Question

Solve the given initial value problems. Find \(\vec{r}(t),\) given that \(\vec{r}^{\prime}(t)=\langle 1 /(t+1), \tan t\rangle\) and \(\vec{r}(0)=\langle 1,2\rangle\)

Short answer

\( \vec{r}(t) = \langle \ln|t+1| + 1, -\ln|\cos t| + 2 \rangle \).

Step-by-step solution

Integrate the First Component of the Vector

The derivative of the first component of the vector is given by \( r_1'(t) = \frac{1}{t+1} \). We need to find \( r_1(t) \) by integrating this expression. The integral of \( \frac{1}{t+1} \) with respect to \( t \) is \( \ln|t+1| + C_1 \).

Integrate the Second Component of the Vector

The derivative of the second component of the vector is given by \( r_2'(t) = \tan t \). Integrating \( \tan t \) with respect to \( t \), we obtain \(-\ln|\cos t| + C_2 \).

Incorporate Initial Conditions to Find Constants

We have the initial condition \( \vec{r}(0) = \langle 1, 2 \rangle \). Substituting \( t = 0 \) into \( r_1(t) = \ln|t+1| + C_1 \), we get \( 1 = \ln(1) + C_1 \), so \( C_1 = 1 \). For \( r_2(t) \), substitute \( t = 0 \) into \(-\ln|\cos t| + C_2 \) to obtain \( 2 = -\ln|\cos(0)| + C_2 \), simplifying to \( C_2 = 2 \).

Write the Solution Including the Constants

Now that we have found the constants, substitute them back into the integrated functions. Therefore, \( r_1(t) = \ln|t+1| + 1 \) and \( r_2(t) = -\ln|\cos t| + 2 \). The vector function is then \( \vec{r}(t) = \langle \ln|t+1| + 1, -\ln|\cos t| + 2 \rangle \).

Key concepts

Integration

Integration is a core concept in calculus that basically helps us reverse the process of differentiation. When we integrate a function, we aim to find a new function whose derivative gives us the original function we started with.
For instance, in the problem, we are given that the derivative of the first component of the vector is \( r_1'(t) = \frac{1}{t+1} \). To find \( r_1(t) \), we integrate the function \( \frac{1}{t+1} \) with respect to \( t \). This results in \( \ln|t+1| + C_1 \).
The constant of integration, denoted as \( C \), is crucial because it accounts for all the constant values that could possibly be added to the function without altering its derivative. Remember, integrating a function tells us about accumulation, area under a curve, or, as used here, finding an original function before it was differentiated.

Vector Calculus

Vector calculus extends traditional calculus to functions of multiple variables, often dealing with vector fields. This allows us to analyze curves, surfaces, and fields in multiple dimensions.
In our exercise, we are working with vector-valued functions, specifically \( \vec{r}(t) \), which is given in terms of its derivative components. The expression \( \vec{r}^{\prime}(t) = \langle \frac{1}{t+1}, \tan t \rangle \) suggests that our vector function has two components, both of which need integrating individually.
Each component of \( \vec{r}(t) \) behaves like a simple function of \( t \), but as a whole, it describes a trajectory in two-dimensional space. Thus, finding \( \vec{r}(t) \) by integrating its derivative components provides us with the complete path or geometrical description.

Differential Equations

Differential equations involve equations containing derivatives of a function. These equations express how changes in one variable relate to changes in another, often involving rates of change.
Our problem presents a system of differential equations via the vector derivative \( \vec{r}^{\prime}(t) \). Solving these involves integrating each component separately to find \( \vec{r}(t) \).
These types of problems are foundational in modeling many real-world systems, like motion paths in physics or population models in biology. Initial value problems specify conditions that help us determine the unique solution to the differential equation.

Constants of Integration

Constants of integration are essential whenever we integrate a function. These constants represent the unknown constant value that disappears when a function is differentiated.
In our exercise, integrating the components yields an indefinite integral, resulting in constants \( C_1 \) and \( C_2 \). Fortunately, initial conditions help us figure out these constants.
For this problem, \( \vec{r}(0) = \langle 1, 2 \rangle \) gives us a starting point. By substituting \( t = 0 \) into our integrated equations, we solve directly for these constants. For example, setting \( t = 0 \) in \( r_1(t) \) helps us find \( C_1 = 1 \), while doing the same for \( r_2(t) \) gives \( C_2 = 2 \). These solutions are unique to the initial conditions provided, demonstrating how initial values influence the final solution of differential equations.