Problem 39 - Calculus Early Transcendentals

Question

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

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: Understand the Problem

We are given a vector function derivative \( \mathbf{r}'(t) = 2t \mathbf{i} + 3t^2 \mathbf{j} + \sqrt{t} \mathbf{k} \) and an initial condition \( \mathbf{r}(1) = \mathbf{i} + \mathbf{j} \). We need to find the original vector function \( \mathbf{r}(t) \).

Step 2: Integrate the Component Functions

Integrate each component of the vector \( \mathbf{r}'(t) = 2t \mathbf{i} + 3t^2 \mathbf{j} + \sqrt{t} \mathbf{k} \) separately. We have:- \( \int 2t\, dt = t^2 + C_1 \) for the \( \mathbf{i} \) component.- \( \int 3t^2\, dt = t^3 + C_2 \) for the \( \mathbf{j} \) component.- \( \int \sqrt{t}\, dt = \frac{2}{3}t^{3/2} + C_3 \) for the \( \mathbf{k} \) component.

Step 3: Form the General Solution

Combine the integrated components to form the general solution:\[ \mathbf{r}(t) = (t^2 + C_1) \mathbf{i} + (t^3 + C_2) \mathbf{j} + \left( \frac{2}{3}t^{3/2} + C_3 \right) \mathbf{k}. \]

Step 4: Apply Initial Condition

Use the initial condition \( \mathbf{r}(1) = \mathbf{i} + \mathbf{j} \) to find \( C_1, C_2, \) and \( C_3 \). Substituting \( t = 1 \):\[ \mathbf{i} + \mathbf{j} = (1 + C_1) \mathbf{i} + (1 + C_2) \mathbf{j} + \left( \frac{2}{3} + C_3 \right) \mathbf{k}. \]Comparing components, we obtain:- For \( \mathbf{i} \): \( 1 + C_1 = 1 \) gives \( C_1 = 0 \).- For \( \mathbf{j} \): \( 1 + C_2 = 1 \) gives \( C_2 = 0 \).- For \( \mathbf{k} \): \( \frac{2}{3} + C_3 = 0 \) gives \( C_3 = -\frac{2}{3} \).

Step 5: Write the Specific Solution

Substitute the constants back into \( \mathbf{r}(t) \):\[ \mathbf{r}(t) = t^2 \mathbf{i} + t^3 \mathbf{j} + \left( \frac{2}{3}t^{3/2} - \frac{2}{3} \right) \mathbf{k}. \]

Key Concepts

Vector Integration

Vector integration is the process of finding a vector function whose derivative is the given vector function. In simpler terms, it's about reconstructing the original function from its rate of change.
This involves integrating each component of the vector separately. In our problem, we start with the vector derivative \( \mathbf{r}'(t) = 2t \mathbf{i} + 3t^2 \mathbf{j} + \sqrt{t} \mathbf{k} \).
Here are the steps to integrate these components individually:

• For the \( \mathbf{i} \)-component: integrate \( \int 2t \, dt = t^2 + C_1 \).
• For the \( \mathbf{j} \)-component: integrate \( \int 3t^2 \, dt = t^3 + C_2 \).
• For the \( \mathbf{k} \)-component: integrate \( \int \sqrt{t} \, dt = \frac{2}{3}t^{3/2} + C_3 \).

Each integration introduces an arbitrary constant \( C_n \), representing any constant shift along the component axes.
The result is a general vector function that satisfies the original rate of change.

Initial Value Problem

An initial value problem involves finding a specific solution to a differential equation that satisfies an initial condition. This gives a particular vector rather than a family of solutions.
In our case, the initial condition given is \( \mathbf{r}(1) = \mathbf{i} + \mathbf{j} \). This means that when \( t = 1 \), the vector \( \mathbf{r}(t) \) must equal \( \mathbf{i} + \mathbf{j} \).
Using this condition, we substitute \( t = 1 \) into the general solution:

• For the \( \mathbf{i} \)-component: the equation \( 1 + C_1 = 1 \) leads to \( C_1 = 0 \).
• For the \( \mathbf{j} \)-component: \( 1 + C_2 = 1 \) gives \( C_2 = 0 \).
• For the \( \mathbf{k} \)-component: \( \frac{2}{3} + C_3 = 0 \) results in \( C_3 = -\frac{2}{3} \).

These constants are then used to refine the general solution into a specific one that fits the initial condition.

Component Functions

Component functions are the individual functions along each axis of a vector field. When dealing with vector calculus, these functions are handled separately to ensure they satisfy the vector as a whole.
For our vector \( \mathbf{r}'(t) = 2t \mathbf{i} + 3t^2 \mathbf{j} + \sqrt{t} \mathbf{k} \), the components \( 2t \), \( 3t^2 \), and \( \sqrt{t} \) belong to the \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) axes, respectively.
Understanding and handling the component functions separately is crucial for integrating them individually:

• \( 2t \) is a linear function integrated to \( t^2 + C_1 \).
• \( 3t^2 \) becomes \( t^3 + C_2 \) after integration - a cubic function.
• \( \sqrt{t} \) transforms into \( \frac{2}{3}t^{3/2} + C_3 \), involving fractional exponents.

This separation allows accurate synthesis of the general vector function by properly integrating along each axis, ensuring precision and completeness in vector calculus.

3D Vectors

3D vectors, as the name implies, exist in a three-dimensional space defined by three axes: \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \). Each axis corresponds to a different dimension - the x, y, and z axes, respectively.
In our problem, the vector function \( \mathbf{r}(t) \) is expressed as a sum of these three axes. Each has independent behavior represented by the component functions. The vector is then described by a combination of these components: \( t^2 \mathbf{i} + t^3 \mathbf{j} + \left( \frac{2}{3}t^{3/2} - \frac{2}{3} \right) \mathbf{k} \).
This illustrates how 3D vectors encapsulate direction and magnitude within a multidimensional frame:

• The \( \mathbf{i} \) component describes horizontal movement.
• The \( \mathbf{j} \) component handles vertical movement.
• The \( \mathbf{k} \) component accounts for depth movement.

Handling vectors of this complexity can initially seem challenging, but by breaking them down into these components makes the calculus more approachable and intuitive.