Problem 15 - Calculus Volume 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 of \( \mathbf{r}(t) \) is \((0, \infty)\setminus\{\frac{\pi}{2}+n\pi\;|\;n\in\mathbb{Z}\}\).

Step 1: Determine the Domain for Each Component

The vector-valued function \( \mathbf{r}(t) = \langle t^2, \tan(t), \ln(t) \rangle \) is composed of three parts: \( t^2 \), \( \tan(t) \), and \( \ln(t) \). Let's analyze the domain for each component: - The function \( t^2 \) is defined for all real numbers, so its domain is \( (-\infty, \infty) \).- The function \( \tan(t) \) is undefined for \( t = \frac{\pi}{2} + n\pi \), where \( n \) is an integer, due to vertical asymptotes.- The function \( \ln(t) \) is only defined for positive values of \( t \), hence its domain is \( (0, \infty) \).

Step 2: Combine Domains for Overall Function

To find the domain of \( \mathbf{r}(t) \), we need \( t \) to satisfy all the restrictions from each component:- The domain of \( t^2 \) is \( (-\infty, \infty) \), allowing any real number \( t \).- For \( \tan(t) \), \( t eq \frac{\pi}{2} + n\pi \) where \( n \) is any integer.- For \( \ln(t) \), \( t > 0 \) must be true.Thus, the overall domain is the intersection of \( (0, \infty) \) and all \( t eq \frac{\pi}{2} + n\pi \).

Step 3: Express the Domain

The final domain combines the interval \( (0, \infty) \) with the exclusion of points \( t = \frac{\pi}{2} + n\pi \). As a result, the domain of \( \mathbf{r}(t) \) is:\[ t \in (0, \infty) \setminus \{ \frac{\pi}{2} + n\pi \mid n \in \mathbb{Z} \} \]This means \( t \) is any positive real number except those that make \( \tan(t) \) undefined.

Key Concepts

Domain of Functions

The domain of a function refers to all the possible input values (typically 't' for many functions) that can be used in the function without causing any mathematical errors or undefined expressions. In this case, we have a vector-valued function made of three parts: \( t^2 \), \( \tan(t) \), and \( \ln(t) \).

• For \( t^2 \), the simplest component, any real number for 't' is valid. So, its domain is all real numbers \((-\infty, \infty)\).
• The function \( \tan(t) \), which is part of trigonometric functions, is a bit more complex. It is undefined where the cosine of 't' equals zero, which happens at intervals \( t = \frac{\pi}{2} + n\pi \). Here, \( n \) is any integer.
• The natural logarithm function \( \ln(t) \) only accepts positive real numbers. Thus, its domain is \((0, \infty)\).

To find the domain of the complete vector \( \mathbf{r}(t) \), consider all these conditions. The key is to find where all components overlap in terms of their allowed input values. This is done by intersecting the individual domains.

Trigonometric Functions

Trigonometric functions, like \( \tan(t) \), play a crucial role in mathematics, providing the link between angles and ratios in triangles. The tangent function, in particular, is periodic and undefined where \( \cos(t) = 0 \).
Key points about \( \tan(t) \):

• It repeats its pattern over an interval of \( \pi \).
• Values of 't' where it goes undefined are specifically \( t = \frac{\pi}{2} + n\pi \), with \( n \) representing any integer.
• This periodic nature with undefined gaps affects the domain of vector-valued functions including \( \tan(t) \).

Understanding trigonometric functions helps in recognizing cycles, periodicities, and can guide correct input values for functions.

Logarithmic Functions

Logarithmic functions, such as \( \ln(t) \), are another cornerstone in mathematics, often used to solve equations involving growth and exponential relationships. The natural logarithm is unique in that it only accepts strictly positive input values.

• Logarithms are undefined for zero and negative values.
• The natural logarithm \( \ln(t) \) is particularly common due to its simple derivative and integration properties.
• The domain of \( \ln(t) \) is \((0, \infty)\), restricting it to positive values for 't'.

This restriction plays a vital role in determining the domain of vector-valued functions that incorporate \( \ln(t) \), as all conditions must be met simultaneously for a valid solution.