Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

\text { If } \mathbf{r}(t)=\sin 3 t \mathbf{i}-\cos 3 t \mathbf{j}, \text { find } D_{t}\left[\mathbf{r}(t) \cdot \mathbf{r}^{\prime}(t)\right]

Step-by-Step Solution

Answer

The derivative is 0.

Step 1: Understand the Given Vectors

We have the vector \( \mathbf{r}(t) = \sin 3t \mathbf{i} - \cos 3t \mathbf{j} \). We need to first find its derivative, \( \mathbf{r}'(t) \).

Step 2: Calculate the Derivative \( \mathbf{r}'(t) \)

Differentiate \( \mathbf{r}(t) = \sin 3t \mathbf{i} - \cos 3t \mathbf{j} \):1. The derivative of \( \sin 3t \) is \( 3\cos 3t \), thus \( \frac{d}{dt}(\sin 3t \mathbf{i}) = 3\cos 3t \mathbf{i} \).2. The derivative of \( -\cos 3t \) is \( 3\sin 3t \), thus \( \frac{d}{dt}(-\cos 3t \mathbf{j}) = 3\sin 3t \mathbf{j} \).Therefore, \( \mathbf{r}'(t) = 3\cos 3t \mathbf{i} + 3\sin 3t \mathbf{j} \).

Step 3: Compute the Dot Product \( \mathbf{r}(t) \cdot \mathbf{r}'(t) \)

The dot product \( \mathbf{r}(t) \cdot \mathbf{r}'(t) \) is calculated as:\[\mathbf{r}(t) \cdot \mathbf{r}'(t) = (\sin 3t)(3\cos 3t) + (-\cos 3t)(3\sin 3t) \].This simplifies to:\[ 3\sin 3t \cos 3t - 3\sin 3t \cos 3t = 0. \]

Step 4: Differentiate the Dot Product Zero Function

Since \( \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0 \) as shown above, its derivative with respect to \( t \) is also zero:\[ D_{t}\left[ \mathbf{r}(t) \cdot \mathbf{r}'(t) \right] = 0. \]

Key Concepts

Understanding the Dot Product

In vector calculus, the dot product is a crucial concept that helps to find the angle between two vectors and to compute the projection of one vector onto another. This operation takes two equal-length sequences of numbers (often in the form of vectors) and returns a single number.

For two vectors, \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} \) and \( \mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated as:

• \( a_1 b_1 + a_2 b_2 \)

In our task, we have two vectors: \( \mathbf{r}(t) \) and its derivative, \( \mathbf{r}'(t) \). The dot product \( \mathbf{r}(t) \cdot \mathbf{r}'(t) \) is used not only in finding angles but also in simplifying expressions during the differentiation process.
Here, it helps to determine whether the vectors are orthogonal, which they often are if their dot product equals zero.

Introduction to Differentiation in Vector Calculus

Differentiation is a technique used to determine the rate at which a function changes. When applied to vector functions, differentiation helps us understand how the vector changes with respect to a variable, usually time or space.

In vector calculus, we differentiate each component of a vector function separately. Given \( \mathbf{r}(t) = \sin 3t \mathbf{i} - \cos 3t \mathbf{j} \), the components are \( \sin 3t \) and \( -\cos 3t \). Differentiating these:

• The derivative of \( \sin 3t \) is \( 3\cos 3t \).
• The derivative of \( -\cos 3t \) is \( 3\sin 3t \).

This gives the derivative vector \( \mathbf{r}'(t) = 3\cos 3t \mathbf{i} + 3\sin 3t \mathbf{j} \), showing how \( \mathbf{r}(t) \) changes over time at each point.

Vector Derivatives Simplified

Understanding vector derivatives involves taking derivatives of all component functions of a vector. The process is straightforward but requires attention to detail with trigonometric and exponential functions.

For a given vector function \( \mathbf{r}(t) \), we differentiate each component separately:

• For \( \mathbf{r}(t) = \sin 3t \mathbf{i} - \cos 3t \mathbf{j} \), differentiate to find: \( \mathbf{r}'(t) = 3 \cos 3t \mathbf{i} + 3 \sin 3t \mathbf{j} \).

This new vector, \( \mathbf{r}'(t) \), represents the velocity of the point over time, reflecting both direction and rate of change. Studying vector derivatives helps to solve many physical problems like those in mechanics and electromagnetism by equating rates of change to forces or other vector fields.

Solving Calculus Problems with Vectors

Calculus problems often involve understanding both the rate of change and the behavior of functions, which is where vectors come into play.

In our original exercise, we used vectors to understand geometric properties and their impact on tasks like finding derivatives. Given vector \( \mathbf{r}(t) = \sin 3t \mathbf{i} - \cos 3t \mathbf{j} \), key tasks included:

• Finding the derivative vector \( \mathbf{r}'(t) \).
• Using the dot product, \( \mathbf{r}(t) \cdot \mathbf{r}'(t) \), to explore the relationships.

The process offers insight into the motion and trajectory of elements defined by vector functions.
By using calculus to differentiate vectors and a dot product, we can solve complex problems including those involving tangent lines and curves, where understanding the "direction" and "magnitude" at any given point is essential.