Problem 24 - Advanced Engineering Mathematics

Question

The motion of a particle in space is described by $$ \mathbf{r}(t)=b \cos t \mathbf{i}+b \sin t \mathbf{j}+c t \mathbf{k}, \quad t \geq 0. $$ (a) Compute \(\|\mathbf{v}(t)\|\). (b) Compute \(s=\int_{0}^{t}\|\mathbf{v}(t)\| d t\) and verify that \(d s / d t\) is the same as the result of part (a). (c) Verify that \(d^{2} s / d t^{2} \neq\|\mathbf{a}(t)\|\).

Step-by-Step Solution

Answer

(a) \(\sqrt{b^2 + c^2}\); (b) \(s = \sqrt{b^2 + c^2} \cdot t\), \(ds/dt = \sqrt{b^2 + c^2}\); (c) \(d^2s/dt^2 = 0\) is not equal to \(b\).

Step 1: Find the Velocity Vector

To find the velocity vector \(\mathbf{v}(t)\), take the derivative of the position vector \(\mathbf{r}(t)\) with respect to time \(t\). The position function is \(\mathbf{r}(t)=b \cos t \mathbf{i}+b \sin t \mathbf{j}+c t \mathbf{k}\). Differentiating each component, we get:\[\mathbf{v}(t) = \frac{d}{dt}[b \cos t] \mathbf{i} + \frac{d}{dt}[b \sin t] \mathbf{j} + \frac{d}{dt}[ct] \mathbf{k} = -b \sin t \mathbf{i} + b \cos t \mathbf{j} + c \mathbf{k}\]

Step 2: Compute Norm of Velocity

The magnitude (norm) of the velocity vector \(\mathbf{v}(t)\) is given by:\[\|\mathbf{v}(t)\| = \sqrt{(-b \sin t)^2 + (b \cos t)^2 + c^2}\]Calculating the expression inside the square root, we have:\[(-b \sin t)^2 + (b \cos t)^2 = b^2 \sin^2 t + b^2 \cos^2 t = b^2 (\sin^2 t + \cos^2 t) = b^2\]Therefore,\[\|\mathbf{v}(t)\| = \sqrt{b^2 + c^2}\]

Step 3: Integrate Velocity Norm

Compute the integral of the velocity norm from 0 to \(t\):\[s = \int_{0}^{t}\|\mathbf{v}(t)\| \, dt = \int_{0}^{t}\sqrt{b^2 + c^2} \, dt = \sqrt{b^2 + c^2} \int_{0}^{t}dt = \sqrt{b^2 + c^2} \cdot t\]

Step 4: Differentiate Arc Length with Respect to Time

Take the derivative of \(s\) with respect to \(t\):\[\frac{ds}{dt} = \frac{d}{dt}\left(\sqrt{b^2 + c^2} \cdot t\right) = \sqrt{b^2 + c^2}\]This matches the result from Step 2, showing that \(\frac{ds}{dt} = \|\mathbf{v}(t)\|\).

Step 5: Find the Acceleration Vector

Differentiate the velocity vector \(\mathbf{v}(t)\) to find the acceleration vector \(\mathbf{a}(t)\):\[\mathbf{a}(t) = \frac{d}{dt}[-b \sin t] \mathbf{i} + \frac{d}{dt}[b \cos t] \mathbf{j} + \frac{d}{dt}[c] \mathbf{k} = -b \cos t \mathbf{i} - b \sin t \mathbf{j}\]

Step 6: Compute Norm of Acceleration

The magnitude of the acceleration vector \(\mathbf{a}(t)\) is given by:\[\|\mathbf{a}(t)\| = \sqrt{(-b \cos t)^2 + (-b \sin t)^2} = \sqrt{b^2 \cos^2 t + b^2 \sin^2 t} = \sqrt{b^2} = b\]

Step 7: Verify Second Derivative of Arc Length

Since the expression for \(\frac{ds}{dt}\) is constant \(\sqrt{b^2 + c^2}\), its second derivative is zero:\[\frac{d^2 s}{dt^2} = 0\]Compare this with \(\|\mathbf{a}(t)\| = b\), confirming \(\frac{d^2 s}{dt^2} eq \|\mathbf{a}(t)\|\).

