Problem 15 - Calculus 3

Question

Find the domain of the vector-valued functions. Domain: \(\mathbf{r}(t)=\left\langle t^{2}, \tan t, \ln t\right\rangle\)

Step-by-Step Solution

Answer

The domain is \(t > 0\) and \(t \neq \frac{\pi}{2} + n\pi\).

Step 1: Determine the domain for each component of the vector

The vector-valued function \(\mathbf{r}(t) = \langle t^2, \tan t, \ln t \rangle\) is composed of three different functions: \(f(t) = t^2\), \(g(t) = \tan t\), and \(h(t) = \ln t\). We need to find the domain for each one separately before determining the overall domain.

Step 2: Find the domain of \(f(t) = t^2\)

The function \(f(t) = t^2\) is a polynomial function, which means it is defined for all real numbers. Hence, the domain is \((-\infty, \infty)\).

Step 3: Find the domain of \(g(t) = \tan t\)

The function \(g(t) = \tan t\) is defined for all real numbers except where \(\tan t\) is undefined, which is at \(t = \frac{\pi}{2} + n\pi\), where \(n\) is any integer. Thus, the domain is \(t eq \frac{\pi}{2} + n\pi\).

Step 4: Find the domain of \(h(t) = \ln t\)

The function \(h(t) = \ln t\) is defined only for \(t > 0\) because the natural logarithm is undefined for non-positive numbers. Hence, the domain is \((0, \infty)\).

Step 5: Determine the intersection of the domains

To find the domain of the vector function \(\mathbf{r}(t)\), we take the intersection of the domains from each component. The overall domain must satisfy \( t > 0 \) and \( t eq \frac{\pi}{2} + n\pi \).

Step 6: Write the final domain

Considering all the restrictions from Steps 2 through 4, the domain of \(\mathbf{r}(t)\) is: \( t \in \mathbb{R}, t > 0, t eq \frac{\pi}{2} + n\pi\) where \(n\) is any integer.

Key Concepts

Polynomial Functions

Polynomial functions are a crucial concept in mathematics and are among the simplest types of functions you'll encounter. A polynomial function has the general form \[ f(t) = a_n t^n + a_{n-1} t^{n-1} + \ + \ + a_1 t + a_0 \] where \(a_n, a_{n-1}, ..., a_0\) are constants. These functions are versatile because they can model a wide range of phenomena.
Now, let's focus on the specific case of \(f(t) = t^2\), as seen in the original exercise. This is a simple polynomial of degree two. One of the best aspects of polynomial functions is that they are defined for all real numbers, meaning their domain is \[ (-\infty, \infty) \].
This wide range is because no real number input will cause the function to be undefined—there's no division by zero or need for square roots of negative numbers. Thus, you can use any real number for \(t\) without restriction, making them lovable in mathematical problems.

Trigonometric Functions

Trigonometric functions like \( \tan t \) are essential, especially in fields like physics and engineering. However, they come with certain limitations within their domains.
The tangent function, \(\tan t\), is the ratio of \( \sin t \) to \( \cos t \) and is unique in that it can become undefined. This occurs whenever \( \cos t = 0 \), meaning the function blows to infinity. This specifically happens when \[ t = \frac{\pi}{2} + n\pi \] for any integer \(n\).
Understanding this is critical for determining where the tangent function is valid, leading to its domain being all real numbers except these points. Therefore, the domain for \( \tan t \) is \[ t eq \frac{\pi}{2} + n\pi \].
Being aware of these limitations helps in managing scenarios where \( \tan t \) might lead to undefined results, which is useful in calculating angles and distances in trigonometry.

Logarithmic Functions

Logarithmic functions, such as \( \ln t \), are another fundamental concept in mathematics. They are the inverse operations of exponential functions.
A logarithmic function \( \ln t \) is only defined for positive values of \(t\). This is because you can't take the logarithm of zero or a negative number in the realm of real numbers. Such restrictions result from the nature of exponents—they can't produce negative or zero results from positive bases.
Thus, the domain is \[ (0, \infty) \] which means that \(t\) must always be greater than zero. This limitation is vital in fields that involve growth models and decay processes, as logarithms are frequently used for transforming exponential data into a linear form. Remembering these constraints helps you solve problems involving growth, scaling, and conversion neatly.

Intersection of Domains

When dealing with vector-valued functions, such as \[ \mathbf{r}(t) = \langle t^2, \tan t, \ln t \rangle \] you combine different component functions into a single vector. This requires finding a common domain where all functions are valid simultaneously.
The key to finding this intersection is considering each component's domain restrictions:

• For \( t^2 \), \( t \in (-\infty, \infty) \).
• For \( \tan t \), \( t eq \frac{\pi}{2} + n\pi \).
• For \( \ln t \), \( t > 0 \).

The overall domain of \( \mathbf{r}(t) \) will be where all these conditions overlap, which requires:

• \( t > 0 \)
• \( t \) is not \(\frac{\pi}{2} + n\pi\)

These combined restrictions define where the vector-valued function can comfortably exist, ensuring each component is properly defined at that point. Understanding this intersection helps in correctly applying the function in real-world and theoretical scenarios, ensuring accurate interpretations.