Problem 13 - Calculus 3

Question

Find the limit of the following vector-valued functions at the indicated value of \(t .\) $$ \lim _{t \rightarrow \infty} \mathbf{r}(t) \text { for } \mathbf{r}(t)=2 e^{-t} \mathbf{i}+e^{-t} \mathbf{j}+\ln (t-1) \mathbf{k} $$

Step-by-Step Solution

Answer

The limit is \( \infty \mathbf{k} \).

Step 1: Identify the Components

The vector function \( \mathbf{r}(t) = 2e^{-t} \mathbf{i} + e^{-t} \mathbf{j} + \ln(t-1) \mathbf{k} \) is composed of three components: \( 2e^{-t} \) along \( \mathbf{i} \), \( e^{-t} \) along \( \mathbf{j} \), and \( \ln(t-1) \) along \( \mathbf{k} \). We need to find the limit of each component as \( t \) approaches infinity.

Step 2: Calculate Limit of the \( \mathbf{i} \)-Component

The \( \mathbf{i} \)-component is \( 2e^{-t} \). As \( t \) approaches infinity, the exponential term \( e^{-t} \) approaches zero. Thus, \( \lim_{t \to \infty} 2e^{-t} = 2 \times 0 = 0 \).

Step 3: Calculate Limit of the \( \mathbf{j} \)-Component

The \( \mathbf{j} \)-component is \( e^{-t} \). As \( t \) approaches infinity, \( e^{-t} \) again approaches zero. Therefore, \( \lim_{t \to \infty} e^{-t} = 0 \).

Step 4: Analyze \( \mathbf{k} \)-Component's Growth

The \( \mathbf{k} \)-component is \( \ln(t-1) \). As \( t \) approaches infinity, \( t-1 \) also reaches infinity, causing \( \ln(t-1) \) to grow without bound. Consequently, \( \lim_{t \to \infty} \ln(t-1) = \infty \).

Step 5: Combine Limits of Components

Combine the limits of each component to find the limit of \( \mathbf{r}(t) \). We have:\[ \lim_{t \to \infty} \mathbf{r}(t) = 0 \mathbf{i} + 0 \mathbf{j} + \infty \mathbf{k} = \infty \mathbf{k} \]Thus, the limit of \( \mathbf{r}(t) \) is unbounded along the \( \mathbf{k} \) direction.

Key Concepts

Limit Calculation

Understanding how to calculate limits is fundamental in analyzing vector-valued functions. When evaluating the limit of a vector function as a variable approaches a specific value, one must consider each of the vector's components individually. The goal is to determine how each component behaves as the variable tends towards infinity, zero, or any finite number.

In the context of the given vector-valued function, \(\mathbf{r}(t) = 2e^{-t} \mathbf{i} + e^{-t} \mathbf{j} + \ln(t-1) \mathbf{k}\), the task is to find the limit as \(t\) approaches infinity. Therefore, we must calculate the limit for each component separately—both exponential terms and the logarithmic term—before combining these results to provide a comprehensive understanding of the vector's behavior.

By breaking the problem into smaller parts, limit calculation becomes a systematic process, helping us understand even complex functions.

Exponential Functions

Exponential functions, like \(e^{-t}\), are critical in calculus due to their unique properties and ubiquitous presence in natural processes such as decay and growth. The base of natural logarithms, \(e\), is approximately 2.718, making functions involving \(e\) particularly important.

For functions in the form of \(e^{-t}\), as \(t\) approaches infinity, the function tends to zero. This behavior results from the negative exponent, which inversely scales the growth of \(e\). Essentially, as the exponent \(-t\) increases positively (given \(-t\) is negative), the fraction that represents \(e^{-t}\) becomes tinier.

This is evident in our exercise through components \(2e^{-t} \mathbf{i}\) and \(e^{-t} \mathbf{j}\). Both components reduce to zero as \(t\) grows indefinitely. Recognizing this characteristic of exponential functions allows us to assess limits precisely.

Natural Logarithm

The natural logarithm, represented by \(\ln(x)\), specifically \(\ln(t-1)\) in our discussion, is a logarithmic function that offers insights into growth, especially when handles transformations involving exponential functions. Unlike exponential functions that level off or approach zero, logarithmic functions reflect unbounded growth, albeit at a declining rate.

In our vector function's \(\mathbf{k}\)-component, \(\ln(t-1)\) demonstrates logarithmic growth as \(t\) approaches infinity. The natural logarithm grows without bounds since \(t-1\) stretches towards infinity. This component helps us understand complexity in dynamics where exponential decay is counterbalanced by logarithmic growth.

Recognizing the properties of logarithms aids in explaining patterns where growth scales differently from linear or exponential changes.

Unbounded Growth

Unbounded growth in mathematical functions addresses situations where a function doesn't converge to a finite limit but instead increases indefinitely. This concept is illustrated in the vector-valued function when observing the \(\mathbf{k}\)-component \(\ln(t-1)\).

While the exponential terms settle into finite limits, the logarithmic term grows larger as \(t\) becomes very large. This is because logs express a steep climb initially but continue to rise as \(x\) grows, albeit slower over time.

Thus, the presence of \(\infty \mathbf{k}\) in our combined limit indicates unbounded growth, portraying a vector that stretches infinitely along the \(\mathbf{k}\) direction. Understanding how different components within vector-valued functions contribute to such outcomes is critical for mastering vector calculus.