Problem 17 - Calculus III

Question

Find the domain of the vector-valued functions. $$ \text { Domain: } \mathbf{r}(t)=\left\langle\csc (t), \frac{1}{\sqrt{t-3}}, \ln (t-2)\right\rangle $$

Step-by-Step Solution

Answer

Domain: \( t > 3 \) and \( t \neq n\pi \) where \( n \) is an integer.

Step 1: Identify the Domain Restrictions of Each Component

The vector-valued function \( \mathbf{r}(t) = \left\langle \csc(t), \frac{1}{\sqrt{t-3}}, \ln(t-2) \right\rangle \) is given. We need to find the domain for each component function separately, then combine these restrictions to find the overall domain of the function. Each component has specific restrictions due to its mathematical properties.

Step 2: Determine Domain of \( \csc(t) \)

The cosecant function \( \csc(t) = \frac{1}{\sin(t)} \) is undefined when \( \sin(t) = 0 \). The values of \( t \) where \( \sin(t) = 0 \) are integer multiples of \( \pi \), i.e., \( t eq n\pi \) where \( n \) is an integer.

Step 3: Determine Domain of \( \frac{1}{\sqrt{t-3}} \)

The expression \( \frac{1}{\sqrt{t-3}} \) is undefined when the square root is zero or negative. So, \( t - 3 > 0 \) which simplifies to \( t > 3 \). This means \( t \) must be greater than 3 for \( \frac{1}{\sqrt{t-3}} \) to be defined.

Step 4: Determine Domain of \( \ln(t-2) \)

The natural logarithm \( \ln(t-2) \) is defined for \( t-2 > 0 \). So, \( t > 2 \). This means \( t \) must be greater than 2 for \( \ln(t-2) \) to be defined.

Step 5: Find the Intersection of Domains

Combine the domain restrictions from Steps 2, 3, and 4:- For \( \csc(t) \), \( t eq n\pi \) where \( n \) is an integer.- For \( \frac{1}{\sqrt{t-3}} \), \( t > 3 \).- For \( \ln(t-2) \), \( t > 2 \).The most restrictive condition is \( t > 3 \). Therefore, the domain of the vector function \( \mathbf{r}(t) \) is all real numbers greater than 3, except where \( t = n\pi \).

Key Concepts

Domain of Functions

The domain of a function is the set of all input values for which the function is defined. For a vector-valued function, which consists of multiple component functions, it is crucial to analyze the domain of each component separately. In this case, we have

• \( \csc(t) \): Cosecant is undefined when the sine of the angle is zero, meaning it is not defined when \( t = n\pi \), where \( n \) is an integer.
• \( \frac{1}{\sqrt{t-3}} \): This expression is undefined when the square root is zero or negative. Thus, \( t > 3 \) ensures it is well-defined.
• \( \ln(t-2) \): A natural logarithm is only defined for positive arguments. Therefore, we must have \( t > 2 \).

To determine the overall domain, intersect these conditions, usually focusing on the strictest. For our example, this means considering \( t > 3 \) while avoiding values \( t = n\pi \). This ensures all components remain defined.

Trigonometric Functions

Trigonometric functions like sine, cosine, and their reciprocals such as cosecant (\( \csc \)) perform key roles in functions. Particularly, \( \csc(t) = \frac{1}{\sin(t)} \) implies it is undefined when \( \sin(t) \) equals zero. Values of \( t \) for which this occurs include multiples of \( \pi \) (zero, \( \pi \), \( 2\pi \), etc.). This periodic nature means trigonometric functions have recurring undefined points, added complexity in finding domains.
In practical terms, whenever dealing with a function involving these components, inspect the fundamental properties of sine and cosine. Evaluate where they are undefined to ensure your domain excludes such points.

Natural Logarithm

The natural logarithm function, \( \ln(x) \), is a crucial tool in advanced mathematics. It operates by only accepting positive real numbers as inputs because a logarithm represents the power to which a base, \( e \) in natural logarithms, must be raised to produce that number. Therefore, \( \ln(t-2) \) implies that \( t-2 > 0 \) or \( t > 2 \).
Understanding logarithms is essential for analyzing how values grow, aiding in calculus and beyond. The base \( e \) (approximately 2.718) is particularly significant as it naturally arises in growth processes in mathematics and nature. Hence, always ascertain that input to a \( \ln \) function is positive as part of domain considerations.

Composite Functions

A composite function is essentially two functions combined, where the output of one becomes the input of another. When finding the domain, it is imperative to ensure the transitions between functions remain valid. For instance, in \( \mathbf{r}(t) = \left\langle \csc(t), \frac{1}{\sqrt{t-3}}, \ln(t-2) \right\rangle \), each component function forms part of the vector-valued function.
Follow these steps to manage composites:

• Identify individual domains: Review each component to establish its domain.
• Combine logically: Consider each piece's restrictions to find the intersection (i.e., the viable domain).
• Maintain continuity: Ensure transitions make sense physically and mathematically, considering undefined points.

This method effectively organizes complex ructions from single entities, simplifying analysis and application. Think of each component as part of a greater narrative in understanding compositional relationships.