals both how fast and in what direction the particle is moving at a particular moment. To determine the velocity vector, we differentiate the position vector with respect to time. For our position vector \( \mathbf{r}(t) = (1+t) \mathbf{i} + \frac{t^2}{\sqrt{2}} \mathbf{j} + \frac{t^3}{3} \mathbf{k} \), the velocity vector \( \mathbf{v}(t) \) becomes the derivative:

• Differentiating \((1+t) \mathbf{i}\), we obtain \(\mathbf{i}\).
• Differentiating \(\frac{t^2}{\sqrt{2}} \mathbf{j}\), yields \(\frac{2t}{\sqrt{2}} \mathbf{j}\).
• Differentiating \(\frac{t^3}{3} \mathbf{k}\), results in \(t^2 \mathbf{k}\).

The cumulative velocity vector is \( \mathbf{v}(t) = \mathbf{i} + \frac{2t}{\sqrt{2}} \mathbf{j} + t^2 \mathbf{k} \). When calculated for \( t = 1 \), the velocity is \( \mathbf{i} + \sqrt{2} \mathbf{j} + \mathbf{k} \). This vector provides both the magnitude and direction of the particle's movement.

Acceleration Vector

Acceleration provides insight into how the velocity of a particle changes over time. To find the acceleration vector, differentiate the velocity vector with respect to time. From our step-by-step solution:

• The velocity vector components are \(\mathbf{i} + \frac{2t}{\sqrt{2}} \mathbf{j} + t^2 \mathbf{k} \).
• Differentiating \(\mathbf{i}\) results in zero since it's a constant.
• Differentiating \(\frac{2t}{\sqrt{2}} \mathbf{j}\) gives \(\frac{2}{\sqrt{2}} \mathbf{j}\).
• Differentiating \(t^2 \mathbf{k}\) yields \(2t \mathbf{k}\).

The acceleration vector \( \mathbf{a}(t) = \frac{2}{\sqrt{2}} \mathbf{j} + 2t \mathbf{k} \) shows that changes in velocity occur along the \( \mathbf{j} \) and \( \mathbf{k} \) directions. At \( t = 1 \), the acceleration is \( \mathbf{a}(1) = \sqrt{2} \mathbf{j} + 2\mathbf{k} \), indicating the rate and direction of change in velocity.

Speed and Direction

Speed, in physics, refers to how fast an object is moving irrespective of its direction, while the velocity vector has both magnitude (speed) and direction. To find speed, we calculate the magnitude of the velocity vector:

• For \( \mathbf{v}(1) = \mathbf{i} + \sqrt{2} \mathbf{j} + \mathbf{k} \), its magnitude is \( \| \mathbf{v}(1) \| = \sqrt{ (1)^2 + (\sqrt{2})^2 + (1)^2 } = 2 \).

The speed is thus \( 2 \). To find the direction, normalize the velocity vector by dividing it by its magnitude:

• The direction is \( \frac{1}{2}(\mathbf{i} + \sqrt{2} \mathbf{j} + \mathbf{k}) \).

This unit vector retains the original vector's direction but scales it to a magnitude of one. This way, the motion's direction is distinct from the speed which signifies how fast the particle moves.

Position Function

The position function provides the particle’s location in space as a function of time. It describes the trajectory that the particle follows, helping us visualize its path. In our example, the position vector is given by \( \mathbf{r}(t) = (1+t) \mathbf{i} + \frac{t^2}{\sqrt{2}} \mathbf{j} + \frac{t^3}{3} \mathbf{k} \). Each component of this vector indicates movement along the \( x \), \( y \), and \( z \) axes respectively.
By differentiating this position function, we move from finding location to determining velocity, revealing how the particle's position changes over time. Differentiating again, we obtain the acceleration, elucidating how the velocity itself changes. Thus, the position function is foundational, serving as the basis from which both the velocity and acceleration vectors are derived.