Problem 98 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The solution is \(s(t) = \sin t - \cos t\).

Step 1: Integrate the Differential Equation

To solve the initial value problem, we need to integrate the differential equation. Given \[\frac{ds}{dt} = \cos t + \sin t\]we will find the antiderivative of each term. The integral of \(\cos t\) is \(\sin t\), and the integral of \(\sin t\) is \(-\cos t\). Thus, the general solution is:\[s(t) = \sin t - \cos t + C\]where \(C\) is a constant of integration.

Step 2: Apply Initial Condition

We are given the initial condition \(s(\pi) = 1\). We use this to find the constant \(C\). Substitute \(t = \pi\) and \(s(t) = 1\) into the equation:\[1 = \sin(\pi) - \cos(\pi) + C\]Since \(\sin(\pi) = 0\) and \(-\cos(\pi) = -(-1) = 1\), the equation becomes:\[1 = 0 + 1 + C \]Simplifying gives \(C = 0\).

Step 3: Write the Specific Solution

Now that we have determined \(C\), we can write the specific solution to the differential equation:\[s(t) = \sin t - \cos t\]This is the solution to the initial value problem, and it satisfies both the differential equation and the initial condition.

Key Concepts

Differential Equations

A differential equation is a mathematical equation that involves functions and their derivatives. In this specific problem, the function is represented by \(s(t)\), and its derivative with respect to \(t\) is given as \(\frac{ds}{dt} = \cos t + \sin t\). This equation tells us how the rate of change of \(s\) with respect to \(t\) is determined by the sum of the cosine and sine functions.

Differential equations are fundamental in modeling real-world systems in physics, engineering, biology, and more as they describe how a system evolves over time or under different conditions. Solving a differential equation means finding a function \(s(t)\) that satisfies the equation for all values of \(t\). In many cases, like our example, this begins with identifying an equation and then integrating it.

Integration

Integration is the mathematical process of finding the antiderivative or the integral of a function. In simpler terms, it's the reverse process of differentiation.

In our exercise, we have the differential equation \(\frac{ds}{dt} = \cos t + \sin t\). Integration is applied to both sides of the equation to find \(s(t)\). As we integrate each term separately, the integral of \(\cos t\) becomes \(\sin t\), and the integral of \(\sin t\) becomes \(-\cos t\).

Here's a key point: when we integrate, we add a constant \(C\), known as the constant of integration. This constant reflects the fact that there are infinitely many functions that could satisfy the differential equation, each differing by a constant shift.

Antiderivative

The antiderivative of a function is another function whose derivative equals the original function. For our problem, we need the antiderivative of the right-hand side of our differential equation, \(\cos t + \sin t\).

The process of finding the antiderivative involves reversing differentiation:

• The antiderivative of \(\cos t\) is \(\sin t\).
• The antiderivative of \(\sin t\) is \(-\cos t\).

So, the antiderivative of \(\cos t + \sin t\) is \(\sin t - \cos t\). We get the general solution \(s(t) = \sin t - \cos t + C\) where \(C\) is a constant.

Boundary Conditions

Boundary conditions, also known as initial conditions in this context, help us find the specific solution from a family of solutions.

In initial value problems, we use given information at a specific point to find unknown constants. In this exercise, we are told \(s(\pi) = 1\).

To find the constant \(C\), we substitute \(t = \pi\) into the general solution \(s(t) = \sin t - \cos t + C\). Knowing that \(\sin(\pi) = 0\) and \(-\cos(\pi) = 1\), we can solve for \(C\).

By applying the initial condition, we determine that \(1 = 0 + 1 + C\), leading to \(C = 0\). Once we know \(C\), we write the specific solution as \(s(t) = \sin t - \cos t\), which satisfies both the differential equation and the initial condition.