Problem 15 - Calculus III

Question

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

Step-by-Step Solution

Answer

The domain is \((0, \infty) \setminus \{ \frac{\pi}{2} + n\pi \} \), \( t > 0, t \neq \frac{\pi}{2} + n\pi \)."

Step 1: Identify the domain for each component function

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

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

The function \( f(t) = t^2 \) is a polynomial. Polynomials do not have restrictions on their domain. Therefore, the domain of \( f(t) = t^2 \) is all real numbers, \((-\infty, \infty)\).

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

The function \( \tan t \) is undefined where \( \cos t = 0 \). This occurs at \( t = \frac{\pi}{2} + n\pi \), where \( n \) is an integer. Thus, the domain of \( g(t) = \tan t \) is all real numbers except for these points, \( t eq \frac{\pi}{2} + n\pi \).

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

The natural logarithm function \( \ln t \) is only defined for positive values of \( t \). Therefore, the domain of \( h(t) = \ln t \) is \( t > 0 \).

Step 5: Find the intersection of all domains

The domain of the vector-valued function \( \mathbf{r}(t) \) is the intersection of the domains of \( f(t), g(t), \) and \( h(t) \). The intersection of \((-\infty, \infty)\), all reals minus \( \frac{\pi}{2} + n\pi \), and \( t > 0 \) results in \( (0, \infty) \setminus \left\{ \frac{\pi}{2} + n\pi \right\} \), where \( t > 0 \) and \( t eq \frac{\pi}{2} + n\pi \).

Key Concepts

Domain of a function

The domain of a function refers to the complete set of possible values of the independent variable for which the function is defined. In simpler terms, it is the collection of all inputs for which the function can produce a valid output. Let's break down finding the domain into easy steps: we look at each function that makes up the vector-valued function separately. For instance, in the given exercise, we have three functions: \( f(t) = t^2 \), \( g(t) = \tan t \), and \( h(t) = \ln t \). Each has its own set of rules for which values of \( t \) it accepts.

Intersection of domains

When dealing with vector-valued functions, we often have to find the intersection of the domains of its component functions. This means finding values for which all functions are simultaneously defined. Let's visualize this:

• Each function's domain is like a separate universe.
• By intersecting, we find common ground where all universes coincide.

For the exercise, this involves intersecting domains of \((-, )\) for \( t^2 \), \( t eq \frac{\pi}{2} + n\pi \) for \( \tan t \), and \( t > 0 \) for \( \ln t \). The overlap or intersection results in \((0, \infty) \setminus \left\{ \frac{\pi}{2} + n\pi \right\}\). This is the shared domain where the vector-valued function is defined.

Trigonometric functions

Trigonometric functions, like sine, cosine, and tangent, have specific characteristics concerning their domains and ranges. The exercise features the tangent function, \( \tan t \), which is vital to understand:

• \( \tan t \) becomes undefined where \( \cos t = 0 \).
• This occurs at \( t = \frac{\pi}{2} + n\pi \), implying the function has breaks or asymptotes at these points.

While \( \tan t \) is periodic and continuously defined between these breaks, these gaps are crucial for determining the domain of the tangent component in any combined function.

Logarithmic functions

Logarithmic functions, like our example \( \ln t \), have their own rules for domains. The natural logarithm is defined only on positive real numbers, which means you cannot take the logarithm of zero or a negative number. Here’s how it affects the domain:

• The domain of \( \ln t \) is \( t > 0 \).
• This condition must be met for the function to be valid.

For the vector-valued function given, any potential domain must satisfy this positivity constraint as part of intersecting with the domains of other functions.