Problem 85 - Thomas Calculus in SI Units

Question

Solve the initial value problems. $$\frac{d^{2} r}{d t^{2}}=\frac{2}{t^{3}} ;\left.\quad \frac{d r}{d t}\right|_{t=1}=1, \quad r(1)=1$$

Step-by-Step Solution

Answer

The solution is \( r(t) = \frac{2}{t} + 3t - 4 \).

Step 1: Integrating the Second Derivative

To solve the differential equation, start by integrating the equation \( \frac{d^2 r}{dt^2} = \frac{2}{t^3} \) with respect to \( t \). When you integrate \( \frac{2}{t^3} \), you get:\[ \frac{dr}{dt} = \int \frac{2}{t^3} \, dt = -\frac{2}{t^2} + C_1 \] This introduces the first constant of integration, \( C_1 \).

Step 2: Apply Initial Condition to the First Derivative

We use the initial condition \( \left. \frac{dr}{dt} \right|_{t=1} = 1 \). Substitute \( t = 1 \) and \( \frac{dr}{dt} = 1 \) into the integrated function:\[ 1 = -\frac{2}{(1)^2} + C_1 \] Simplifying gives:\[ 1 = -2 + C_1 \] Therefore, \( C_1 = 3 \).

Step 3: Integrate the First Derivative

Now integrate the expression \( \frac{dr}{dt} = -\frac{2}{t^2} + 3 \) to find \( r(t) \):\[ r(t) = \int \left(-\frac{2}{t^2} + 3\right) \, dt \] This yields:\[ r(t) = 2t^{-1} + 3t + C_2 = \frac{2}{t} + 3t + C_2 \] where \( C_2 \) is the second constant of integration.

Step 4: Apply Initial Condition to the Function

Use the initial condition \( r(1) = 1 \) to solve for \( C_2 \). Substitute \( t = 1 \) and \( r = 1 \) into the function:\[ 1 = \frac{2}{1} + 3 \times 1 + C_2 \] Simplify the equation to find \( C_2 \):\[ 1 = 2 + 3 + C_2 \] This reduces to:\[ 1 = 5 + C_2 \] and therefore, \( C_2 = -4 \).

Step 5: Write the Final Solution

Combine the results to write the solution for \( r(t) \), incorporating \( C_1 \) and \( C_2 \):\[ r(t) = \frac{2}{t} + 3t - 4 \] Thus, this is the function that satisfies the given initial value problem.

Key Concepts

Initial Value Problem

An initial value problem is a kind of differential equation that includes extra conditions called initial conditions. These are specified values that the solution must satisfy at a particular point. In this exercise, for example, the initial conditions are given as \( \left.\frac{dr}{dt}\right|_{t=1} = 1 \) and \( r(1) = 1 \). These conditions are crucial because they help us determine the exact values for any constants that come up while integrating the differential equation. This process ensures that the solution is specific to the problem at hand.When solving an initial value problem:

• Identify the differential equation and its derivatives involved.
• Integrate to find the general solution.
• Use the initial conditions to find the constants of integration.

Each step is vital as it helps to pinpoint a unique solution rather than a family of solutions.

Integration

Integration is essential in solving differential equations, especially ordinary differential equations (ODEs), like in the given exercise. It's the opposite operation of differentiation and is used to find functions given their derivatives. Here, we started with a second derivative equation and found its first integral to get \( \frac{dr}{dt} \) and then a second integral to determine \( r(t) \).In the integration process:

• Identify the function you need to integrate.
• Apply the basic integration rules, like power rule or substitution if necessary.
• Don't forget to add a constant of integration (\( C \)) for indefinite integrals.

Integration helps to construct a function iteratively from its rate of change, or derivative, while the constant represents any unknown accumulated change.

Constant of Integration

The constant of integration is a fundamental concept in integration, reflecting the fact that there are infinitely many antiderivatives for a given function. In indefinite integration, every time you integrate, you add a constant \( C \), which accounts for any constant value that could have been differentiated away in the original function.In our exercise, after integrating the second derivative, we introduced the first constant \( C_1 \). When we solved the initial condition \( \left.\frac{dr}{dt}\right|_{t=1} = 1 \), it helped us find \( C_1 = 3 \). Similarly, when integrating the first derivative, we introduced \( C_2 \) and used \( r(1) = 1 \) to find \( C_2 = -4 \).To remember:

• Every indefinite integral should have a constant \( C \).
• Initial conditions are used to solve for these constants.
• Each equation and step can introduce its own integration constant.

These constants tailor the solution to fit not just any curve, but the precise curve that satisfies the initial conditions.