Problem 20 - Calculus 3

Question

Find the domain of \(\mathbf{r}(t)=2 e^{-t} \mathbf{i}+e^{-t} \mathbf{j}+\ln (t-1) \mathbf{k}\)

Step-by-Step Solution

Answer

The domain is \( t > 1 \).

Step 1: Determine Function Components

The given vector function is \( \mathbf{r}(t) = 2e^{-t} \mathbf{i} + e^{-t} \mathbf{j} + \ln(t-1) \mathbf{k} \). We must analyze each component separately to determine any restrictions on \( t \).

Step 2: Analyze Exponential Components

The components \( 2e^{-t} \) and \( e^{-t} \) involve the exponential function, which is defined for all real numbers. Therefore, \( t \) does not have restrictions from these components.

Step 3: Analyze Logarithmic Component

The component \( \ln(t-1) \) is defined only when its argument is greater than zero. Thus, we have the inequality \( t-1 > 0 \) which simplifies to \( t > 1 \). This imposes a restriction on the domain.

Step 4: Combine Conditions

Since \( t \) must satisfy conditions from all components of the vector function, the domain of \( \mathbf{r}(t) \) is \( t > 1 \) because this is the only restriction identified.

Key Concepts

Domain of a Function

In mathematics, the domain of a function is a crucial concept that determines where the function is defined. For vector functions like the one given, the domain lies where each of its components is valid. This involves checking each part to see if it has any restrictions. The domain tells us what values the input variable can take.
Understanding the domain is like knowing the "life zone" of a function, ensuring it works right where it's supposed to.
Here’s how we examined it for our vector function:

• The exponential parts, such as those with terms like \( e^{-t} \), are defined for all real numbers.
• The logarithmic part, \( \ln(t-1) \), requires the inside to be positive, implying \( t > 1 \).

Combining these considerations, the domain of the entire function is defined where all parts are valid. For this vector function, since only the logarithmic component imposes a limitation, \( t > 1 \) defines its domain.

Exponential Function

Exponential functions are mathematical functions of the form \( a \cdot e^{bt} \), where \( a \) and \( b \) are constants, and \( e \) is Euler’s number, approximately 2.718. They are characterized by a consistent rate of growth or decay.
This makes exponential functions incredibly versatile, appearing in real-world situations, such as population growth, radioactive decay, and interest calculations.
Here’s what you need to know about them:

• Exponential functions are always defined for all real numbers. This includes both positive and negative values.
• They graph typically increases or decreases rapidly, depending on whether they represent growth or decay.

In our vector function problem, the components like \( 2e^{-t} \) and \( e^{-t} \) indicate exponential decay. This means they don't put any restrictions on \( t \), as they smoothly accept every real number as input.

Logarithmic Function

Logarithmic functions are the inverses of exponential functions. For a function \( y = \ln(x) \), it means \( e^y = x \). They require their arguments to be positive to be defined. This makes them a bit special compared to other functions that are defined over all real numbers.
Understanding these restrictions helps in determining where the function is "active" or "alive."
Consider the following characteristics:

• Defined only for positive arguments: \( x > 0 \).
• Commonly used to solve exponential equations.
• Grows continuously, but very slowly, as \( x \) increases.

In our example, the logarithmic component \( \ln(t-1) \) requires \( t - 1 > 0 \), simplifying to \( t > 1 \). This is the constraint on the domain of the vector function, ensuring the logarithm's argument stays positive.