Problem 66 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Function c solves the initial value problem.

Step 1: Analyze the Initial Conditions

Given the problem, we want to find which function corresponds to the initial value problem with the differential equation \( y' = \sec x \) and initial condition \( y(-1) = 4 \). This requires finding the function whose derivative and initial condition match those specified.

Step 2: Consider Function Derivatives

Identify the derivatives of the provided functions to determine which matches the derivative of \( y' = \sec x \).- For function (a) \( y = \int_{1}^{x} \frac{1}{t} \, dt - 3 \), the derivative is \( y' = \frac{1}{x} \).- For function (b) \( y = \int_{0}^{x} \sec t \, dt + 4 \), the derivative is \( y' = \sec x \).- For function (c) \( y = \int_{-1}^{x} \sec t \, dt + 4 \), the derivative is also \( y' = \sec x \).- For function (d) \( y = \int_{\pi}^{x} \frac{1}{t} \, dt - 3 \), the derivative is \( y' = \frac{1}{x} \).Therefore, functions (b) and (c) have derivatives \( y' = \sec x \), matching the differential equation.

Step 3: Check Initial Conditions for Matching Functions

Now, we will test functions (b) and (c) against the initial condition \( y(-1) = 4 \).- For function (b) with \( y = \int_{0}^{x} \sec t \, dt + 4 \), evaluating at \( x = -1 \): \[ y(-1) = \int_{0}^{-1} \sec t \, dt + 4 \]. Calculating this integral would require specifics of \( \sec t \) over \([0, -1]\), which generally doesn't simplify straightforwardly.- For function (c) with \( y = \int_{-1}^{x} \sec t \, dt + 4 \), evaluating at \( x = -1 \): \[ y(-1) = \int_{-1}^{-1} \sec t \, dt + 4 = 0 + 4 = 4 \].Function (c) exactly satisfies \( y(-1) = 4 \).

Step 4: Conclusion of Matching Function

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

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve derivatives of functions. They are vital in describing phenomena like motion, growth, and decay. A differential equation specifies the rate of change of a function.

• This rate is expressed as a derivative.
• The equation may include constants or terms of the function.

In the given exercise, we deal with a differential equation expressed as \( y' = \sec x \). This notation signifies that the rate of change of function \( y \) with respect to \( x \) is equal to \( \sec x \). The challenge is to find a function \( y \) that satisfies this equation along with other specified conditions.

Function Derivatives

Derivatives are fundamental in calculus, representing how a function changes as its input changes. The derivative \( y' \) of a function \( y \) with respect to \( x \) is the limit of the function's slope as the interval approaches zero.

• Function derivatives indicate instantaneous rates of change.
• They help us understand the behavior of functions.

In this problem, we examine the derivatives of several functions to see which matches \( y' = \sec x \). Function (b) and function (c) both lead to this derivative naturally, pointing us toward further checking initial conditions for selection.

Initial Conditions

Initial conditions are values given for the function and its derivatives at specific points. These values are crucial as they define a particular solution from all possible solutions to a differential equation.

• Initial conditions specify unique scenarios for equations.
• They often involve known values at certain input numbers.

In assessing functions (b) and (c), it's the initial condition \( y(-1) = 4 \) that determines the match. This condition is used as a criterion to choose the function \( y \) that meets both the derivative and initial condition, narrowing down to function (c).

Integral Calculus

Integral calculus is the inverse process of differentiation. It involves calculating integrals, giving the area under a curve or recovering a function from its derivative. This method is used to solve initial value problems like the one in this scenario.

• Integrals can find functions based on their rate of change.
• They can calculate accumulated values, like areas under curves.

Solving the problem relies heavily on examining integrals. Function (c) \( y = \int_{-1}^{x} \sec t \, dt + 4 \) satisfies both integral attributes and initial conditions, resulting in a correct solution to the equation.