Problem 21 - Calculus III

Question

For what values of \(t\) is \(\mathbf{r}(t)=2 e^{-t} \mathbf{i}+e^{-t} \mathbf{j}+\ln (t-1) \mathbf{k}\) continuous?

Step-by-Step Solution

Answer

\( \mathbf{r}(t) \) is continuous for \( t > 1 \).

Step 1: Identify Continuity Conditions

For a vector function \( \mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} + h(t) \mathbf{k} \) to be continuous, each of its component functions \( f(t) \), \( g(t) \), and \( h(t) \) must also be continuous. We'll look at each component separately to determine their continuity.

Step 2: Analyze the First Component

The first component is \( f(t) = 2e^{-t} \). The exponential function \( e^{-t} \) is continuous for all real numbers. Therefore, \( f(t) = 2e^{-t} \) is continuous for all \( t \in \mathbb{R} \).

Step 3: Analyze the Second Component

The second component is \( g(t) = e^{-t} \). Like the first component, this exponential function is continuous for all real numbers. So, \( g(t) = e^{-t} \) is continuous for all \( t \in \mathbb{R} \).

Step 4: Analyze the Third Component

The third component is \( h(t) = \ln(t-1) \). The natural logarithm \( \ln(x) \) is continuous for \( x > 0 \). Since it's \( \ln(t-1) \), the expression \( t-1 > 0 \) must hold, which means \( t > 1 \). Thus, \( h(t) = \ln(t-1) \) is continuous for \( t > 1 \).

Step 5: Determine Overall Continuity Interval

The overall vector function \( \mathbf{r}(t) \) is continuous where all components are continuously defined. Since the first two components are continuous for all \( t \), and the third component is continuous for \( t > 1 \), the vector function \( \mathbf{r}(t) \) is continuous for \( t > 1 \).

Key Concepts

Component Functions

In the study of vector functions, each vector component is represented by an individual function. These are known as component functions. For example, in a vector function \( \mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} + h(t) \mathbf{k} \), - \( f(t) \) is the component function for the \( \mathbf{i} \) direction.- \( g(t) \) is for the \( \mathbf{j} \) direction.- \( h(t) \) is for the \( \mathbf{k} \) direction.
The overall behavior of the vector function depends on the behavior of these component functions. If all component functions are continuous, then the vector function itself is continuous.
Understanding each component function individually allows us to piece together the behavior of the entire vector function.

Exponential Function

Exponential functions are among the most common functions in calculus. They have the form \( a e^{bt} \) where \( a \) and \( b \) are constants. These functions are inherently smooth and continuous over all real numbers.
In our exercise, we have the component functions \( f(t) = 2e^{-t} \) and \( g(t) = e^{-t} \). Both functions fit the general form of an exponential function, \( e^{-t} \), which is known to be continuous everywhere.
As such:

• \( e^{-t} \) maintains a steady, smooth curve without breaks or gaps.
• Multiplying by a constant, like \( 2 \), does not affect this continuity.

Thus, any function in the form \( ae^{bt} \) can be confidently said to be continuous across the entire set of real numbers.

Natural Logarithm

Natural logarithm functions are another key concept in calculus. They are written as \( \ln(x) \) and represent the power to which the base \( e \) must be raised to obtain \( x \). Unlike exponential functions, natural logarithms are only continuous for positive values of \( x \).
In the exercise, the third component is \( h(t) = \ln(t-1) \). For continuity:

• \( t-1 \) must be greater than zero, meaning \( t > 1 \).
• If \( t = 1 \), \( \ln(0) \) becomes undefined, leading to discontinuity.

To sum up, while natural logarithms are continuous and smooth within their domain \( x > 0 \), we must carefully consider the domain in vector functions. For \( \ln(t-1) \), continuity is achieved for \( t > 1 \).

Condition for Continuity

Continuity is a fundamental concept in calculus. For a function to be continuous at a point, it must be defined at that point, limit to that point, and the limit must equal the function's value there.
For vector functions, this rule applies to each component function. Therefore, a vector function like \( \mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} + h(t) \mathbf{k} \) will be continuous provided:

• Each component function is continuous in its defined domain.
• The overall domain is where all components are simultaneously continuous.

For our specific function,

• \(f(t) = 2e^{-t} \) and \( g(t) = e^{-t} \) are continuous for all real \( t \).
• \( h(t) = \ln(t-1) \) demands \( t > 1 \) for continuity.

Thus, the entire vector function is continuous for \( t > 1 \). This is the region where all component functions overlap in continuity.