Problem 81 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(67-86\). $$ \frac{d^{2} r}{d t^{2}}=\frac{2}{t^{3}} ;\left.\quad \frac{d r}{d t}\right|_{t=1}=1, \quad r(1)=1 $$

Step-by-Step Solution

Answer

The solution to the initial value problem is \( r(t) = \frac{1}{t} + 2t - 2 \).

Step 1: Integrate the Differential Equation

First, we need to integrate the given second derivative of \( r \) to find the first derivative. The differential equation is given as \( \frac{d^2r}{dt^2} = \frac{2}{t^3} \). By integrating with respect to \( t \), we obtain:\[ \frac{dr}{dt} = \int \frac{2}{t^3} \, dt = -\frac{1}{t^2} + C_1 \]Where \( C_1 \) is a constant to be determined.

Step 2: Apply Initial Condition to Find Constant C1

We have the initial condition \( \left.\frac{dr}{dt}\right|_{t=1} = 1 \). Substitute \( t = 1 \) and \( \frac{dr}{dt} = 1 \) into the integrated equation:\[ 1 = -\frac{1}{1^2} + C_1 \]This simplifies to:\[ 1 = -1 + C_1 \]Solving for \( C_1 \), we find \( C_1 = 2 \). The first derivative is now:\[ \frac{dr}{dt} = -\frac{1}{t^2} + 2 \].

Step 3: Integrate Again to Find r(t)

Next, integrate \( \frac{dr}{dt} = -\frac{1}{t^2} + 2 \) to find \( r(t) \):\[ r(t) = \int \left( -\frac{1}{t^2} + 2 \right) \, dt \]This results in:\[ r(t) = \frac{1}{t} + 2t + C_2 \]Where \( C_2 \) is another constant to be determined.

Step 4: Apply Initial Condition to Find Constant C2

Using the initial condition \( r(1) = 1 \), substitute \( t = 1 \) and \( r = 1 \) into the integrated function:\[ 1 = \frac{1}{1} + 2 \cdot 1 + C_2 \]Simplify to solve for \( C_2 \):\[ 1 = 1 + 2 + C_2 \]\[ C_2 = -2 \]So, \( r(t) = \frac{1}{t} + 2t - 2 \).

Step 5: Verify the Solution

We can verify our solution by checking if it satisfies the original differential equation and initial conditions. Using \( r(t) = \frac{1}{t} + 2t - 2 \):1. Compute \( \frac{dr}{dt} \): \[ \frac{dr}{dt} = -\frac{1}{t^2} + 2 \]2. Compute \( \, \frac{d^2 r}{dt^2} \): \[ \frac{d^2r}{dt^2} = \frac{2}{t^3} \]This matches the original differential equation. Verify initial conditions:- \( \left.\frac{dr}{dt}\right|_{t=1} = 1 \) and \( r(1) = 1 \). Both conditions are satisfied.

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In this context, we are dealing with a second-order differential equation, which means we need to find a function, referred to as a solution, that satisfies the given equation. The main goal is to determine this function by integrating the differential equation.
The given differential equation is a second derivative equation of the form \( \frac{d^2 r}{dt^2} = \frac{2}{t^3} \). To solve it, we will tackle it by integrating back from the second derivative to find the first derivative, and eventually, the original function. By solving these equations, we can model physical systems where the rate of change itself changes, such as in motion under the influence of variable forces.

Integration

Integration is a crucial mathematical process used to find a function from its derivative. When solving differential equations, we often integrate to "backtrack" from the derivatives to the original function.
In our problem, we started with the integral of \( \frac{d^2 r}{dt^2} = \frac{2}{t^3} \) to find the first derivative. Here are the steps:

• Integrate \( \frac{2}{t^3} \) with respect to \( t \) to get \( \frac{dr}{dt} = -\frac{1}{t^2} + C_1 \), where \( C_1 \) is a constant of integration.
• We integrate once more to find \( r(t) = \int \left( -\frac{1}{t^2} + 2 \right) \, dt = \frac{1}{t} + 2t + C_2 \), with \( C_2 \) as another constant.

The act of integration adds a "+ C" because many functions can have the same derivative. This "C" represents an infinite range of possible vertical shifts on the graph, which we later define with initial conditions.

Initial Conditions

Initial conditions are essential to solving differential equations because they help us determine the constant terms that arise from integration. These are often known as boundaries or starting parameters that specify a particular solution from a general family of solutions provided by integration.
For the solution to our problem:

• We used the initial condition \( \left.\frac{dr}{dt}\right|_{t=1} = 1 \) to solve for \( C_1 \). Substituting this into our integrated equation helped us find \( C_1 = 2 \).
• Similarly, with the initial condition \( r(1) = 1 \), we determined \( C_2 = -2 \).

By applying these initial conditions, we could ascertain a specific solution \( r(t) \) that satisfies not only the differential equation but the specific scenario described by these initial values.

Second Derivative

The second derivative of a function is simply the derivative of the derivative of that function. It provides insight into the curvature or concavity of the original function. In physical terms for motion, the second derivative of position with respect to time is acceleration.
For the given problem, the second derivative \( \frac{d^2r}{dt^2} = \frac{2}{t^3} \) tells us about the rate of change of the rate of change of \( r \) as a function of \( t \).

• This function \( \frac{2}{t^3} \) indicates that the acceleration decreases as \( t \) increases, due to the \( t^3 \) in the denominator.
• Re-integrating this to find the expression for \( r \), we clarified how these changes influence \( r \), integrating twice to find \( r(t) = \frac{1}{t} + 2t - 2 \).

Understanding and calculating the second derivative helps frame the broader dynamics described in the differential equation, crucially helping to understand the behavior modeled by \( r(t) \).