Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

. If \(\mathbf{r}(t)=\sin 2 t \mathbf{i}+\cosh t \mathbf{j}\) and \(h(t)=\ln (3 t-2)\), find \(D_{t}[h(t) \mathbf{r}(t)]\).

Step-by-Step Solution

Answer

Use the product rule to find the derivative: \( 2\cos(2t)\ln(3t-2)\mathbf{i} + \sinh(t)\ln(3t-2)\mathbf{j} + \frac{3\sin(2t)}{3t-2}\mathbf{i} + \frac{3\cosh(t)}{3t-2}\mathbf{j} \).

Step 1: Identify the expression to differentiate

We need to find the derivative of the product \(h(t) \mathbf{r}(t)\), where \(h(t) = \ln(3t-2)\) and \(\mathbf{r}(t) = \sin(2t)\mathbf{i} + \cosh(t)\mathbf{j}\). To differentiate a product, we will apply the product rule.

Step 2: Apply the Product Rule

The product rule states that \(D_{t}[h(t) \mathbf{r}(t)] = h(t) D_{t}[\mathbf{r}(t)] + \mathbf{r}(t) D_{t}[h(t)]\). Here, we will derive each part separately.

Step 3: Differentiate \(\mathbf{r}(t)\)

Differentiate \(\mathbf{r}(t) = \sin(2t)\mathbf{i} + \cosh(t)\mathbf{j}\):- The derivative of \(\sin(2t)\) is \(2 \cos(2t)\), so the \(\mathbf{i}\) component is \(2 \cos(2t)\mathbf{i}\).- The derivative of \(\cosh(t)\) is \(\sinh(t)\), so the \(\mathbf{j}\) component is \(\sinh(t)\mathbf{j}\).Thus, \(D_{t}[\mathbf{r}(t)] = 2 \cos(2t)\mathbf{i} + \sinh(t)\mathbf{j}\).

Step 4: Differentiate \(h(t)\)

The derivative of \(h(t) = \ln(3t-2)\) is found using the chain rule: \[D_{t}[h(t)] = \frac{1}{3t-2} \cdot 3 = \frac{3}{3t-2}\].

Step 5: Substitute derivatives into the Product Rule

Substitute the derived expressions back into the product rule:The expression becomes:\[D_{t}[h(t) \mathbf{r}(t)] = \ln(3t-2)(2 \cos(2t)\mathbf{i} + \sinh(t)\mathbf{j}) + (\sin(2t)\mathbf{i} + \cosh(t)\mathbf{j})\frac{3}{3t-2}\]

Step 6: Simplify the expression

Combine and distribute each term:- First part: \(\ln(3t-2) \cdot 2\cos(2t)\mathbf{i} + \ln(3t-2) \cdot \sinh(t)\mathbf{j}\).- Second part: \(\frac{3 \sin(2t)}{3t-2} \mathbf{i} + \frac{3 \cosh(t)}{3t-2} \mathbf{j}\).Combine the terms to get:\[2\cos(2t)\ln(3t-2)\mathbf{i} + \sinh(t)\ln(3t-2)\mathbf{j} + \frac{3\sin(2t)}{3t-2}\mathbf{i} + \frac{3\cosh(t)}{3t-2}\mathbf{j}\]

Key Concepts

Understanding the Product Rule in Calculus

When you are working with derivatives, there are several rules to help you differentiate various functions. One of the most useful is the Product Rule. This is especially important when dealing with the multiplication of two functions, as it allows you to handle their derivative efficiently.
This rule states that if you have two functions, say \( u(t) \) and \( v(t) \), then the derivative of their product \( u(t) v(t) \) can be derived using the following formula:

• \( D_t [u(t) v(t)] = u(t) D_t[v(t)] + v(t) D_t[u(t)] \)

In our problem, this involves applying the product rule to \( h(t) \) and \( \mathbf{r}(t) \)..
The idea is simple: you take the derivative of the first function, then multiply by the second function as it is. Then, you swap the process by keeping the first function as it is and take the derivative of the second one. Make sure to add the results together.
This technique is particularly useful in vector calculus, where functions can represent vectors and still interact under the product rule.

The Chain Rule for Derivatives

The Chain Rule is another essential technique in calculus, especially when you need to differentiate compositions of functions. It provides a systematic way to tackle derivatives when one function is nested within another. Simply put, the chain rule helps us find the derivative of a composite function.
Imagine you have a function \( f(g(t)) \), which means that function \( g \) is inside function \( f \). The chain rule tells us to multiply the derivative of the outer function by the derivative of the inner function. This is represented as:

• \( D_t[f(g(t))] = f'(g(t)) \cdot g'(t) \)

In our scenario, the function \( h(t) = \ln(3t-2) \) is a perfect example of when the chain rule comes handy since it's a composition of the natural logarithm and a linear function.
Following the chain rule, we first look at the derivative of the natural log which gives us \( \frac{1}{3t-2} \), and then multiply by the derivative of the inside function, \( 3 \). This gives \( \frac{3}{3t-2} \) as the result.

Introduction to Vector Functions and Their Differentiation

Vector functions are quite common in calculus, involving vectors as the output of given equations. These functions are crucial in representing physical quantities like velocity and force in physics, where quantities have both magnitude and direction.
For a vector function \( \mathbf{r}(t) \), differentiation means finding how the function changes over time. The differentiation is done component-wise, meaning each vector direction is treated separately.
In our example, the vector function \( \mathbf{r}(t) = \sin(2t)\mathbf{i} + \cosh(t)\mathbf{j} \) has components along the \( \mathbf{i} \) and \( \mathbf{j} \) directions.
To find the derivative \( D_t[\mathbf{r}(t)] \), you differentiate each part independently:

• The \( \sin(2t) \) becomes \( 2\cos(2t) \) along \( \mathbf{i} \).
• The \( \cosh(t) \) becomes \( \sinh(t) \) along \( \mathbf{j} \).

The result illustrates how changes in \( t \) affect \( \mathbf{r}(t) \) and showcases the beauty of vector calculus.