Problem 8 - Advanced Engineering Mathematics

Question

In Problems \(7-16, \mathbf{r}(t)\) is the position vector of a moving particle. Find the tangential and normal components of the acceleration at any \(t\). $$ \mathbf{r}(t)=3 \cos t \mathbf{i}+2 \sin t \mathbf{j}+t \mathbf{k} $$

Step-by-Step Solution

Answer

Tangential component: \(a_T = \frac{5 \sin t \cos t}{\sqrt{5 + 5\sin^2 t}}\), normal component: \(a_N = \sqrt{9 \cos^2 t + 4 \sin^2 t - \frac{25 \cos^2 t \sin^2 t}{5 + 5\sin^2 t}}\).

Step 1: Find Velocity Vector

To find the velocity vector, differentiate the position vector \( \mathbf{r}(t) \) with respect to \( t \).\[ \mathbf{v}(t) = \frac{d}{dt} [3 \cos t \, \mathbf{i} + 2 \sin t \, \mathbf{j} + t \, \mathbf{k}] = -3 \sin t \, \mathbf{i} + 2 \cos t \, \mathbf{j} + \mathbf{k} \]

Step 2: Find Acceleration Vector

Differentiate the velocity vector \( \mathbf{v}(t) \) to find the acceleration vector \( \mathbf{a}(t) \).\[ \mathbf{a}(t) = \frac{d}{dt} [-3 \sin t \, \mathbf{i} + 2 \cos t \, \mathbf{j} + \mathbf{k}] = -3 \cos t \, \mathbf{i} - 2 \sin t \, \mathbf{j} \]

Step 3: Magnitude of Velocity

Calculate the magnitude of the velocity vector, \( \|\mathbf{v}(t)\| \).\[ \|\mathbf{v}(t)\| = \sqrt{(-3 \sin t)^2 + (2 \cos t)^2 + 1^2} = \sqrt{9 \sin^2 t + 4 \cos^2 t + 1} \]

Step 4: Calculate \(\|\mathbf{v}(t)\| \) Simplified

Further simplify the magnitude of the velocity, \(\|\mathbf{v}(t)\|\).\[ \|\mathbf{v}(t)\| = \sqrt{9 \sin^2 t + 4 \cos^2 t + 1} = \sqrt{5 + 5\sin^2 t} \]

Step 5: Tangential Component of Acceleration

The tangential component \( a_T \) is given by \( a_T = \frac{\mathbf{a}(t) \cdot \mathbf{v}(t)}{\|\mathbf{v}(t)\|} \).Calculate the dot product: \[ \mathbf{a}(t) \cdot \mathbf{v}(t) = (-3 \cos t)(-3 \sin t) + (-2 \sin t)(2 \cos t) = 9 \cos t \sin t - 4 \sin t \cos t = 5 \cos t \sin t \]Thus, \[ a_T = \frac{5 \sin t \cos t}{\sqrt{5 + 5\sin^2 t}} \]

Step 6: Normal Component of Acceleration

The normal component \( a_N \) can be found using the formula \[ a_N = \sqrt{ \|\mathbf{a}(t)\|^2 - a_T^2 } \]First, find \( \|\mathbf{a}(t)\| \):\[ \|\mathbf{a}(t)\| = \sqrt{(-3 \cos t)^2 + (-2 \sin t)^2} = \sqrt{9 \cos^2 t + 4 \sin^2 t} \]Now calculate \( a_N \):\[ a_N = \sqrt{9 \cos^2 t + 4 \sin^2 t - \left( \frac{5 \cos t \sin t}{\sqrt{5 + 5\sin^2 t}} \right)^2} \]

Step 7: Simplify \(a_N\)

Further simplify the expression for \( a_N \):Compute\[ \left( \frac{5 \cos t \sin t}{\sqrt{5 + 5\sin^2 t}} \right)^2 = \frac{25 \cos^2 t \sin^2 t}{5 + 5\sin^2 t} \]Then\[ a_N = \sqrt{9 \cos^2 t + 4 \sin^2 t - \frac{25 \cos^2 t \sin^2 t}{5 + 5\sin^2 t}} \]

Key Concepts

Position Vector

The position vector \( \mathbf{r}(t) \) describes the location of a particle in space relative to the origin at any time \( t \). It's a fundamental concept in vector calculus. For our exercise, the position vector is given by \( \mathbf{r}(t) = 3 \cos t \mathbf{i} + 2 \sin t \mathbf{j} + t \mathbf{k} \). This equation uses parameter \( t \) to indicate motion:

• \( 3 \cos t \mathbf{i} \) establishes the x-coordinate movement with time, creating a wave-like motion on a horizontal plane.
• \( 2 \sin t \mathbf{j} \) shows the y-coordinate, resembling the vertical component's periodic movement.
• \( t \mathbf{k} \) reveals the z-coordinate's growth at a constant rate, indicative of uniform motion along the z-axis.

