Question

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

Short answer

Question: Given the vector-valued function's derivative $$\mathbf{r}'(t) = \langle e^{2t}, 1-2e^{-t}, 1-2e^t \rangle$$ and the value of the function at \(t = 0\): $$\mathbf{r}(0) = \langle 1, 1, 1 \rangle$$, find the original vector-valued function \(\mathbf{r}(t)\). Answer: The original vector-valued function \(\mathbf{r}(t)\) that satisfies the given conditions is: $$\mathbf{r}(t) = \langle \frac{1}{2}e^{2t} + \frac{1}{2}, t + 2e^{-t} - 1, t - 2e^t + 3 \rangle.$$

Step-by-step solution

Integrate each component of the derivative function

To find the function \(\mathbf{r}(t)\), we need to integrate the derivative function component-wise. The given derivative function is: $$\mathbf{r}'(t) = \langle e^{2t}, 1-2e^{-t}, 1-2e^t \rangle.$$ We will find the antiderivative of each component independently: 1. Integrate \(e^{2t}\) with respect to \(t\): $$\int e^{2t} dt = \frac{1}{2}e^{2t} + c_1$$ 2. Integrate \((1 - 2e^{-t})\) with respect to \(t\): $$\int (1 - 2e^{-t}) dt = t + 2e^{-t} + c_2$$ 3. Integrate \((1 - 2e^t)\) with respect to \(t\): $$\int (1 - 2e^t) dt = t - 2e^t + c_3$$ Now we have found the antiderivative for each component: $$\mathbf{r}(t) = \langle \frac{1}{2}e^{2t} + c_1, t + 2e^{-t} + c_2, t - 2e^t + c_3 \rangle.$$

Use the initial condition to find the constant values

We are given the value of the function at \(t = 0\): $$\mathbf{r}(0) = \langle 1, 1, 1 \rangle.$$ We will now use this condition to find the values of \(c_1, c_2,\) and \(c_3\): 1. \(\frac{1}{2}e^{2(0)} + c_1 = 1 \Rightarrow c_1 = \frac{1}{2}\) 2. \(0 + 2e^{-0} + c_2 = 1 \Rightarrow c_2 = -1\) 3. \(0 - 2e^0 + c_3 = 1 \Rightarrow c_3 = 3\) So, we have found the constant values for each component: $$c_1 = \frac{1}{2}, \quad c_2 = -1,\quad c_3 = 3.$$

Write the complete vector-valued function

Substitute the values of the constants \(c_1, c_2\), and \(c_3\) into the antiderivative function we found in Step 1: $$\mathbf{r}(t) = \langle \frac{1}{2}e^{2t} + \frac{1}{2}, t + 2e^{-t} - 1, t - 2e^t + 3 \rangle.$$ Therefore, the vector-valued function \(\mathbf{r}(t)\) that satisfies the given conditions is: $$\mathbf{r}(t) = \langle \frac{1}{2}e^{2t} + \frac{1}{2}, t + 2e^{-t} - 1, t - 2e^t + 3 \rangle.$$

Key concepts

Integration Techniques

The process of integration is an essential mathematical tool that involves finding a function whose derivative is the given function, otherwise known as finding the antiderivative. In this exercise, we deal with vector-valued functions where each component of the vector must be integrated separately.

**Component-wise Integration:**

• To find the function \( \mathbf{r}(t) \), we must integrate each component of the given derivative vector \( \mathbf{r}'(t) = \langle e^{2t}, 1-2e^{-t}, 1-2e^t \rangle \).
• Instead of dealing with the entire vector at once, we focus on one component at a time: \( e^{2t} \), \( 1-2e^{-t} \), and \( 1-2e^t \).
• Each of these components is considered a separate function of \( t \), and we find its antiderivative independently.

**Antiderivatives of Basic Functions:**

• For the component \( e^{2t} \), the antiderivative involves a simple exponential rule: \( \int e^{kt} dt = \frac{1}{k} e^{kt} + C \), where \( C \) is the integration constant.
• Logarithmic and constant integration rules assist in finding the antiderivative of the remaining components, such as \( \int (1 - 2e^{-t}) dt \) and \( \int (1 - 2e^t) dt \).

Vector-Valued Functions

Vector-valued functions are multifunctional, with each component representing a separate function of a single variable, typically time \( t \).

**Understanding Vector-Valued Functions:**

• These functions map a single input into a vector of several outputs. The example given maps the scalar \( t \) into a 3-dimensional vector \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \).
• Each component function \( x(t), y(t), z(t) \) represents different attributes depending on the context. Commonly, they could represent position coordinates in space.

**Applications of Vector-Valued Functions:**

• They are widely used in physics and engineering, particularly in describing motion along a path where you need to consider multiple dimensions.
• Your differentiation gives a velocity vector, and integration can retrieve the position vector from the velocity.

Initial Conditions

Initial conditions are the specific values provided to solve for any unknown constants resulting from integration. In differential equations, these conditions are fundamental for finding unique solutions.

**Using Initial Conditions Effectively:**

• In this exercise, the initial condition is given as \( \mathbf{r}(0)=\langle 1,1,1 \rangle\). It's crucial because without it, the constants \( c_1, c_2, c_3 \) remain undetermined.
• These initial values describe the state of the system at the beginning (i.e., \( t = 0 \)).

**Solving for Constants:**

• To determine the constants, we substitute \( t = 0 \) into the antiderivative functions.
• This substitution alongside the initial condition lets us form simultaneous equations: \( \frac{1}{2}e^{2\cdot0} + c_1 = 1 \), etc., which we solve to find each \( c \).