Problem 65 - Essential Calculus Early Transcendentals

Question

Find \(\mathbf{r}(t)\) if \(\mathbf{r}^{\prime}(t)=2 t \mathbf{i}+3 t^{2} \mathbf{j}+\sqrt{t} \mathbf{k}\) and \(\mathbf{r}(1)=\mathbf{i}+\mathbf{j}\)

Step-by-Step Solution

Answer

\( \mathbf{r}(t) = t^2 \mathbf{i} + t^3 \mathbf{j} + \left( \frac{2}{3} t^{3/2} - \frac{2}{3} \right) \mathbf{k} \)

Step 1: Understanding the Problem

We need to find the vector function \( \mathbf{r}(t) \) given its derivative \( \mathbf{r}^{\prime}(t) = 2t \mathbf{i} + 3t^2 \mathbf{j} + \sqrt{t} \mathbf{k} \), and an initial condition \( \mathbf{r}(1) = \mathbf{i} + \mathbf{j} \). This requires integrating each component of the derivative vector to find the original vector function.

Step 2: Integrate Each Component

To find \( \mathbf{r}(t) \), we integrate each component of the derivative vector separately:- \( \mathbf{i} \)-component: \( \int 2t \, dt = t^2 + C_1 \)- \( \mathbf{j} \)-component: \( \int 3t^2 \, dt = t^3 + C_2 \)- \( \mathbf{k} \)-component: \( \int \sqrt{t} \, dt = \frac{2}{3} t^{3/2} + C_3 \)Thus, initially, \( \mathbf{r}(t) = (t^2 + C_1) \mathbf{i} + (t^3 + C_2) \mathbf{j} + \left( \frac{2}{3} t^{3/2} + C_3 \right) \mathbf{k} \).

Step 3: Apply the Initial Condition

We use the initial condition \( \mathbf{r}(1) = \mathbf{i} + \mathbf{j} \) to solve for the constants:- From \( \mathbf{i} \)-component: \( 1^2 + C_1 = 1 \) implies \( C_1 = 0 \).- From \( \mathbf{j} \)-component: \( 1^3 + C_2 = 1 \) implies \( C_2 = 0 \).- From \( \mathbf{k} \)-component: \( \frac{2}{3}(1)^{3/2} + C_3 = 0 \) implies \( C_3 = -\frac{2}{3} \).

Step 4: Construct the Final Function

Using the values of \( C_1 \), \( C_2 \), and \( C_3 \), the vector function \( \mathbf{r}(t) \) becomes:\[ \mathbf{r}(t) = t^2 \mathbf{i} + t^3 \mathbf{j} + \left( \frac{2}{3} t^{3/2} - \frac{2}{3} \right) \mathbf{k} \].

Key Concepts

Integration

Integration is a fundamental concept in calculus involving the process of finding antiderivatives or the original function given its derivative. In this problem, we use integration to determine the function \( \mathbf{r}(t) \) from its derivative \( \mathbf{r}^{\prime}(t) \).
We achieve this by integrating each vector component separately:

• For the \( \mathbf{i} \)-component: Start with the integrand \( 2t \), integrating it yields \( \int 2t \, dt = t^2 + C_1 \).
• For the \( \mathbf{j} \)-component: The integrand is \( 3t^2 \), so \( \int 3t^2 \, dt = t^3 + C_2 \).
• For the \( \mathbf{k} \)-component: Begin with \( \sqrt{t} \), integrating gives \( \int \sqrt{t} \, dt = \frac{2}{3} t^{3/2} + C_3 \).

Each constant \( C_n \) represents an integration constant, which helps adjust the function to satisfy initial conditions later.

Vector Functions

Vector functions are mathematical expressions that describe a vector's magnitude and direction as functions of a parameter, usually time \( t \). In this exercise, \( \mathbf{r}(t) \) is our vector function. It is a position vector that includes three components corresponding to the \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) directions.
Each component is a scalar function of \( t \) and is determined by the integration of the corresponding component of the derivative \( \mathbf{r}^{\prime}(t) \). Thus, the combined equation of the vector function is initially expressed as:

• \( \mathbf{r}(t) = (t^2 + C_1) \mathbf{i} + (t^3 + C_2) \mathbf{j} + \left( \frac{2}{3} t^{3/2} + C_3 \right) \mathbf{k} \)

These functions are employed in physics to describe motion, including translation and rotation of objects over time.

Initial Conditions

Initial conditions are particular constraints that are used to determine the specific solution of a differential equation among a family of possible solutions. In this context, the initial condition helps us find the values of the constants of integration, ensuring our solution satisfies certain real-world constraints or reference points.
We have the initial condition \( \mathbf{r}(1) = \mathbf{i} + \mathbf{j} \), which denotes the vector's specific value at \( t=1 \). This leads to the following individual equations by plugging in \( t = 1 \):

• For the \( \mathbf{i} \)-component: \( 1 + C_1 = 1 \) gives \( C_1 = 0 \).
• For the \( \mathbf{j} \)-component: \( 1 + C_2 = 1 \) yields \( C_2 = 0 \).
• For the \( \mathbf{k} \)-component: \( \frac{2}{3}(1)^{3/2} + C_3 = 0 \), resulting in \( C_3 = -\frac{2}{3} \).

With these constants, the specific vector function \( \mathbf{r}(t) \) is fully determined and accurately maps the desired trajectory or path.