Problem 40 - Essential Calculus Early Transcendentals

Question

\(39-44=\) Find the derivative of the vector function. $$ \mathbf{r}(t)=\left\langle\tan t, \sec t, 1 / t^{2}\right\rangle $$

Step-by-Step Solution

Answer

The derivative is \( \mathbf{r}'(t) = \left\langle \sec^2 t, \sec t \tan t, -\frac{2}{t^3} \right\rangle \).

Step 1: Differentiate Each Component

The vector function is \( \mathbf{r}(t) = \left\langle \tan t, \sec t, \frac{1}{t^2} \right\rangle \). We need to find the derivative of each component with respect to \( t \):\[ \frac{d}{dt} \tan t, \frac{d}{dt} \sec t, \frac{d}{dt} \frac{1}{t^2} \]- The derivative of \( \tan t \) with respect to \( t \) is \( \sec^2 t \).- The derivative of \( \sec t \) with respect to \( t \) is \( \sec t \tan t \) using the chain rule.- The derivative of \( \frac{1}{t^2} \) with respect to \( t \) is \( -\frac{2}{t^3} \) using the power rule.

Step 2: Write the Derivative as a Vector

Combine the derivatives of each component into a new vector. The derivative of \( \mathbf{r}(t) \), which is \( \mathbf{r}'(t) \), is:\[ \mathbf{r}'(t) = \left\langle \sec^2 t, \sec t \tan t, -\frac{2}{t^3} \right\rangle \]

Key Concepts

Derivative of Vector Functions

In vector calculus, dealing with functions that have multiple components is common. A vector function, like \( \mathbf{r}(t) = \langle \tan t, \sec t, 1/t^2 \rangle \), assigns a vector to each value of \( t \). To find the derivative of such a function, we differentiate each component individually.
Let's break down this process. Each component of the vector function can be thought of as a separate, ordinary function of \( t \):

• \( \tan t \)
• \( \sec t \)
• \( \frac{1}{t^2} \)

Each of these functions can be differentiated using the standard rules of calculus, which we'll explore in detail.
Once these individual derivatives are calculated, they are combined back into a single vector. Therefore, the derivative of the whole vector function—denoted \( \mathbf{r}'(t) \)—is formed by these new components.

Differentiation Techniques

Differentiation is the process of finding the derivative of a function, which gives us the rates of change of the function with respect to its independent variable. When dealing with the vector function \( \mathbf{r}(t) = \langle \tan t, \sec t, 1/t^2 \rangle \), there are certain techniques used to differentiate each component correctly.
For the component \( \tan t \):

• We typically use the trigonometric derivative rule: the derivative of \( \tan t \) is \( \sec^2 t \).

For \( \sec t \):

• The chain rule comes into play, providing us with \( \sec t \tan t \) as its derivative.

Finally, for \( \frac{1}{t^2} \):

• The power rule is applied here. Converting \( \frac{1}{t^2} \) to \( t^{-2} \) allows us to use the formula \( d/dt[t^n] = n \cdot t^{n-1} \), resulting in \( -2/t^3 \).

These differentiation techniques are essential tools in vector calculus that allow us to navigate through complex problems with ease and find solutions efficiently.

Vector Functions in Calculus

Vector functions are an integral part of calculus. They extend calculus concepts to higher dimensions by considering vectors instead of just scalar functions. A vector function assigns a vector—an entity with both magnitude and direction—to each value in its domain.
Understanding vector functions is crucial in fields like physics and engineering, where multidimensional systems frequently appear. When working with vector functions, one might encounter tasks such as differentiation and integration, each with its unique rules and approaches.
In the vector function \( \mathbf{r}(t) = \langle \tan t, \sec t, 1/t^2 \rangle \), the aim is often to describe changes or motions within a system, like the path or velocity of a moving object. Such descriptions require comprehensive analysis including the derivatives of each component.
Thus, knowing how to handle vector functions enhances one's ability to model and resolve real-world problems, making them a powerful tool in the broader mathematical toolkit.