Problem 100

Question

Solve the initial value problems in Exercises. $$\frac{d v}{d t}=\frac{1}{2} \sec t \tan t, \quad v(0)=1$$

Step-by-Step Solution

Verified

Answer

The solution is $ v(t) = \frac{1}{2} \sec t + \frac{1}{2} $.

1Step 1: Identify the Differential Equation and Initial Condition

The given differential equation is $ \frac{d v}{d t} = \frac{1}{2} \sec t \tan t $ with an initial condition $ v(0) = 1 $. Our goal is to find the function $ v(t) $ that satisfies both the differential equation and the initial condition.

2Step 2: Integrate the Differential Equation

To solve the differential equation, integrate both sides with respect to $ t $. The integral of $ \frac{d v}{d t} $ with respect to $ t $ is $ v(t) $, and the integral of $ \frac{1}{2} \sec t \tan t $ with respect to $ t $ is $ \frac{1}{2} \int \sec t \tan t \, dt $.

3Step 3: Integrate $ \sec t \tan t $

Recall that the derivative of $ \sec t $ is $ \sec t \tan t $, so $ \int \sec t \tan t \, dt = \sec t + C $. Thus, $ \frac{1}{2} \int \sec t \tan t \, dt = \frac{1}{2} \sec t + C $.

4Step 4: Solve for the Constant of Integration Using Initial Condition

Substitute $ v(0) = 1 $ into the equation from Step 3, i.e., $ v(t) = \frac{1}{2} \sec t + C $. At $ t = 0 $, $ \sec 0 $ is 1, so $ v(0) = \frac{1}{2} \times 1 + C = 1 $. Solve for $ C $ to get $ \frac{1}{2} + C = 1 $, thus $ C = \frac{1}{2} $.

5Step 5: Write the Solution

Now substituting $ C $ back into the expression for $ v(t) $, we have $ v(t) = \frac{1}{2} \sec t + \frac{1}{2} $. This is the solution to the initial value problem.

Key Concepts

Initial Value ProblemIntegrationTrigonometric Functions

Initial Value Problem

An initial value problem is a common type of differential equation that comes with an additional piece of information called the initial condition. This information, usually given as a point through which the solution curve must pass, is crucial for pinpointing the exact solution from a family of potential solutions.
In our exercise, the differential equation is $ \frac{d v}{d t} = \frac{1}{2} \sec t \tan t $ with the initial condition $ v(0) = 1 $.

Here's a simplified explanation:

The differential equation tells you how the function $ v(t) $ changes with respect to time $ t $.
The initial condition specifies the value of the function $ v(t) $ at a specific point, here $ t = 0 $, where $ v(0) = 1 $.

This initial condition allows us to find the constant of integration and hence determine a particular solution instead of a general one. Without this condition, you would only know the general form of the solution, which isn't as useful for practical problems.

Integration

Integration is a powerful mathematical technique often used to solve differential equations. In this context, it is the process of finding a function, given its derivative.
In the step-by-step solution provided, integration is used to find the function $ v(t) $ from its derivative $ \frac{1}{2} \sec t \tan t $.

Here's a step-by-step breakdown:

The integral of $ \frac{d v}{d t} $ with respect to $ t $ is straightforwardly $ v(t) $.
For the integral of $ \frac{1}{2} \sec t \tan t $, a known trigonometric identity is used; since the derivative of $ \sec t $ is $ \sec t \tan t $, this integral evaluates to $ \frac{1}{2} \sec t + C $, where $ C $ is the constant of integration.

Integration transforms the problem from finding rates of change back to finding the original function. This stage is crucial for converting a differential equation into an algebraic equation, allowing us to apply the initial condition and solve for any unknown constants. Remember, the constant of integration arises because an indefinite integral encompasses a whole family of solutions.

Trigonometric Functions

Trigonometric functions like secant ($ \sec $) and tangent ($ \tan $) play a vital role in solving differential equations involving periodic or oscillating behaviors.
In this exercise, we dealt with $ \sec t \tan t $. To solve the integral $ \int \sec t \tan t \, dt $, we rely on the fact that the derivative of $ \sec t $ is $ \sec t \tan t $. This connection is crucial because it allows us to integrate by recognizing these functions.

Here are some key reminders about trigonometric functions:

The secant function, $ \sec t $, is the reciprocal of the cosine function: $ \sec t = 1/\cos t $.
The tangent function, $ \tan t $, is the ratio of the sine and cosine functions: $ \tan t = \sin t/\cos t $.
Understanding and recognizing the derivative identities of trigonometric functions help simplify the process of integration significantly.

Using these functions correctly allows for more manageable calculations and solutions when dealing with differential equations.

Problem 99

Problem 101

Other exercises in this chapter

Problem 98

Solve the initial value problems in Exercises. $$\frac{d s}{d t}=\cos t+\sin t, \quad s(\pi)=1$$

View solution

Problem 99

Solve the initial value problems in Exercises. $$\frac{d r}{d \theta}=-\pi \sin \pi \theta, \quad r(0)=0$$

View solution

Problem 101

Solve the initial value problems in Exercises. $$\frac{d v}{d t}=\frac{1}{2} \sec t \tan t, \quad v(0)=1$$

View solution

Problem 102

Solve the initial value problems in Exercises. $$\frac{d v}{d t}=8 t+\csc ^{2} t, \quad v\left(\frac{\pi}{2}\right)=-7$$

View solution