Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { If } \mathbf{r}(t)=e^{-t} \mathbf{i}-\ln \left(t^{2}\right) \mathbf{j} \text {, find } D_{t}\left[\mathbf{r}(t) \cdot \mathbf{r}^{\prime \prime}(t)\right] \text {. } $$

Step-by-Step Solution

Answer

The derivative is \( -2e^{-2t} + \frac{4 \ln(t)}{t^3} - \frac{4}{t^3} \).

Step 1: Differentiate the Vector Function

Calculate the first derivative of \( \mathbf{r}(t) \), denoted as \( \mathbf{r}'(t) \). For \( \mathbf{r}(t) = e^{-t} \mathbf{i} - \ln(t^2) \mathbf{j} \), differentiate each component. The derivative of \( e^{-t} \) with respect to \( t \) is \( -e^{-t} \), while the derivative of \( -\ln(t^2) \) is \( -\frac{2}{t} \). Thus, \( \mathbf{r}'(t) = -e^{-t} \mathbf{i} - \frac{2}{t} \mathbf{j} \).

Step 2: Differentiate the First Derivative

Calculate the second derivative, \( \mathbf{r}''(t) \). Differentiate \( \mathbf{r}'(t) = -e^{-t} \mathbf{i} - \frac{2}{t} \mathbf{j} \) with respect to \( t \). The derivative of \( -e^{-t} \) is \( e^{-t} \), and the derivative of \( -\frac{2}{t} \) is \( \frac{2}{t^2} \). Therefore, \( \mathbf{r}''(t) = e^{-t} \mathbf{i} + \frac{2}{t^2} \mathbf{j} \).

Step 3: Compute the Dot Product

Find the dot product \( \mathbf{r}(t) \cdot \mathbf{r}''(t) \). Compute this as \((e^{-t} \mathbf{i} - \ln(t^2) \mathbf{j}) \cdot (e^{-t} \mathbf{i} + \frac{2}{t^2} \mathbf{j})\). This results in \( (e^{-t})(e^{-t}) + (-\ln(t^2))(\frac{2}{t^2}) = e^{-2t} - \frac{2 \ln(t^2)}{t^2} \).

Step 4: Differentiate the Dot Product

Differentiate the expression \( e^{-2t} - \frac{2 \ln(t^2)}{t^2} \) with respect to \( t \). The derivative of \( e^{-2t} \) is \( -2e^{-2t} \). For \( -\frac{2 \ln(t^2)}{t^2} \), use the quotient rule or simplify to differentiate, yielding \( \frac{4 \ln(t)}{t^3} - \frac{4}{t^3} \). The complete derivative is \( -2e^{-2t} + \frac{4 \ln(t)}{t^3} - \frac{4}{t^3} \).

Key Concepts

Derivatives of Vector Functions

When we talk about derivatives of vector functions, we are referring to how each component of a vector changes with respect to a variable—in this case, time \( t \). A vector function like \( \mathbf{r}(t) = e^{-t} \mathbf{i} - \ln(t^2) \mathbf{j} \) is composed of scalar functions for each of its coordinate directions: \( \mathbf{i} \) and \( \mathbf{j} \). Calculating the derivative of \( \mathbf{r}(t) \) involves differentiating these individual components:

• The derivative of \( e^{-t} \) in the x-direction (\( \mathbf{i} \)) results in \( -e^{-t} \).
• In the y-direction (\( \mathbf{j} \)), the derivative of \( -\ln(t^2) \) is determined using the chain rule, producing \( -\frac{2}{t} \).

If you compute these derivatives correctly, you combine them to get \( \mathbf{r}'(t) = -e^{-t} \mathbf{i} - \frac{2}{t} \mathbf{j} \). It’s clear that each component responds uniquely to the change, emphasizing the importance of component-wise differentiation.

Dot Product Differentiation

The dot product in vector calculus is a way of multiplying two vectors to produce a scalar. In this problem, you need to calculate the dot product of \( \mathbf{r}(t) \) and \( \mathbf{r}''(t) \). This product is computed by multiplying corresponding components and summing up:

• For the \( \mathbf{i} \) component, \( e^{-t} \times e^{-t} = e^{-2t} \).
• And for the \( \mathbf{j} \) component, \( -\ln(t^2) \times \frac{2}{t^2} = -\frac{2\ln(t^2)}{t^2} \).

The resulting expression \( e^{-2t} - \frac{2\ln(t^2)}{t^2} \) outlines the dot product. This is suitable for differentiation since it’s a scalar, not a vector anymore, simplifying further analysis.

Quotient Rule in Calculus

When you encounter a quotient, like \( \frac{2\ln(t^2)}{t^2} \), and you need its derivative, you apply the quotient rule. The quotient rule is stated as: if \( u(t) \) and \( v(t) \) are functions of \( t \) such that \( y = \frac{u(t)}{v(t)} \), then:\[\frac{dy}{dt} = \frac{v(t) \frac{du}{dt} - u(t) \frac{dv}{dt}}{(v(t))^2}\]Utilizing this formula, you find the derivative of \( -\frac{2\ln(t^2)}{t^2} \) as follows:

• Let \( u(t) = 2\ln(t^2) \) and \( v(t) = t^2 \).
• The derivative \( \frac{du}{dt} \) simplifies to \( \frac{4}{t} \), and \( \frac{dv}{dt} = 2t \).
• Substituting into the quotient rule gives \( \frac{4\ln(t)}{t^3} - \frac{4}{t^3} \).

This methodical approach helps uncover complexities within the scalar expression \( -\frac{2\ln(t^2)}{t^2} \) when treated as a differentiable quotient.