Problem 4 - Calculus III

Question

Given the following position functions, find the velocity, acceleration, and speed in terms of the parameter t. $$ \mathbf{r}(t)=e^{-t} \mathbf{i}+t^{2} \mathbf{j}+\tan t \mathbf{k} $$

Step-by-Step Solution

Answer

Velocity: \(-e^{-t} \mathbf{i} + 2t \mathbf{j} + \sec^2 t \mathbf{k}\); Acceleration: \(e^{-t} \mathbf{i} + 2 \mathbf{j} + 2 \sec^2 t \tan t \mathbf{k}\); Speed: \(\sqrt{e^{-2t} + 4t^2 + \sec^4 t}\).

Step 1: Understand the Problem

We have a position vector function in terms of the parameter \( t \). The position function is \( \mathbf{r}(t)=e^{-t} \mathbf{i}+t^{2} \mathbf{j}+\tan t \mathbf{k} \). We need to find the velocity, acceleration, and speed of the object described by this function.

Step 2: Find the Velocity Vector

The velocity vector is the derivative of the position vector function \( \mathbf{r}(t) \) with respect to \( t \). The derivative is calculated component-wise:\[ \mathbf{v}(t) = \frac{d}{dt}(e^{-t}) \mathbf{i} + \frac{d}{dt}(t^2) \mathbf{j} + \frac{d}{dt}(\tan t) \mathbf{k} \]Calculating each derivative, we get:- Derivative of \( e^{-t} \) is \( -e^{-t} \)- Derivative of \( t^2 \) is \( 2t \)- Derivative of \( \tan t \) is \( \sec^2 t \)Thus, the velocity vector is:\[ \mathbf{v}(t) = -e^{-t} \mathbf{i} + 2t \mathbf{j} + \sec^2 t \mathbf{k} \]

Step 3: Find the Acceleration Vector

The acceleration vector is the derivative of the velocity vector function \( \mathbf{v}(t) \) with respect to \( t \). The derivative is again calculated component-wise:\[ \mathbf{a}(t) = \frac{d}{dt}(-e^{-t}) \mathbf{i} + \frac{d}{dt}(2t) \mathbf{j} + \frac{d}{dt}(\sec^2 t) \mathbf{k} \]Calculating each derivative, we get:- Derivative of \( -e^{-t} \) is \( e^{-t} \)- Derivative of \( 2t \) is \( 2 \)- Derivative of \( \sec^2 t \) is \( 2 \sec^2 t \tan t \) (using the chain rule)Thus, the acceleration vector is:\[ \mathbf{a}(t) = e^{-t} \mathbf{i} + 2 \mathbf{j} + 2 \sec^2 t \tan t \mathbf{k} \]

Step 4: Calculate the Speed

The speed is the magnitude of the velocity vector. It is calculated as:\[ \|\mathbf{v}(t)\| = \sqrt{ [-e^{-t}]^2 + [2t]^2 + [\sec^2 t]^2 } \]Simplifying gives:\[ \|\mathbf{v}(t)\| = \sqrt{ e^{-2t} + 4t^2 + \sec^4 t } \]

Key Concepts

Position Vector Function

In vector calculus, a position vector function is a crucial concept as it describes the position of a point in space as a function of one or more variables, typically time. For instance, let’s consider the position vector function given by \[ \mathbf{r}(t) = e^{-t} \mathbf{i} + t^2 \mathbf{j} + \tan t \mathbf{k} \] This vector describes how an object's position changes over time. The components of this vector function are

• \( e^{-t} \) in the direction of \( \mathbf{i} \): Represents exponential decay as the object moves along the x-axis.
• \( t^2 \) in the direction of \( \mathbf{j} \): Shows quadratic growth along the y-axis.
• \( \tan t \) in the direction of \( \mathbf{k} \): Indicates a tangent function affecting movement along the z-axis.

Understanding these components helps predict the behavior of the particle or object over time in a 3D space.

Velocity Vector

The velocity vector is derived by taking the derivative of the position vector function with respect to time \( t \). This derivative provides the rate of change of position, effectively describing the object's velocity at any point in time. For our vector function, the velocity \( \mathbf{v}(t) \) is calculated as \[ \mathbf{v}(t) = \frac{d}{dt}(e^{-t}) \mathbf{i} + \frac{d}{dt}(t^2) \mathbf{j} + \frac{d}{dt}(\tan t) \mathbf{k} \]This results in:\[ \mathbf{v}(t) = -e^{-t} \mathbf{i} + 2t \mathbf{j} + \sec^2 t \mathbf{k} \] Here’s what each term represents:

• \(-e^{-t} \) in \( \mathbf{i} \): Represents how the decay rate affects the velocity in the x direction.
• \(2t \) in \( \mathbf{j} \): Indicates a linear increase over time along the y direction.
• \( \sec^2 t \) in \( \mathbf{k} \): Shows how a tangent-related growth affects the velocity in the z direction with time.

Each component tells us how quickly the position is changing at any moment.

Acceleration Vector

Finding the acceleration vector involves taking the derivative of the velocity vector with respect to time \( t \). This second derivative measures how the velocity changes over time, providing insights into an object's acceleration. Given the velocity \( \mathbf{v}(t) \), we calculate the acceleration \( \mathbf{a}(t) \) as \[ \mathbf{a}(t) = \frac{d}{dt}(-e^{-t}) \mathbf{i} + \frac{d}{dt}(2t) \mathbf{j} + \frac{d}{dt}(\sec^2 t) \mathbf{k} \]Calculating these, we have:\[ \mathbf{a}(t) = e^{-t} \mathbf{i} + 2 \mathbf{j} + 2 \sec^2 t \tan t \mathbf{k} \]Breaking down each term:

• \( e^{-t} \) in \( \mathbf{i} \): Shows acceleration linked to the rate of decay in the x direction.
• \(2 \) in \( \mathbf{j} \): Represents a constant acceleration along the y direction.
• \(2 \sec^2 t \tan t \) in \( \mathbf{k} \): Illustrates how the tangent-related speed is accelerating in the z direction.

Understanding the acceleration components is key to predicting future movements and identifying changes in speed.

Speed Calculation

Speed, being a scalar quantity, differs from velocity as it represents only how fast an object is moving, without regard to its direction. We calculate the speed as the magnitude of the velocity vector, simplifying our 3D velocity vector to a measure of how quickly the distance changes. For the given velocity vector \[ \mathbf{v}(t) = -e^{-t} \mathbf{i} + 2t \mathbf{j} + \sec^2 t \mathbf{k} \]The speed at time \( t \) is:\[ \|\mathbf{v}(t)\| = \sqrt{ (-e^{-t})^2 + (2t)^2 + (\sec^2 t)^2 } \] This simplifies to:\[ \|\mathbf{v}(t)\| = \sqrt{ e^{-2t} + 4t^2 + \sec^4 t } \]What this expression tells us is:

• \(e^{-2t}\): Contributes to slowing down or speeding up along the x direction.
• \(4t^2\): Reflects changing speed in the y direction due to quadratic growth.
• \(\sec^4 t\): Relates to rapid speed variations due to the tangent function in the z direction.

Understanding speed helps us appreciate how quickly a position changes over time, regardless of the path taken.