Problem 41 - Essential Calculus Early Transcendentals

Question

\(39-44=\) Find the derivative of the vector function. $$ \mathbf{r}(t)=e^{t^{2}} \mathbf{i}-\mathbf{j}+\ln (1+3 t) \mathbf{k} $$

Step-by-Step Solution

Answer

\( \mathbf{r}'(t) = 2t e^{t^2} \mathbf{i} + \frac{3}{1+3t} \mathbf{k} \).

Step 1: Understanding the Vector Function

The vector function \( \mathbf{r}(t) \) is given by \( \mathbf{r}(t) = e^{t^2} \mathbf{i} - \mathbf{j} + \ln(1+3t) \mathbf{k} \). It is composed of three components: \( e^{t^2} \) along the \( \mathbf{i} \) direction, a constant \( -1 \) along the \( \mathbf{j} \) direction, and \( \ln(1+3t) \) along the \( \mathbf{k} \) direction. We need to find the derivative of each component separately.

Step 2: Derivative of the \( \mathbf{i} \) Component

The \( \mathbf{i} \) component of \( \mathbf{r}(t) \) is \( e^{t^2} \). To differentiate \( e^{t^2} \), use the chain rule. The derivative of the exponent \( t^2 \) with respect to \( t \) is \( 2t \). Thus, the derivative of \( e^{t^2} \) is \( 2t e^{t^2} \).

Step 3: Derivative of the \( \mathbf{j} \) Component

The \( \mathbf{j} \) component is a constant \(-1\), which does not depend on \( t \). Therefore, its derivative is 0.

Step 4: Derivative of the \( \mathbf{k} \) Component

The \( \mathbf{k} \) component is \( \ln(1+3t) \). The derivative of \( \ln(u) \) is \( \frac{1}{u} \cdot \frac{du}{dt} \). Here, \( u = 1+3t \), so \( \frac{du}{dt} = 3 \). Thus, the derivative of \( \ln(1+3t) \) is \( \frac{3}{1+3t} \).

Step 5: Combine the Derivatives

Combine the derivatives found in each component. The derivative \( \mathbf{r}'(t) \) is \( \left(2t e^{t^2}\right) \mathbf{i} + 0 \mathbf{j} + \left(\frac{3}{1+3t}\right) \mathbf{k} \). This simplifies to \( \mathbf{r}'(t) = 2t e^{t^2} \mathbf{i} + \frac{3}{1+3t} \mathbf{k} \).

Key Concepts

Derivative of Vector Functions

In vector calculus, finding the derivative of a vector function is a fundamental task. It involves taking the derivative of each of its components separately. A vector function is often represented as \( \mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} + h(t) \mathbf{k} \). Here, \( f(t) \), \( g(t) \), and \( h(t) \) are the component functions of the vector along the \( \mathbf{i}, \mathbf{j}, \) and \( \mathbf{k} \) directions, respectively.

To find the derivative \( \mathbf{r}'(t) \), we compute the derivative of each component:

• \( f'(t) \mathbf{i} \)
• \( g'(t) \mathbf{j} \)
• \( h'(t) \mathbf{k} \)

This gives us a new vector, which represents the rate of change of the original vector function with respect to \( t \). For example, if we have \( \mathbf{r}(t) = e^{t^{2}} \mathbf{i} - \mathbf{j} + \ln(1+3t) \mathbf{k} \), we simply differentiate each term to get \( \mathbf{r}'(t) = 2te^{t^{2}} \mathbf{i} + 0 \mathbf{j} + \frac{3}{1+3t} \mathbf{k} \).

Thus, the derivatives of vector functions provide insight into how their components change and result in a new vector indicative of this change.

Chain Rule in Calculus

The chain rule is a vital tool in calculus used to differentiate composite functions. This rule allows us to take the derivative of a function nested within another function. It becomes especially handy when dealing with vector functions, as seen in our previous example.

For a composite function \( f(g(t)) \), the chain rule states that its derivative is \( f'(g(t)) \cdot g'(t) \). In simple terms, we first differentiate the outer function, keeping the inner function as it is, then multiply by the derivative of the inner function.

Applying the chain rule helps simplify complex differentiations. Let's consider the component \( e^{t^2} \) from our earlier vector function example. The derivative of \( e^{t^2} \) involves finding the derivative of the exponential function \( e^u \) with \( u = t^2 \):

• The derivative of \( e^u \) is \( e^u \).
• The derivative of \( t^2 \) using basic rules is \( 2t \).

Thus, applying the chain rule gives \( 2t e^{t^2} \).

This process demonstrates the efficiency and utility of the chain rule, particularly when dealing with exponential and logarithmic components in vector functions.

Components of Vector Functions

The components of a vector function are the individual functions that describe a vector's behavior in different spatial directions. A vector function like \( \mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} + h(t) \mathbf{k} \) consists of three primary components:

• \( f(t) \), controlling the vector's behavior along the \( \mathbf{i} \)-axis
• \( g(t) \), dictating the vector's influence along the \( \mathbf{j} \)-axis
• \( h(t) \), impacting the vector in the \( \mathbf{k} \)-axis direction

The understanding of each component separately is crucial when differentiating vector functions. Each function might have its distinctive features, such as being a polynomial, an exponential function, or a logarithmic function.

In our example, the vector function \( \mathbf{r}(t) = e^{t^2} \mathbf{i} - \mathbf{j} + \ln(1+3t) \mathbf{k} \) has distinct components:

• \( e^{t^2} \) as the \( \mathbf{i} \)-component, signifying exponential growth based on \( t^2 \)
• -1 as the \( \mathbf{j} \)-component, a constant component indicating no change
• \( \ln(1+3t) \) representing the \( \mathbf{k} \)-component, which models a logarithmic change

Recognizing and working with each component plays a critical role in applying calculus concepts like differentiation accurately.