Problem 37 - Calculus Early Transcendentals

Question

\(33-38\) Find the tangential and normal components of the acceleration vector. $$\mathbf{r}(t)=e^{t} \mathbf{i}+\sqrt{2} t \mathbf{j}+e^{-t} \mathbf{k}$$

Step-by-Step Solution

Answer

Tangential and Normal components are \( a_T \) and \( a_N \) respectively as calculated.

Step 1: Find Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to \( t \). So, compute the derivative:\[ \mathbf{v}(t) = \frac{d}{dt}(e^t \mathbf{i} + \sqrt{2}t \mathbf{j} + e^{-t} \mathbf{k}) = e^t \mathbf{i} + \sqrt{2} \mathbf{j} - e^{-t} \mathbf{k}\]

Step 2: Find Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \). So, compute the derivative:\[ \mathbf{a}(t) = \frac{d}{dt}(e^t \mathbf{i} + \sqrt{2} \mathbf{j} - e^{-t} \mathbf{k}) = e^t \mathbf{i} + e^{-t} \mathbf{k}\]

Step 3: Magnitude of Velocity Vector

Compute the magnitude of the velocity vector \( \mathbf{v}(t) \):\[ \| \mathbf{v}(t) \| = \sqrt{(e^t)^2 + (\sqrt{2})^2 + (-e^{-t})^2} = \sqrt{e^{2t} + 2 + e^{-2t}}\]

Step 4: Tangential Component of Acceleration

The tangential component of acceleration \( a_T \) is given by the derivative of the magnitude of the velocity:\[ a_T = \frac{d}{dt}\| \mathbf{v}(t) \|\]Compute by differentiating the magnitude expression:\[ a_T = \frac{d}{dt} \sqrt{e^{2t} + 2 + e^{-2t}} = \frac{e^{2t} - e^{-2t}}{\sqrt{e^{2t} + 2 + e^{-2t}}}\]

Step 5: Normal Component of Acceleration

The normal component of acceleration \( a_N \) can be computed using the formula:\[ a_N = \sqrt{\| \mathbf{a}(t) \|^2 - a_T^2}\]First, compute the magnitude of the acceleration vector \( \mathbf{a}(t) \):\[ \| \mathbf{a}(t) \| = \sqrt{(e^t)^2 + (e^{-t})^2} = \sqrt{e^{2t} + e^{-2t}}\]Then compute \( a_N \):\[ a_N = \sqrt{e^{2t} + e^{-2t} - \left(\frac{e^{2t} - e^{-2t}}{\sqrt{e^{2t} + 2 + e^{-2t}}}\right)^2}\]

Key Concepts

Tangential Components

The tangential component of acceleration refers to how much of an object's acceleration is directed along the path or trajectory. In simpler terms, it indicates how fast the object speeds up or slows down along its path.

• To find this, first determine the magnitude of the velocity vector.
• The tangential component, denoted as \( a_T \), is calculated by taking the derivative of this magnitude.

For our problem, the velocity magnitude \( \| \mathbf{v}(t) \| = \sqrt{e^{2t} + 2 + e^{-2t}} \).
We then derive this function with respect to time \( t \) to get \( a_T = \frac{d}{dt} \sqrt{e^{2t} + 2 + e^{-2t}} = \frac{e^{2t} - e^{-2t}}{\sqrt{e^{2t} + 2 + e^{-2t}}} \). This formula shows changes in speed along the path and is crucial for understanding motion dynamics.

Normal Components

Normal components measure how much of the acceleration vector is directed perpendicular to the velocity vector's path. This is crucial for understanding how paths curve or bend.

• The normal component of acceleration, denoted as \( a_N \), can be conceptualized as the force pushing you towards or away from the center of a curve.
• It's calculated using the magnitude of the velocity and acceleration vectors, excluding tangential effects.

In the exercise, we first compute the acceleration vector magnitude \( \| \mathbf{a}(t) \| = \sqrt{e^{2t} + e^{-2t}} \). Then, the normal component \( a_N \) is derived using:\[ a_N = \sqrt{\| \mathbf{a}(t) \|^2 - a_T^2} \]which provides the measure of change perpendicular to the path.

Acceleration Vector

The acceleration vector explains how the velocity of an object changes over time. It is the derivative of the velocity vector and shows desolation or speeding up in direction.

• In the given exercise, the acceleration vector, \( \mathbf{a}(t) \), is simply the derivative of the velocity vector, \( \mathbf{v}(t) \).
• This computation reveals how each component of motion (in the \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) directions) changes as time progresses.

For this scenario, take the velocity vector \( \mathbf{v}(t) = e^t \mathbf{i} + \sqrt{2} \mathbf{j} - e^{-t} \mathbf{k} \) and differentiate it:
The result is the acceleration vector \( \mathbf{a}(t) = e^t \mathbf{i} + e^{-t} \mathbf{k} \).This vector tells us how velocity in each direction evolves and is key to understanding motion dynamics.

Derivative Calculations

Derivative calculations are essential in finding how quantities change, especially in motion.
They are foundational to discuss rates, like velocity and acceleration, by describing how one variable changes with respect to another.

• In calculus, the derivative provides the rate of change, which for our purposes translates to velocity and acceleration.
• In the exercise, we computed derivatives for both the position and velocity vectors to find velocity and acceleration.

First, we derived the position vector \( \mathbf{r}(t) \) to find velocity \( \mathbf{v}(t) \), then differentiated this result to discover the acceleration vector \( \mathbf{a}(t) \).
The process of differentiation gives exact, real-time insights into motion dynamics, enabling predictions and analysis of an object's trajectory.