The position vector simply combines these into a path that the particle follows, making it crystal clear where the particle is at any given \( t \).

Velocity Vector

The velocity vector \( \mathbf{v}(t) \) represents the rate of change of the position vector, showing the particle's speed and direction at any point in time. To find it, we differentiate the position vector \( \mathbf{r}(t) \) with respect to \( t \): \[ \mathbf{v}(t) = \frac{d}{dt} [3 \cos t \, \mathbf{i} + 2 \sin t \, \mathbf{j} + t \, \mathbf{k}] = -3 \sin t \, \mathbf{i} + 2 \cos t \, \mathbf{j} + \mathbf{k} \]Interpretation of each component:

• \(-3 \sin t \mathbf{i} \) drives the horizontal change, pointing in the opposite direction of x.
• \(2 \cos t \mathbf{j} \) dictates vertical speed, modulating based on the sine wave graph.
• \( \mathbf{k} \) symbolizes uniform speed in the z-direction.

The vector's direction constantly changes, following the path prescribed by the original function. Examining the velocity vector helps in understanding how the particle's position changes with time.

Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is found by differentiating the velocity vector, telling us how the velocity changes over time. This provides insight into any alterations in the particle’s velocity, such as speeding up or slowing down:\[ \mathbf{a}(t) = \frac{d}{dt} [-3 \sin t \, \mathbf{i} + 2 \cos t \, \mathbf{j} + \mathbf{k}] = -3 \cos t \, \mathbf{i} - 2 \sin t \, \mathbf{j} \]Explaining each component:

• The \( -3 \cos t \mathbf{i} \) term shows periodic acceleration oppositely along the x-axis.
• The \(-2 \sin t \mathbf{j} \) implies varying acceleration vertically, oscillating with sine function characteristics.
• Absence of a \( \mathbf{k} \) term reflects constant speed in the z-direction with no acceleration.

Grasping the acceleration vector is essential for predicting changes in velocity due to forces like gravity or friction. This vector provides key tools to analyze dynamics in space or on a plane.

Tangential Component

The tangential component of acceleration, denoted as \( a_T \), measures the portion of acceleration that causes the speed of the particle along its path to change. Calculating \( a_T \) involves using the dot product of acceleration and velocity vectors divided by the magnitude of velocity:\[ a_T = \frac{\mathbf{a}(t) \cdot \mathbf{v}(t)}{\|\mathbf{v}(t)\|} \]In practice:\[ a_T = \frac{5 \sin t \cos t}{\sqrt{5 + 5\sin^2 t}} \]Key elements in understanding the solution:

• The dot product \( \mathbf{a}(t) \cdot \mathbf{v}(t) = 5 \sin t \cos t \) reveals interaction between acceleration and velocity.
• \( \|\mathbf{v}(t)\| \) is the velocity magnitude which standardizes the effect of acceleration to the path's tangent.

By focusing on this component, one can evaluate how acceleration impacts the particle's speed along its path, crucial for understanding energy changes.

Normal Component

The normal component of acceleration, \( a_N \), describes how the particle's path curves, pertaining to the change in direction. It is computed using:\[ a_N = \sqrt{ \|\mathbf{a}(t)\|^2 - a_T^2 } \]Breaking it down:

• First find \( \|\mathbf{a}(t)\| = \sqrt{9 \cos^2 t + 4 \sin^2 t} \), giving the overall magnitude of acceleration.
• Subtract the squared tangential component \( a_T^2 \), calculated previously.

The result,:\[ a_N = \sqrt{9 \cos^2 t + 4 \sin^2 t - \frac{25 \cos^2 t \sin^2 t}{5 + 5\sin^2 t}} \]This component is critical in understanding centripetal forces and how they cause the path to bend, offering valuable insight into motion kinetics especially on curved tracks.