Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

If \(\mathbf{r}(t)=\sqrt{t-1} \mathbf{i}+\ln \left(2 t^{2}\right) \mathbf{j}\) and \(h(t)=e^{-3 t}\), find \(D_{r}[h(t) \mathbf{r}(t)]\)

Step-by-Step Solution

Answer

Use product rule and simplify: \(-3e^{-3t}\sqrt{t-1} \mathbf{i} + \frac{e^{-3t}}{2\sqrt{t-1}} \mathbf{i} - 3e^{-3t}\ln(2t^2) \mathbf{j} + \frac{e^{-3t}}{t} \mathbf{j}\)."

Step 1: Understand the Expression

We are given the expression \(h(t) \mathbf{r}(t)\), where \(\mathbf{r}(t)=\sqrt{t-1} \mathbf{i}+\ln \left(2 t^{2}\right) \mathbf{j}\) and \(h(t)=e^{-3 t}\). We need to find the derivative \(D_{r}[h(t) \mathbf{r}(t)]\).

Step 2: Recall the Product Rule for Derivatives

To differentiate the product of two functions, we'll use the product rule: if \(u(t)\) and \(v(t)\) are functions, then \(\frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t)\).

Step 3: Apply the Product Rule

Using the product rule, we need to differentiate the expression \(h(t)\mathbf{r}(t) = e^{-3t} (\sqrt{t-1} \mathbf{i} + \ln \left(2 t^{2}\right) \mathbf{j})\).

Step 4: Differentiate the Scalar Function \(h(t)\)

Differentiate \(h(t) = e^{-3t}\). The derivative is \(h'(t) = -3e^{-3t}\).

Step 5: Differentiate the Vector Function \(\mathbf{r}(t)\)

Differentiate each component of \(\mathbf{r}(t) = \sqrt{t-1} \mathbf{i} + \ln \left(2 t^{2}\right) \mathbf{j}\). - For \(\sqrt{t-1}\): The derivative is \(\frac{1}{2\sqrt{t-1}}\).- For \(\ln(2t^2)\): Using the chain rule, the derivative is \(\frac{1}{2t^2}2t = \frac{1}{t}\).

Step 6: Compute the Derivative

Substitute these into the product rule:\[ D_{r}[h(t) \mathbf{r}(t)] = h'(t)\mathbf{r}(t) + h(t) \frac{d}{dt}\mathbf{r}(t) \]This becomes:- \(-3e^{-3t}(\sqrt{t-1} \mathbf{i} + \ln(2t^2) \mathbf{j})\) (from \(h'(t)\mathbf{r}(t)\))- \( e^{-3t}(\frac{1}{2\sqrt{t-1}}\mathbf{i} + \frac{1}{t}\mathbf{j}) \) (from \(h(t)\frac{d}{dt}\mathbf{r}(t)\))

Step 7: Simplify the Expression

Combine both parts from Step 6:\[ D_{r}[h(t) \mathbf{r}(t)] = -3e^{-3t}\sqrt{t-1} \mathbf{i} - 3e^{-3t}\ln(2t^2) \mathbf{j} + \frac{e^{-3t}}{2\sqrt{t-1}} \mathbf{i} + \frac{e^{-3t}}{t} \mathbf{j} \].Simplify further by collecting like terms.

Key Concepts

Product Rule

The product rule is a fundamental technique in calculus used to differentiate functions that are multiplied together. In simpler terms, if you have two functions of a variable, say \(u(t)\) and \(v(t)\), and you multiply them, the product rule tells you how to find the derivative of that product. The formula for the product rule is:

• \(\frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t)\)

This means you take the derivative of the first function, multiply it by the second function, then add the product of the first function and the derivative of the second function. It's a way to systematically handle the derivatives of products, which come up frequently in mathematics and physics.

In our exercise, we apply the product rule to differentiate the expression \(h(t) \mathbf{r}(t) = e^{-3t} (\sqrt{t-1} \mathbf{i} + \ln(2t^{2}) \mathbf{j})\). Each component of the vector function gets the product rule applied to it, ensuring that the scalar and vector components are treated appropriately.

Vector Calculus

Vector calculus is an area of mathematics concerned with vector fields and operations on them. Vectors have both magnitude and direction, which means they are very useful for describing physical phenomena such as velocity and force. In vector calculus, we extend differentiation and integration to vector fields, handling scenarios that involve multiple dimensions.

Vector functions like \(\mathbf{r}(t) = \sqrt{t-1} \mathbf{i} + \ln(2t^2) \mathbf{j}\) express vectors that change with some parameter \(t\). Each vector component is a function of \(t\), and to differentiate, you manage each component separately. Comprehensive understanding of vector calculus functions is crucial for solving complex physics and engineering problems where real-world dynamics are modeled.

In our exercise, vector calculus principles help us understand how the vector function \(\mathbf{r}(t)\) evolves over time and how it interacts with the scalar function \(h(t)\). This interaction is expressed and managed using differentiation rules applied to vector fields.

Differentiation

Differentiation is the process of finding the derivative of a function, which measures how the function changes as its input changes. Derivatives are essential tools in calculus, providing the rate of change of quantities. For a scalar function like \(h(t) = e^{-3t}\), the derivative \(h'(t)\) represents how quickly \(e^{-3t}\) changes with respect to \(t\).

In vector functions, differentiation extends to each component of the vector independently. For \(\mathbf{r}(t)\), the vector derivative \(\mathbf{r}'(t)\) is computed by differentiating each component of \(\mathbf{r}(t)\):

• For \(\sqrt{t-1}\), the derivative is \(\frac{1}{2\sqrt{t-1}}\)
• For \(\ln(2t^2)\), the derivative is \(\frac{1}{t}\), using the chain rule

Differentiation allows us to analyze and predict changes, unlocks deeper insights into function behaviors over time, and equips us to tackle multi-dimensional problems efficiently.

Chain Rule

The chain rule is a powerful tool in calculus for finding the derivative of composite functions. If you have a nested function, like \(f(g(t))\), the chain rule provides a way to differentiate it by working 'inside out'. The rule is expressed as:

• \(\frac{d}{dt}[f(g(t))] = f'(g(t))g'(t)\)

The idea is that you differentiate the outer function \(f\) as if its argument was just \(t\), then multiply by the derivative of the inner function \(g(t)\).

In the expression \(\ln(2t^2)\), the chain rule is essential. The derivative \(\frac{1}{t}\) emerges because the chain rule allows us to manage the \(2t^2\) eye adroitly. First, you differentiate \(\ln(u)\) with respect to \(u\) (which is \(\frac{1}{u}\) when \(u = 2t^2\)), and then multiply by the derivative of \(u\) with respect to \(t\), which is \(4t\). Combined, this simplifies our derivatives effectively.

Mastering the chain rule is crucial for tackling complex derivatives and significantly expands the repertoire of functions we can handle easily, as it facilitates breaking down the layering of composite functions.