Problem 41 - Calculus Early Transcendentals

Question

The position function of a spaceship is $$\mathbf{r}(t)=(3+t) \mathbf{i}+(2+\ln t) \mathbf{j}+\left(7-\frac{4}{t^{2}+1}\right) \mathbf{k}$$ and the coordinates of a space station are \((6,4,9) .\) The captain wants the spaceship to coast into the space station. When should the engines be turned off?

Step-by-Step Solution

Answer

Turn off the engines at \( t = 3 \).

Step 1: Understanding the Problem

The spaceship's path is described by the position function \( \mathbf{r}(t)=(3+t) \mathbf{i}+(2+\ln t) \mathbf{j}+\left(7-\frac{4}{t^{2}+1}\right) \mathbf{k} \). We need to find the time \( t \) such that \( \mathbf{r}(t)=(6,4,9) \), the coordinates of the space station.

Step 2: Setting the Equations

We compare each component of the position function with the corresponding coordinate of the space station:1. \( 3+t = 6 \)2. \( 2+\ln t = 4 \)3. \( 7-\frac{4}{t^2+1} = 9 \) We will solve these equations to find \( t \).

Step 3: Solving the First Equation

The equation \( 3+t = 6 \) simplifies to \( t = 3 \).

Step 4: Solving the Second Equation

The equation \( 2+\ln t = 4 \) simplifies to \( \ln t = 2 \). Solve for \( t \) by exponentiating: \( t = e^2 \approx 7.39 \).

Step 5: Solving the Third Equation

The third equation \( 7-\frac{4}{t^2+1} = 9 \) simplifies to \( -\frac{4}{t^2+1} = 2 \), which further simplifies to \( \frac{4}{t^2+1} = -2 \). This equation has no real solutions because the left side cannot be negative.

Step 6: Finding a Consistent Solution

From the analyzed components, only the equation \( t = 3 \) is consistent without involving impossible conditions (like a negative logarithm or division). Therefore, turning the engines off at \( t = 3 \) aligns with a possible relationship between the spaceship's position and the space station.

Key Concepts

Spacecraft Trajectory

Understanding the trajectory of a spacecraft is crucial for its navigation through space. A trajectory is represented by a position function, like in our exercise,\[ \mathbf{r}(t) = (3+t) \mathbf{i} + (2+\ln t) \mathbf{j} + \left(7-\frac{4}{t^{2}+1}\right) \mathbf{k} \].This function describes the position of a spaceship along its path over time.

In this context:

• \( t \) represents time.
• The components \( (3+t) \mathbf{i}, (2+\ln t) \mathbf{j}, \text{and} \left(7-\frac{4}{t^{2}+1}\right) \mathbf{k} \) represent the position in the \( x \), \( y \), and \( z \) directions, respectively.

For the spaceship to reach a specific destination, such as a space station, you need to find the moment \( t \) when the position function matches the space station's coordinates.
By solving the component equations, we determined that turning off the engines at \( t = 3 \) would allow the spacecraft to coast into the space station's location without further propulsion.

Exponential and Logarithmic Equations

Exponential and logarithmic relationships are common in mathematical modeling, including the study of spacecraft trajectories. In the given exercise,we encounter the logarithmic component within the position function: \( 2+\ln t = 4 \).

This equation requires understanding of logarithms:

• \( \ln t \) signifies the natural logarithm of \( t \).
• Solving \( 2+\ln t = 4 \) involves isolating \( \ln t \) to get \( \ln t = 2 \).
• By exponentiating, you find \( t = e^2 \approx 7.39 \).

Logarithms are the inverses of exponential functions and are useful for solving equations where the variable is inside a logarithm.
They are essential tools, particularly when dealing with growth processes or decay, such as signal strength loss over distance in space communications.
In our scenario, although \( t = e^2 \) sizes up realistically, spatial consistency in the other components confirms \( t = 3 \) as the appropriate time for engine shutdown.

Component Comparison in Vectors

Comparing vector components is an essential technique in vector mathematics, allowing you to find precise solutions to problems.

In our exercise, the position of the spaceship is expressed as a vector, and each component is matched against the space station's corresponding coordinates.This involves:

• Equating each dimension's value of the position vector to the respective coordinate of the station.
• Solving these resultant equations independently to find \( t \).

For example, addressing the components:

• \( 3+t = 6 \) solves to give \( t = 3 \).
• \( 2+\ln t = 4 \) attempts to give \( t = e^2 \approx 7.39 \).
• \( 7-\frac{4}{t^2+1} = 9 \) results in a no real solution condition.

By comparing each component and finding consistent solutions, it’s possible to confirm the appropriate action.
This approach ensures that trajectory planning meets the desired destination accurately and efficiently, minimizing the need for corrections or alternate maneuver planning.