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Chapter 3: Problem 46

Compute the derivatives of the vector-valued functions. \(\mathbf{r}(t)=t e^{t} \mathbf{i}+t \ln (t) \mathbf{j}+\sin (3 t) \mathbf{k}\)

Short Answer

Expert verified
\(\mathbf{r}'(t) = (t e^t + e^t) \mathbf{i} + (\ln(t) + 1) \mathbf{j} + 3 \cos(3t) \mathbf{k}\)

Step by step solution

01

Understand the Function Components

The vector-valued function \(\mathbf{r}(t)\) is given as follows with three components: \( t e^{t} \mathbf{i} + t \ln(t) \mathbf{j} + \sin(3t) \mathbf{k} \). Each function component needs its own derivative. Let's break them down into individual functions: \( x(t) = t e^{t} \), \( y(t) = t \ln(t) \), and \( z(t) = \sin(3t) \).
02

Derivative of First Component \( x(t) = t e^{t} \)

Apply the product rule to find the derivative of the function \( x(t) = t e^t \):\[\frac{d}{dt}[x(t)] = \frac{d}{dt}[t e^t] = e^t (t) + e^t (1) = t e^t + e^t\]
03

Derivative of Second Component \( y(t) = t \ln(t) \)

Use the product rule to differentiate the component \( y(t) = t \ln(t) \):\[\frac{d}{dt}[y(t)] = \frac{d}{dt}[t \ln(t)] = 1 \cdot \ln(t) + t \cdot \frac{1}{t} = \ln(t) + 1\]
04

Derivative of Third Component \( z(t) = \sin(3t) \)

To differentiate \( z(t) = \sin(3t) \), use the chain rule:\[\frac{d}{dt}[z(t)] = \cos(3t) \cdot \frac{d}{dt}(3t) = 3 \cos(3t)\]
05

Combine the Derivatives

Gather the derivatives of all three components to find the derivative of the vector-valued function \( \mathbf{r}(t) \):\[\mathbf{r}'(t) = (t e^t + e^t) \mathbf{i} + (\ln(t) + 1) \mathbf{j} + 3 \cos(3t) \mathbf{k}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector-Valued Functions
A vector-valued function is a function that assigns a vector to each value in its domain, frequently used to describe paths or curves in space. Think of a vector-valued function as a combination of separate functions that come together to form a vector.
  • Each component of a vector-valued function is a real function of the same variable.
  • For example, the function \( \mathbf{r}(t) = t e^{t} \mathbf{i} + t \ln(t) \mathbf{j} + \sin(3t) \mathbf{k} \) in the original exercise uses components reliant on the variable \(t\).
  • These components allow vector-valued functions to describe quick movements and directions in a multi-dimensional space.
Understanding vector-valued functions is fundamental in physics and engineering since they can describe motion, forces, and many other physical phenomena.
Derivatives
A derivative, in simple terms, is the rate at which a function changes at a particular point. When dealing with vector-valued functions, derivatives allow us to understand how each component changes concerning its variable.
  • The process involves differentiating each component of the vector-valued function separately.
  • In our example, we derived the components \(x(t)\), \(y(t)\), and \(z(t)\), respectively, using standard differentiation rules.
Derivatives of vector-valued functions are crucial in understanding the velocity and acceleration of objects following those paths.
Product Rule
The product rule is a formula used to find the derivative of a product of two functions. It is essential when the function you're working with is a product of simpler functions.
  • The rule is expressed as: \( \frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t) \).
  • For \( x(t) = t e^t \), the derivative was calculated as \( e^t(t) + t(e^t) \), producing \( t e^t + e^t \).
  • Similarly, in the component \( y(t) = t \ln(t) \), the rule is applied to get \( \ln(t) + 1 \).
In essence, the product rule is a key tool in managing the complexity of differentiating multiplied functions.
Chain Rule
The chain rule is a method for differentiating compositions of functions. This comes in handy when a function is nested inside another.
  • It allows you to break down the derivative of a composite function into simpler steps.
  • For instance, in the component \( z(t) = \sin(3t) \), it required taking the derivative of the outer function, \( \cos(u) \), and multiply by the derivative of the inner function, \( 3t \).
  • Thus, the derivative turned out to be \( 3 \cos(3t) \).
The chain rule simplifies working with complex, multi-layered functions, particularly in vector calculus, helping to understand intricate changes of rate.

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