Key Concepts

Velocity Vector

In the study of particle motion, the velocity vector is a crucial component. It describes the rate of change of a particle's position with respect to time. The velocity vector, given by \( \mathbf{v}(t) \), is the derivative of the position vector \( \mathbf{r}(t) \). For the provided motion description \( \mathbf{r}(t) = b \cos t \mathbf{i} + b \sin t \mathbf{j} + ct \mathbf{k} \), the velocity vector becomes:

• \( \frac{d}{dt}[b \cos t] \mathbf{i} = -b \sin t \mathbf{i} \)
• \( \frac{d}{dt}[b \sin t] \mathbf{j} = b \cos t \mathbf{j} \)
• \( \frac{d}{dt}[ct] \mathbf{k} = c \mathbf{k} \)

Thus, the resultant velocity vector is \( \mathbf{v}(t) = -b \sin t \mathbf{i} + b \cos t \mathbf{j} + c \mathbf{k} \). This vector not only indicates the direction of motion but also the speed at which the particle is travelling along its path.
The magnitude, or norm, of this velocity vector, \( \|\mathbf{v}(t)\| \), is calculated as the square root of the sum of the squares of its components. In this case:\[\|\mathbf{v}(t)\| = \sqrt{(-b \sin t)^2 + (b \cos t)^2 + c^2} = \sqrt{b^2 + c^2}\]This result demonstrates that the speed is constant, determined by \( b^2 + c^2 \). Understanding the velocity vector is fundamental in analyzing how fast and in what direction particles are moving.

Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) measures how the velocity of a particle changes over time. It is the derivative of the velocity vector, \( \mathbf{v}(t) \). For the given motion, after determining the velocity vector \( \mathbf{v}(t) = -b \sin t \mathbf{i} + b \cos t \mathbf{j} + c \mathbf{k} \), we find the acceleration vector by differentiating each of its components:

• \( \frac{d}{dt}[-b \sin t] \mathbf{i} = -b \cos t \mathbf{i} \)
• \( \frac{d}{dt}[b \cos t] \mathbf{j} = -b \sin t \mathbf{j} \)
• \( \frac{d}{dt}[c] \mathbf{k} = 0 \)

Thus, the acceleration vector is \( \mathbf{a}(t) = -b \cos t \mathbf{i} - b \sin t \mathbf{j} \).
This calculation shows the acceleration affects only the x and y components, not the z component.
The magnitude of this acceleration vector is:\[\|\mathbf{a}(t)\| = \sqrt{(-b \cos t)^2 + (-b \sin t)^2} = b\]This constant magnitude indicates that the particle's velocity is changing at a constant rate with respect to the components \( x \) and \( y \). Understanding acceleration helps illustrate how forces might be acting on the moving particle.

Arc Length

Arc length is the total distance a particle travels along its path over a certain period. The formula for arc length \( s \) in this context, integrates the velocity magnitude over time. Specifically, it is measured as:\[ s = \int_{0}^{t} \|\mathbf{v}(t)\| \, dt = \int_{0}^{t} \sqrt{b^2 + c^2} \, dt = \sqrt{b^2 + c^2} \cdot t\]This integration reveals the arc length to be a constant velocity multiple of time, showing that the distance covered is directly proportional to the time.
Computing the derivative of arc length with respect to time \( \frac{ds}{dt} \) yields the speed \( \sqrt{b^2 + c^2} \), confirming it is consistent with the velocity magnitude calculated earlier.
Upon further differentiation \( \frac{d^2s}{dt^2} \), the result is zero, showing the speed is uniform as expected.
Understanding arc length is critical in determining how far a particle travels, which can also assist in predicting future position based on current speed and direction.