Problem 48 - Thomas Calculus

Question

Each of the following functions solves one of the initial value problems in Exercises \(47-50 .\) Which function solves which problem? Give brief reasons for your answers. $$ \begin{array}{ll}{\text { a. } y=\int_{1}^{x} \frac{1}{t} d t-3} & {\text { b. } y=\int_{0}^{x} \sec t d t+4} \\ {\text { c. } y=\int_{-1}^{x} \sec t d t+4} & {\text { d. } y=\int_{\pi}^{x} \frac{1}{t} d t-3}\end{array} $$ $$ y^{\prime}=\sec x, \quad y(-1)=4 $$

Step-by-Step Solution

Answer

Function (c) solves the initial value problem.

Step 1: Understand the Initial Value Problem

We are given an initial value problem (IVP) with a differential equation \( y' = \sec x \) and an initial condition \( y(-1) = 4 \). This means we need to find a function \( y \) that has a derivative of \( \sec x \) and satisfies the initial condition \( y(-1) = 4 \).

Step 2: Differentiate Functions to Find \( y' \)

Differentiate each given function to find \( y' \): - For \( y = \int_1^x \frac{1}{t} dt - 3 \), \( y' = \frac{1}{x} \).- For \( y = \int_0^x \sec t dt + 4 \), \( y' = \sec x \).- For \( y = \int_{-1}^x \sec t dt + 4 \), \( y' = \sec x \).- For \( y = \int_\pi^x \frac{1}{t} dt - 3 \), \( y' = \frac{1}{x} \).Thus, functions (b) and (c) have \( y' = \sec x \).

Step 3: Test Initial Condition

For functions (b) and (c), check which satisfies \( y(-1) = 4 \):- For \( y = \int_0^x \sec t dt + 4 \), substitute \( x = -1 \): \( y(-1) = \int_0^{-1} \sec t dt + 4 \) (Value needs actual calculation).- For \( y = \int_{-1}^x \sec t dt + 4 \), substitute \( x = -1 \), giving: \( y(-1) = \int_{-1}^{-1} \sec t dt + 4 = 0 + 4 = 4 \).Function (c) satisfies the initial condition.

Step 4: Conclusion

Function (c), \( y = \int_{-1}^x \sec t dt + 4 \), satisfies both the derivative \( y' = \sec x \) and the initial condition \( y(-1) = 4 \). Therefore, it is the solution to the given initial value problem.

Key Concepts

Differential Equations

Differential equations form the mathematical foundation for expressing the way quantities change. An initial value problem (IVP) involves finding a particular solution to a differential equation that satisfies a given initial condition. The general form of a differential equation is given by determining a function and its derivatives, like in the problem where we have the derivative of the function given as \( y' = \sec x \). The task in these exercises is to find a function that not only satisfies the derivative condition but also meets a specified value at a given point (called the initial condition), such as \( y(-1) = 4 \). This connection is crucial for solving real-world problems, such as those in physics and engineering, where starting values often dictate entire system behaviors.With differential equations, it's about understanding the relationships between the rates of change rather than the actual numbers at each point. This is why we differentiate functions as part of solving the problem – this calculation tells us whether the rate of change conforms to eigenvalues like \( \sec x \).

Integration

Integration is central to solving differential equations, as it essentially helps to reverse the process of differentiation. Here, integration enables us to go from a derivative back to a function, providing us with an expression for \( y \) that we can test against the initial condition.In our solution, we deal with functions like \( y = \int_{-1}^x \sec t \, dt + 4 \), where the integral is evaluated over a range and then adjusted to match specific initial conditions. The indefinite integral introduces a constant of integration, which is why we add or subtract a number to fit the initial condition given.

• The integral \( \int_{-1}^x \sec t \, dt + 4 \) represents the area under the curve of \( \sec t \) over the interval from \(-1\) to \(x\).
• By evaluating this integral, we can verify that when \( x = -1 \), the function gives the required initial value of 4.

This process of solving for the initial condition by directly manipulating the bounds of the integral illustrates how integration allows us to fine-tune a general solution to meet specific criteria.

Calculus Solution Methods

Understanding calculus involves mastering both differentiation and integration, which are pivotal methods in determining the solutions to initial value problems.When approaching a problem with differential equations:

• First, find the derivative condition given in the problem. This reveals the behavior of the function’s slope over its domain.
• Second, integrate the derivative to determine the functions whose derivative equals the given expression, in our case, \( \sec x \). This step helps in generating a general solution from the differential equation.
• Finally, apply the initial condition to the general function, ensuring it matches the given value exactly at that point, thus identifying the specific solution to the IVP.

For students, the challenge lies in performing these steps methodically and verifying their results against the initial conditions. This structured approach ensures precision in finding accurate and applicable solutions, whether in academics or practical applications in the real world.