Problem 207 - Calculus Volume 3

Question

Find the domains of the vector-valued functions. \(\quad \mathbf{r}(t)=\left\langle e^{t}, \frac{1}{\sqrt{4-t}}, \sec (t)\right\rangle\)

Step-by-Step Solution

Answer

The domain is \((-\infty, 4)\) excluding odd multiples of \(\frac{\pi}{2}\).

Step 1: Identify individual component functions

The vector-valued function \( \mathbf{r}(t) = \langle e^t, \frac{1}{\sqrt{4-t}}, \sec(t) \rangle \) has three components: \( e^t \), \( \frac{1}{\sqrt{4-t}} \), and \( \sec(t) \). We will find the domain of each separately and then combine the restrictions to find the overall domain.

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

The exponential function \( e^t \) is defined for all real numbers, so there are no restrictions imposed by this component. Its domain is \( (-\infty, \infty) \).

Step 3: Find the domain of \( \frac{1}{\sqrt{4-t}} \)

The function \( \frac{1}{\sqrt{4-t}} \) is defined when the denominator is not zero and the expression under the square root is positive. Hence, \( 4-t > 0 \) which simplifies to \( t < 4 \). Therefore, the domain for this component is \( (-\infty, 4) \).

Step 4: Find the domain of \( \sec(t) \)

The secant function \( \sec(t) = \frac{1}{\cos(t)} \) is undefined when \( \cos(t) = 0 \). This occurs at odd multiples of \( \frac{\pi}{2} \), i.e., \( t = (2k+1)\frac{\pi}{2} \), where \( k \) is an integer. Thus, the domain of \( \sec(t) \) is all real numbers except these points.

Step 5: Combine restrictions

The overall domain of \( \mathbf{r}(t) \) is the intersection of the individual domains. It must satisfy all component restrictions: no odd multiples of \( \frac{\pi}{2} \) and \( t < 4 \). Hence, the domain is \( \{ t \in \mathbb{R} \,|\ t < 4, \, t eq (2k+1)\frac{\pi}{2} \text{ for any integer } k \} \).

Key Concepts

Domain of Vector-Valued Functions

When dealing with vector-valued functions, determining the domain is all about finding where the entire vector is well-defined. This means ensuring each component of the vector function is defined at the same time. A vector-valued function, like \[ \mathbf{r}(t) = \langle e^t, \frac{1}{\sqrt{4-t}}, \sec(t) \rangle \]consists of several component functions: an exponential, a rational square root, and a secant function in this case. The overall domain is a combination of conditions ensuring that all these components are defined.

• First list each component function.
• Then determine the domain for each.
• The intersection of these domains will give us the domain of the entire vector function.

To work through this process, consider the individual domains for the functions before combining them for a final set of valid inputs.

Component Functions

Each piece of a vector-valued function is called a component function. For our function \[ \mathbf{r}(t) = \langle e^t, \frac{1}{\sqrt{4-t}}, \sec(t) \rangle \] there are three component functions: \(e^t\), \(\frac{1}{\sqrt{4-t}}\), and \(\sec(t)\).
Each of these components contributes its own rules and restrictions to the overall vector function:

• \(e^t\) is an exponential function, usually covering all real numbers without problem.
• \(\frac{1}{\sqrt{4-t}}\) involves a square root and rational expression, meaning it is only defined when \(t < 4\).
• \(\sec(t)\) depends on \(\cos(t)\) as \(\sec(t) = \frac{1}{\cos(t)}\) and is undefined where \(\cos(t)=0\), specifically at odd multiples of \(\frac{\pi}{2}\).

Together, these ensure the vector-valued function is completely defined only within certain bounds and conditions on \(t\).

Exponential Function Domains

Exponential functions like \(e^t\) are quite friendly when it comes to domains. An exponential function grows continuously for all real inputs, meaning there are no restrictions on \(t\). Therefore, the domain is \[ (-\infty, \infty) \].
This is particularly significant when combined with other more restricted functions. It means the exponential component does not restrict the vector function; it only remains for the other components to drive domain limitation, if any. Thus, \(e^t\) allows us to focus on where its partners struggle, such as with fractional or trigonometric conditions.

Secant Function Restrictions

The secant function, \(\sec(t)\), brings more specific challenges because it is defined based on \(\cos(t)\). Remember, \(\sec(t) = \frac{1}{\cos(t)}\), which is naturally problematic whenever \(\cos(t) = 0\). These occur at values:

• \( t = (2k+1)\frac{\pi}{2} \), where \( k \) is any integer.

The catch is that these are irregular interruptions along the continuous path of real numbers. They mean impurities exist unless avoided specifically. In vector-valued functions, such discontinuities need excluding, demanding sometimes complicated conditions like intersecting these exceptions with other conditions—say from \(\frac{1}{\sqrt{4-t}}\). The need for excluding specific points, such as odd multiples of half-\(\pi\), makes the real analysis here intricate.