Problem 61 - Calculus 3

Question

Let $$\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}-t^{4} \mathbf{k}$$ and $$s(t)=\sin (t) \mathbf{i}+e^{t} \mathbf{j}+\cos (t) \mathbf{k}$$ Here is the graph of the function: Find the following. $$\frac{d}{d t}[r(t) \cdot s(t)]$$

Step-by-Step Solution

Answer

The derivative is \((1+t)\cos(t) + (2t + t^2)e^t + t^4 \sin(t)\).

Step 1: Understanding the Dot Product

The dot product between two vectors \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} - t^4 \mathbf{k} \) and \( s(t) = \sin(t) \mathbf{i} + e^t \mathbf{j} + \cos(t) \mathbf{k} \) is given by \( \mathbf{r}(t) \cdot s(t) = (t \cdot \sin(t)) + (t^2 \cdot e^t) + (-t^4 \cdot \cos(t)) \). We'll find the derivative with respect to \( t \).

Step 2: Dot Product Derivative Rule

Using the derivative rule for a dot product, \( \frac{d}{dt} [\mathbf{a}(t) \cdot \mathbf{b}(t)] = \mathbf{a}'(t) \cdot \mathbf{b}(t) + \mathbf{a}(t) \cdot \mathbf{b}'(t) \), we'll compute \( \mathbf{r}'(t) \) and \( s'(t) \) individually.

Step 3: Finding Derivative of \( \mathbf{r}(t) \)

The derivative of \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} - t^4 \mathbf{k} \) with respect to \( t \) is \( \mathbf{r}'(t) = \mathbf{i} + 2t \mathbf{j} - 4t^3 \mathbf{k} \).

Step 4: Finding Derivative of \( s(t) \)

The derivative of \( s(t) = \sin(t) \mathbf{i} + e^t \mathbf{j} + \cos(t) \mathbf{k} \) with respect to \( t \) is \( s'(t) = \cos(t) \mathbf{i}+ e^t \mathbf{j} - \sin(t) \mathbf{k} \).

Step 5: Applying Dot Product Derivative Rule

Now substitute \( \mathbf{r}'(t) \) and \( s'(t) \) into the dot product derivative rule. Calculate \( \mathbf{r}'(t) \cdot s(t) = (\mathbf{i} + 2t \mathbf{j} - 4t^3 \mathbf{k}) \cdot (\sin(t) \mathbf{i} + e^t \mathbf{j} + \cos(t) \mathbf{k}) \) leading to \( \cos(t) + 2t \mathbf{e}^{t} - 4t^3 \cos(t) \).

Step 6: Completing the Computation

Next, calculate \( \mathbf{r}(t) \cdot s'(t) = (t \mathbf{i} + t^2 \mathbf{j} - t^4 \mathbf{k}) \cdot (\cos(t) \mathbf{i} + e^t \mathbf{j} - \sin(t) \mathbf{k}) \), leading to \( t \cos(t) + t^2 e^t + t^4 \sin(t) \).

Step 7: Summing Up the Results

Sum the results from the previous step: \( \frac{d}{dt} [\mathbf{r}(t) \cdot s(t)] = ( \cos(t) + 2t e^t - 4t^3 \cos(t) ) + ( t \cos(t) + t^2 e^t + t^4 \sin(t) ) \). After combining terms, the answer is \( (1+t)\cos(t) + (2t + t^2)e^t + t^4 \sin(t) \).

Key Concepts

Vector Calculus

Vector calculus is an important branch of mathematics that deals with vector fields and operations on them. It extends the concepts of calculus to multiple dimensions where vectors represent quantities that have both magnitude and direction, such as force or velocity.

In vector calculus, the basic objects are vectors and their derivatives. Two primary calculations involving vectors include the dot product and cross product. The dot product, in particular, results in a scalar and is expressed as the sum of the products of corresponding components of two vectors.

For the vectors \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} - t^4 \mathbf{k} \) and \( s(t) = \sin(t) \mathbf{i} + e^t \mathbf{j} + \cos(t) \mathbf{k} \), their dot product is \( \mathbf{r}(t) \cdot s(t) = t \sin(t) + t^2 e^t - t^4 \cos(t) \). This expression is key in applying calculus operations like differentiation on vector-valued functions.

Derivative

A derivative, in simple terms, measures how a function changes as its input changes. It's a fundamental concept in calculus and is used to understand rates of change and slopes of curves.

When working with vector-calculus, we often need to find derivatives of vector-valued functions. The derivative of a vector function like \( \mathbf{r}(t) \), which represents a position vector, involves differentiating each component separately. Similarly, the derivative here of \( s(t) \) is computed by finding the derivative of each component function.

The product rule for derivatives extends to vector-valued functions in the form: \( \frac{d}{dt} [\mathbf{a}(t) \cdot \mathbf{b}(t)] = \mathbf{a}'(t) \cdot \mathbf{b}(t) + \mathbf{a}(t) \cdot \mathbf{b}'(t) \). This rule lets us differentiate the dot product of two vector functions efficiently by breaking it down into simpler parts.

Parameterization

Parameterization refers to expressing a set of quantities as explicit functions of one or more independent variables called parameters. In vector calculus, using parameterization allows one to describe geometric shapes and paths using vector functions.

For the given vector functions, \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} - t^4 \mathbf{k} \) and \( s(t) = \sin(t) \mathbf{i} + e^t \mathbf{j} + \cos(t) \mathbf{k} \), the variable \( t \) serves as the parameter. Parameterizing curves or surfaces enables easier calculation of derivatives, integrations, and applications of vector calculus operations.

By parameterizing these vectors as a function of \( t \), we turn them into functions that are simpler to analyze and differentiate.

Functions of a Single Variable

Functions of a single variable are a foundational concept in calculus where outputs depend on only one input variable. In the context of vector functions, this idea is extended where each component of the vector is a function of the single variable.

In our exercise, each component of the vectors \( \mathbf{r}(t) \) and \( s(t) \) is a distinct function of the variable \( t \). For instance, \( t \), \( t^2 \), and \(-t^4 \) in the vector \( \mathbf{r}(t) \) are functions of \( t \), as are \( \sin(t) \), \( e^t \), and \( \cos(t) \) in \( s(t) \).

Understanding how each component varies with \( t \) is crucial for calculating dots and derivatives as shown in the solution process where we find the derivative of their dot product by considering each function separately.