Problem 50 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

At \( t = 2 \), the tangential component \( a_T = \frac{8}{\sqrt{17}} \) and the normal component \( a_N = \frac{6}{\sqrt{17}} \).

Step 1: Find the Velocity Vector

First, we compute the velocity vector \( \mathbf{v}(t) \) by differentiating the position vector \( \mathbf{r}(t) \) with respect to time \( t \). The position vector is \( \mathbf{r}(t)=(t-2)^{2} \mathbf{i}-t^{2} \mathbf{j}+t \mathbf{k} \). Differentiating, we find:\[ \mathbf{v}(t) = \frac{d}{dt}((t-2)^{2}) \mathbf{i} + \frac{d}{dt}(-t^{2}) \mathbf{j} + \frac{d}{dt}(t) \mathbf{k} = 2(t-2) \mathbf{i} - 2t \mathbf{j} + \mathbf{k}. \]

Step 2: Find the Acceleration Vector

Next, we differentiate the velocity vector \( \mathbf{v}(t) \) to get the acceleration vector \( \mathbf{a}(t) \). Differentiating \( \mathbf{v}(t) = 2(t-2) \mathbf{i} - 2t \mathbf{j} + \mathbf{k} \), we have:\[ \mathbf{a}(t) = \frac{d}{dt}(2(t-2)) \mathbf{i} + \frac{d}{dt}(-2t) \mathbf{j} + \frac{d}{dt}(1) \mathbf{k} = 2 \mathbf{i} - 2 \mathbf{j}. \]

Step 3: Find the Magnitude of the Velocity

Calculate the magnitude of the velocity vector, \( ||\mathbf{v}(t)|| \), which is needed to find the tangential and normal components. We have:\[ ||\mathbf{v}(t)|| = \sqrt{(2(t-2))^2 + (-2t)^2 + 1^2} = \sqrt{4(t-2)^2 + 4t^2 + 1}. \]

Step 4: Evaluate at \( t = t_1 = 2 \)

Plug \( t = 2 \) into \( \mathbf{v}(t) \), \( \mathbf{a}(t) \), and \( ||\mathbf{v}(t)|| \):- \( \mathbf{v}(2) = 2(2-2) \mathbf{i} - 2(2) \mathbf{j} + \mathbf{k} = -4 \mathbf{j} + \mathbf{k} \).- \( \mathbf{a}(2) = 2 \mathbf{i} - 2 \mathbf{j} \). - \( ||\mathbf{v}(2)|| = \sqrt{4 \cdot 0^2 + 4 \cdot 2^2 + 1} = \sqrt{16 + 1} = \sqrt{17} \).

Step 5: Calculate the Tangential Component \( a_T \)

The tangential component of acceleration, \( a_T \), is given by the formula:\[ a_T = \frac{\mathbf{v} \cdot \mathbf{a}}{||\mathbf{v}||}. \]At \( t = 2 \), we calculate:\( \mathbf{v}(2) \cdot \mathbf{a}(2) = (0 \mathbf{i} - 4 \mathbf{j} + \mathbf{k}) \cdot (2 \mathbf{i} - 2 \mathbf{j} + 0 \mathbf{k}) = 0 - 4(-2) + 0 = 8. \)Thus, \( a_T = \frac{8}{\sqrt{17}}. \)

Step 6: Calculate the Normal Component \( a_N \)

The normal component of acceleration, \( a_N \), is found using the formula:\[ a_N = \sqrt{||\mathbf{a}||^2 - a_T^2}. \]First, find \( ||\mathbf{a}(2)|| \):\[ ||\mathbf{a}(2)|| = \sqrt{2^2 + (-2)^2} = 2\sqrt{2}. \]Then, calculate \( a_N \):\[ a_N = \sqrt{(2\sqrt{2})^2 - \left(\frac{8}{\sqrt{17}}\right)^2} = \sqrt{8 - \frac{64}{17}} = \sqrt{\frac{72}{17}} = \frac{6}{\sqrt{17}}. \]

Key Concepts

Tangential Component of Acceleration

The tangential component of acceleration, often denoted as \( a_T \), is an essential part of understanding motion in calculus. It refers to the rate at which the speed of an object changes along its path. In other words, it is the component of acceleration that acts in the direction of the velocity vector.This component can be computed using the formula:

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

Here, \( \mathbf{v} \) is the velocity vector, \( \mathbf{a} \) is the acceleration vector, and \( ||\mathbf{v}|| \) is the magnitude of the velocity.Understanding \( a_T \) is crucial for analyzing how an object's velocity changes over time. For instance, if \( a_T \) is positive, the object's speed is increasing, while a negative \( a_T \) indicates a decrease in speed. Calculating this component helps in comprehensively analyzing the dynamics of moving objects.

Normal Component of Acceleration

The normal component of acceleration, identified as \( a_N \), is as important as the tangential component for a comprehensive understanding of acceleration. Unlike the tangential component, \( a_N \) is responsible for changing the direction of velocity rather than its magnitude.In order to find \( a_N \), use the formula:

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

This formula helps us differentiate between linear and angular changes within a motion.One can interpret \( a_N \) as describing how rapidly an object changes direction as it moves along its path. A larger \( a_N \) indicates a sharper turn or a high rate of directional change. Understanding \( a_N \) gives deeper insights into the overall behavior of an object in motion, particularly in circular or curved trajectories.

Velocity Vector in Calculus

The concept of a velocity vector is central to calculus when describing motion. A velocity vector, \( \mathbf{v}(t) \), not only states the speed of an object at any given moment but also its direction of travel.To determine the velocity vector, differentiate the position vector, \( \mathbf{r}(t) \), with respect to time \( t \):

• \( \mathbf{v}(t) = \frac{d}{dt}\mathbf{r}(t) \)

This derivative provides both components of speed (magnitude) and constant directional information within the vector form.A velocity vector is fundamental for further calculations in dynamics, such as finding acceleration by differentiating it again. Understanding velocity vectors also facilitates the visualization of motion paths and aids in predicting future motion based on current velocities.

Differentiation in Calculus

Differentiation is one of the most powerful tools in calculus. It allows one to determine the rate of change of a function with respect to a variable. In physics, this concept is essential to ascertain how motion progresses over time.When we differentiate a function, we obtain its derivative, providing information about its instantaneous rate of change or slope. For motion, differentiating the position vector \( \mathbf{r}(t) \) results in the velocity vector \( \mathbf{v}(t) \), and subsequently differentiating \( \mathbf{v}(t) \) gives the acceleration vector \( \mathbf{a}(t) \).Differentiation in calculus is indispensable for:

• Analyzing rates of change in various scenarios
• Understanding acceleration and velocity in dynamic systems
• Solving real-world problems involving motion

Mastering differentiation creates a strong foundation for further studies in mathematics and physics, offering detailed insights into the mechanics of movement.