Problem 78
Question
Solve the initial value problems. $$\frac{d s}{d t}=\cos t+\sin t, \quad s(\pi)=1$$
Step-by-Step Solution
Verified Answer
The solution is \( s(t) = \sin t - \cos t \).
1Step 1: Integrate the Differential Equation
The differential equation is given by \( \frac{d s}{d t} = \cos t + \sin t \). To find \( s(t) \), we need to integrate both sides with respect to \( t \). Thus, \( s(t) = \int (\cos t + \sin t) \, dt \).
2Step 2: Find the Antiderivative
Calculate the integral \( \int (\cos t + \sin t) \, dt \). The antiderivative is \( \sin t - \cos t + C \), where \( C \) is the constant of integration.
3Step 3: Apply the Initial Condition
We use the initial condition \( s(\pi) = 1 \) to determine the constant \( C \). Substitute \( t = \pi \) into the antiderivative equation: \( \sin(\pi) - \cos(\pi) + C = 1 \).
4Step 4: Solve for the Constant C
Since \( \sin(\pi) = 0 \) and \( \cos(\pi) = -1 \), the equation becomes \( 0 - (-1) + C = 1 \), simplifying to \( 1 + C = 1 \). Solve to find \( C = 0 \).
5Step 5: Write the Final Solution
Replace \( C \) into the general solution to obtain \( s(t) = \sin t - \cos t \). This is the solution to the initial value problem.
Key Concepts
Differential EquationsAntiderivativeIntegrationConstant of Integration
Differential Equations
A differential equation is like a bridge linking two quantities. It's an equation that includes a function and its derivative. In simple terms, it shows how one thing changes with respect to something else.
In our exercise, the differential equation is \( \frac{d s}{d t} = \cos t + \sin t \). This tells us how the function \( s(t) \) changes as time \( t \) changes. The task is to solve this equation to find an expression for \( s(t) \).
In our exercise, the differential equation is \( \frac{d s}{d t} = \cos t + \sin t \). This tells us how the function \( s(t) \) changes as time \( t \) changes. The task is to solve this equation to find an expression for \( s(t) \).
- The left side \( \frac{d s}{d t} \) represents the rate of change of \( s \) with respect to \( t \).
- The right side \( \cos t + \sin t \) represents the combination of how these trigonometric functions affect this rate.
Antiderivative
The antiderivative is like an undo button for differentiation. If differentiating a function gives you a result, the antiderivative helps you find the original function before it was differentiated.
To solve the differential equation, we need to find the antiderivative of \( \cos t + \sin t \). The antiderivative tells us the function \( s(t) \) whose rate of change (derivative) matches our differential equation.
To solve the differential equation, we need to find the antiderivative of \( \cos t + \sin t \). The antiderivative tells us the function \( s(t) \) whose rate of change (derivative) matches our differential equation.
- For \( \cos t \), the antiderivative is \( \sin t \).
- For \( \sin t \), the antiderivative is \( -\cos t \).
Integration
Integration is a process that combines or accumulates values, essentially the reverse of differentiation. By integrating a function, we can find related functions whose derivatives match the given function.
In our problem, integration helps us turn the differential equation into a more understandable form of \( s(t) \). The process involves evaluating integrals, like \( \int (\cos t + \sin t) \, dt \), to derive an expression for \( s(t) \). This gives us a form that includes an arbitrary constant.
In our problem, integration helps us turn the differential equation into a more understandable form of \( s(t) \). The process involves evaluating integrals, like \( \int (\cos t + \sin t) \, dt \), to derive an expression for \( s(t) \). This gives us a form that includes an arbitrary constant.
- Integration sums up infinitesimally small quantities.
- It is useful in finding areas under curves, among many other applications.
Constant of Integration
The constant of integration \( C \) comes up every time we find an antiderivative. It's essentially an unknown number added to the function, allowing for all possible antiderivative solutions.
In the context of an initial value problem like ours, the constant of integration ensures the solution fits a specific starting point or boundary condition.
When given an initial condition like \( s(\pi) = 1 \), we use this condition to solve for the exact value of \( C \). First, substitute \( t = \pi \) into the function \( s(t) = \sin t - \cos t + C \). From the step-by-step solution, we see that \( 1 + C = 1 \), leading to \( C = 0 \). This step ensures that among all the possible functions \( s(t) \) could take, we find the one that exactly fits our initial condition.
In the context of an initial value problem like ours, the constant of integration ensures the solution fits a specific starting point or boundary condition.
When given an initial condition like \( s(\pi) = 1 \), we use this condition to solve for the exact value of \( C \). First, substitute \( t = \pi \) into the function \( s(t) = \sin t - \cos t + C \). From the step-by-step solution, we see that \( 1 + C = 1 \), leading to \( C = 0 \). This step ensures that among all the possible functions \( s(t) \) could take, we find the one that exactly fits our initial condition.
- The constant makes the equation fit real-world scenarios or specific problem constraints.
- It personalizes the general solution derived from integration to a unique one adhering to starting points.
Other exercises in this chapter
Problem 77
Solve the initial value problems. $$\frac{d s}{d t}=1+\cos t, \quad s(0)=4$$
View solution Problem 77
Graph the functions. Then find the extreme values of the function on the interval and say where they occur. $$f(x)=|x-2|+|x+3|, \quad-5 \leq x \leq 5$$
View solution Problem 78
Graph the functions. Then find the extreme values of the function on the interval and say where they occur. $$g(x)=|x-1|-|x-5|, \quad-2 \leq x \leq 7$$
View solution Problem 79
Solve the initial value problems. $$\frac{d r}{d \theta}=-\pi \sin \pi \theta, \quad r(0)=0$$
View solution