Question

Find \(D_{t} \mathbf{r}(t)\) and \(D_{t}^{2} \mathbf{r}(t)\) for each of the following: (a) \(\mathbf{r}(t)=(3 t+4)^{3} \mathbf{i}+e^{i^{2}} \mathbf{j}+\mathbf{k}\) (b) \(\mathbf{r}(t)=\sin ^{2} t \mathbf{i}+\cos 3 t \mathbf{j}+t^{2} \mathbf{k}\)

Short answer

(a) First: \(27(3t+4)^2 \mathbf{i}\); Second: \(162(3t+4) \mathbf{i}\). (b) First: \(\sin(2t)\mathbf{i} - 3\sin(3t)\mathbf{j} + 2t\mathbf{k}\); Second: \(2\cos(2t)\mathbf{i} - 9\cos(3t)\mathbf{j} + 2\mathbf{k}\).

Step-by-step solution

Understand the Problem

We need to find the first derivative, \( D_t \mathbf{r}(t) \), and the second derivative, \( D_t^2 \mathbf{r}(t) \), of the vector functions given for parts (a) and (b). The components of \( \mathbf{r}(t) \) are functions of \( t \).

Differentiate Component-wise for (a)

For \( \mathbf{r}(t) = (3t+4)^3 \mathbf{i} + e^{i^2} \mathbf{j} + \mathbf{k} \), differentiate each component:- \( (3t+4)^3 \): Derivative is \( 3(3t+4)^2 \cdot 3 = 27(3t+4)^2 \)- \( e^{i^2} \): Derivative is 0 as \( e^{i^2} \) is constant w.r.t. \( t \)- \( \mathbf{k} \): Derivative is 0 as \( \mathbf{k} \) is constant w.r.t. \( t \)

Calculate First Derivative for (a)

The first derivative is:\[ D_t \mathbf{r}(t) = 27(3t+4)^2 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k} = 27(3t+4)^2 \mathbf{i} \]

Calculate Second Derivative for (a)

Differentiate \( D_t \mathbf{r}(t) \) again:- \( 27(3t+4)^2 \): Derivative is \( 2 \cdot 27 \cdot (3t+4)^{1} \cdot 3 = 162(3t+4) \)The second derivative is:\[ D_t^2 \mathbf{r}(t) = 162(3t+4) \mathbf{i} \]

Differentiate Component-wise for (b)

For \( \mathbf{r}(t) = \sin^2(t) \mathbf{i} + \cos(3t) \mathbf{j} + t^2 \mathbf{k} \), differentiate each component:- \( \sin^2(t) \): Derivative is \( 2\sin(t)\cos(t) = \sin(2t) \)- \( \cos(3t) \): Derivative is \( -3\sin(3t) \)- \( t^2 \): Derivative is \( 2t \)

Calculate First Derivative for (b)

The first derivative is:\[ D_t \mathbf{r}(t) = \sin(2t) \mathbf{i} - 3\sin(3t) \mathbf{j} + 2t \mathbf{k} \]

Calculate Second Derivative for (b)

Differentiate \( D_t \mathbf{r}(t) \) again:- \( \sin(2t) \): Derivative is \( 2\cos(2t) \)- \( -3\sin(3t) \): Derivative is \( -9\cos(3t) \)- \( 2t \): Derivative is \( 2 \)The second derivative is:\[ D_t^2 \mathbf{r}(t) = 2\cos(2t) \mathbf{i} - 9\cos(3t) \mathbf{j} + 2 \mathbf{k} \]

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus that allows us to determine the rate of change of a function. When we differentiate a function, we are essentially finding its derivative. This process involves taking the derivative of the function with respect to a particular variable, often time (denoted as \( t \) in our case). In vector calculus, differentiation extends to vector functions, where each component of the vector is differentiated separately with respect to the given variable.

In the given exercise, differentiation helps in finding the first derivative \( D_t \mathbf{r}(t) \) and second derivative \( D_t^2 \mathbf{r}(t) \) for vector functions. The exercise breaks down into differentiating the individual scalar functions that make up each component of the vector \( \mathbf{r}(t) \). This step-by-step approach simplifies complex differentiation tasks.

Vector Functions

Vector functions are fundamental in representing multi-dimensional systems in vector calculus. A vector function assigns a vector to each point in its domain, often parameterized by a variable such as time \( t \). These functions are widely used in physics and engineering to model motion and fields, capturing not just magnitude but direction as well.

In our exercise, the vector function \( \mathbf{r}(t) \) has components expressed as functions of \( t \), specifically functions assigned to \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \). For example, in part (a), the vector function is \( \mathbf{r}(t) = (3t+4)^3 \mathbf{i} + e^{i^2} \mathbf{j} + \mathbf{k} \), representing how each direction (or dimension) independently varies with \( t \).

This means to evaluate these functions, a separate calculus process is applied to each component, respecting the unique characteristics of vector operations.

Derivatives

Derivatives measure how a function changes as its input changes. In vector calculus, derivatives apply to each component of a vector function separately, which offers insight into how the vector's magnitude and direction change over time.

In part (a) of the exercise, finding the derivatives of the vector function \( \mathbf{r}(t) \) involves solving for the derivatives of each component, such as \( (3t+4)^3 \), \( e^{i^2} \), and constants, and then integrating these into a holistic change represented by \( D_t \mathbf{r}(t) \).

Similarly, in part (b), the derivative requires dealing with trigonometric and polynomial components, shown by finding expressions like \( \sin(2t) \), demonstrating derivative rules like the chain rule on \( \sin^2(t) \) and familiar product and power rules.

Calculus Solutions

Solving calculus problems often involves a systematic approach leveraging differentiation rules, understanding, and manipulating vector components. In the original exercise, calculus solutions are applied to determine the rates of change both at a first and second derivative level. This reflects the solving phase in calculus problems, often involving clear identification of each component to apply correct calculus principles.

The calculus solution process in step-by-step format ensures components like polynomials and trigonometric functions are accurately differentiated, and secondary efforts like taking the derivative of a derivative are handled precisely. By breaking down components, the process illustrates problem-solving in calculus, helping students grasp the progression from understanding the problem, through differentiation, to finding solutions, both initial and ongoing, within a clear and structured framework.