Problem 34 - Essential Calculus Early Transcendentals

Question

Find \(f\) $$f^{\prime \prime}(t)=2 e^{t}+3 \sin t, \quad f(0)=0, \quad f(\pi)=0$$

Step-by-Step Solution

Answer

Find \(f(t) = 2e^{t} - 3\sin t + \left(\frac{2 - 2e^{\pi}}{\pi}\right)t - 2\).

Step 1: Integrate the Second Derivative

To find the original function \(f(t)\), we start by integrating the second derivative \(f''(t) = 2e^{t} + 3\sin t\). The first integral will give us the first derivative \(f'(t)\). Integrating, we have:\[ f'(t) = \int (2e^{t} + 3\sin t) \, dt = 2e^{t} - 3\cos t + C_1 \]where \(C_1\) is a constant of integration.

Step 2: Integrate Again to Find the Function

Integrate \(f'(t)\) to find \(f(t)\):\[ f(t) = \int (2e^{t} - 3\cos t + C_1) \, dt \]This results in:\[ f(t) = 2e^{t} - 3\sin t + C_1t + C_2 \]where \(C_2\) is another constant of integration.

Step 3: Use Initial Condition \(f(0) = 0\)

Substitute \(t = 0\) into \(f(t)\) and solve for \(C_2\).\[ f(0) = 2e^{0} - 3\sin 0 + C_1 \cdot 0 + C_2 = 0 \]\[ 2 + C_2 = 0 \Rightarrow C_2 = -2 \]

Step 4: Use Boundary Condition \(f(\pi) = 0\)

Now use the boundary condition \(f(\pi) = 0\) to solve for \(C_1\).\[ f(\pi) = 2e^{\pi} - 3\sin \pi + C_1\pi - 2 = 0 \]\[ 2e^{\pi} + C_1\pi - 2 = 0 \]\[ C_1\pi = 2 - 2e^{\pi} \]\[ C_1 = \frac{2 - 2e^{\pi}}{\pi} \]

Step 5: Formulate the Final Function

With \(C_1\) and \(C_2\) determined, substitute them back into the function \(f(t)\).\[ f(t) = 2e^{t} - 3\sin t + \left(\frac{2 - 2e^{\pi}}{\pi}\right)t - 2 \]

Key Concepts

Integration

Integration is a crucial operation in calculus that helps us find original functions when we are given their derivatives. Think of it like playing detective; given pieces of clues (derivatives), we work backwards to find the entire story (original function). When integrating, each step adds a constant of integration because integration reverses differentiation, which erases information about constants.

In our problem, we start with the second derivative, \(f''(t) = 2e^{t} + 3\sin t\). To find the function itself, we integrate this expression twice. The first integration gives us \(f'(t)\), the first derivative of the function. Then, integrating \(f'(t)\) gives us \(f(t)\), the original function. Each integration step introduces a new constant of integration, represented as \(C_1\) and \(C_2\) in this exercise.

Thus, every integration step does not just yield an antiderivative but additionally includes an undefined constant that accounts for any horizontal shift in the function.

Differential Equations

A differential equation involves unknown functions and their derivatives, essentially describing how a function changes. They form a fundamental part of calculus and mathematical modeling. In our exercise, we start with a second-order differential equation: \(f''(t) = 2e^{t} + 3\sin t\).

To solve this differential equation, we integrate it step by step. The goal is to find a function \(f(t)\) that satisfies the equation. We first find the first derivative \(f'(t)\) through integration and then find \(f(t)\) by integrating \(f'(t)\).

• Step 1: Integrate to obtain \(f'(t)\).
• Step 2: Integrate \(f'(t)\) to find \(f(t)\).

Using differential equations, we saw methods to transform a second-order differential equation into a problem of integration, gradually unwinding it back to the original function with the help of initial conditions.

Initial Conditions

Initial conditions help determine the exact value of constants we introduced during integration. Without these, an entire family of solutions is possible, making our solution incomplete.
In the exercise, we are provided with two conditions: \(f(0) = 0\) and \(f(\pi) = 0\). These conditions allow us to solve for \(C_1\) and \(C_2\), the constants from integrating.

• Substituting \(f(0) = 0\) allows us to find \(C_2\).
• Using \(f(\pi) = 0\), we can solve for \(C_1\).

These initial conditions narrow the multiple potential solutions down to the specific one that describes our original problem precisely. It's much like using a map and a compass to pinpoint exactly where on a route you are, rather than wandering indefinitely.