Question

Find the function \(\mathbf{r}\) that satisfies the given conditions. $$\mathbf{r}^{\prime}(t)=\left\langle e^{2 t}, 1-2 e^{-t}, 1-2 e^{t}\right\rangle ; \mathbf{r}(0)=\langle 1,1,1\rangle$$

Short answer

Question: Determine the vector-valued function $\mathbf{r}(t)$ given its derivative $\mathbf{r}^{\prime}(t) = \left\langle e^{2 t}, 1-2 e^{-t}, 1-2 e^{t}\right\rangle$ and initial condition $\mathbf{r}(0) = \langle 1, 1, 1\rangle$. Answer: The vector-valued function $\mathbf{r}(t)$ is given by: $$\mathbf{r}(t) = \left\langle \frac{1}{2}e^{2t} + \frac{1}{2}, t + 2e^{-t} - 1, t - 2e^{t} + 3\right\rangle$$

Step-by-step solution

Identify the components

Given the derivative of the vector-valued function, $$\mathbf{r}^{\prime}(t)=\left\langle e^{2 t}, 1-2 e^{-t}, 1-2 e^{t}\right\rangle$$ Let's identify the components of the given vector-valued function: - \(x(t)' = e^{2t}\) - \(y(t)' = 1-2e^{-t}\) - \(z(t)' = 1-2e^{t}\)

Integrate each component

Integrate each component to find \(\mathbf{r}(t)\): - \(x(t) = \int x(t)' dt = \int e^{2t} dt = \frac{1}{2}e^{2t} + C_x\) - \(y(t) = \int y(t)' dt = \int (1-2e^{-t}) dt = t + 2e^{-t} + C_y\) - \(z(t) = \int z(t)' dt = \int (1-2e^{t}) dt = t - 2e^{t} + C_z\) Now, we have: $$\mathbf{r}(t) = \left\langle \frac{1}{2}e^{2t} + C_x, t + 2e^{-t} + C_y, t - 2e^{t} + C_z\right\rangle$$

Apply the initial condition

Using the initial condition \(\mathbf{r}(0)=\langle 1,1,1\rangle\), we can find the constants \(C_x\), \(C_y\), and \(C_z\): - For \(x(t)\): \(1 = \frac{1}{2}e^{2(0)} + C_x \Rightarrow C_x = \frac{1}{2}\) - For \(y(t)\): \(1 = 0 + 2e^{0} + C_y \Rightarrow C_y = -1\) - For \(z(t)\): \(1 = 0 - 2e^{0} + C_z \Rightarrow C_z = 3\)

Write the final expression for the vector-valued function

Now that we have the constants of integration, we can write the final expression for \(\mathbf{r}(t)\): $$\mathbf{r}(t) = \left\langle \frac{1}{2}e^{2t} + \frac{1}{2}, t + 2e^{-t} - 1, t - 2e^{t} + 3\right\rangle$$

Key concepts

Integration of Vector Functions

Learning how to integrate vector-valued functions is essential in fields like physics and engineering, where vectors are used to describe motion and forces. When given a vector function's derivative, such as \( \mathbf{r}'(t) = \langle e^{2t}, 1-2e^{-t}, 1-2e^{t} \rangle \), the process of integration involves finding the original vector function \( \mathbf{r}(t) \). This process consists of integrating each component of the vector independently.

For example, integrating each component of \( \mathbf{r}'(t) \) gives you:\( x(t) = \frac{1}{2}e^{2t} + C_x \), \( y(t) = t + 2e^{-t} + C_y \), and \( z(t) = t - 2e^{t} + C_z \), where \( C_x \), \( C_y \), and \( C_z \) are constants of integration. These constants play a crucial role and are determined by initial conditions, which we'll discuss next.

Bear in mind, the integration of vector functions is not just stacking integrals of the components; it also requires understanding the interplay of these components in physical space.

Initial Conditions in Calculus

In calculus, initial conditions allow us to find the particular solution to a differential equation from the family of all possible solutions. By applying the initial condition, we effectively determine the specific constants of integration that make the solution unique to the problem at hand. Initial conditions are usually given as values of the function at a certain point, such as \( \mathbf{r}(0) = \langle 1,1,1 \rangle \).

For \(\mathbf{r}(t)\), we use the initial condition to solve for \(C_x\), \(C_y\), and \(C_z\). After substituting \( t=0 \) into our integrated components and setting them equal to the respective values provided in \( \mathbf{r}(0) \), we find exact numerical values for our constants. This transforms our general solution into one that precisely fits the problem's context.

Derivative of Vector-Valued Functions

The derivative of a vector-valued function provides the rate of change of each component with respect to the independent variable, which in physics can correspond to time. When dealing with \( \mathbf{r}'(t) \), the goal is to describe the behavior of the function's components: \( x(t) \), \( y(t) \), and \( z(t) \).

For instance, \( \mathbf{r}'(t) \) represents the velocity of a particle moving in space, and each component of this derivative can be understood as the velocity in the direction of the corresponding axis. In our exercise, we determined these components to be \( e^{2t} \), \( 1-2e^{-t} \), and \( 1-2e^{t} \), which represent how rapidly the position changes over time along the x, y, and z axes. Understanding derivatives in this vector context bridges the gap between pure mathematics and its application to real-world problems.

Calculus Exponential Functions

Exponential functions, such as \( e^{kt} \), where \( k \) is a constant, are pervasive in calculus because of their distinctive properties, especially in growth and decay models. When these functions are part of a vector-valued function, as seen with \( x(t) = \frac{1}{2}e^{2t} \), \( y(t) = 2e^{-t} \), and \( z(t) = -2e^{t} \), they add a layer of complexity. However, their derivatives and integrals follow the same general rules as scalar functions.

The key property to remember is that the derivative of \( e^{kt} \) is \( ke^{kt} \), and the integral is \( \frac{1}{k}e^{kt} \), subject to \( k eq 0 \). This makes dealing with exponential expressions within vector components straightforward, as long as one remembers to apply standard calculus rules in a vector context, tailoring the calculation to each component's exponential form.