Problem 107 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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

Step 1: Integrate the Differential Equation

Start by integrating the differential equation \(\frac{d^{2} r}{d t^{2}}=\frac{2}{t^{3}}\) with respect to \(t\). This will give the first derivative, \(\frac{d r}{d t}\): \[ \frac{d r}{d t} = \int \frac{2}{t^3} \, dt = -\frac{2}{t^2} + C_1 \] where \(C_1\) is the constant of integration.

Step 2: Apply Initial Condition for First Derivative

Use the initial condition \(\left.\frac{d r}{d t} \right|_{t=1} = 1\) to solve for \(C_1\): \[ 1 = -\frac{2}{1^2} + C_1 \]Solve for \(C_1\): \[ C_1 = 1 + 2 = 3 \] Thus, \(\frac{d r}{d t} = -\frac{2}{t^2} + 3\).

Step 3: Integrate to Find \(r(t)\)

Integrate the equation \(\frac{d r}{d t} = -\frac{2}{t^2} + 3\) with respect to \(t\) to find \(r(t)\): \[ r(t) = \int \left(-\frac{2}{t^2} + 3\right) \, dt = \int -\frac{2}{t^2} \, dt + \int 3 \, dt \]The integrals are: \[ \int -\frac{2}{t^2} \, dt = \frac{2}{t}, \quad \int 3 \, dt = 3t \] Combine and add a new constant of integration \(C_2\): \[ r(t) = \frac{2}{t} + 3t + C_2 \]

Step 4: Apply Initial Condition for \(r(t)\)

Use the initial condition \(r(1) = 1\) to solve for \(C_2\): \[ 1 = \frac{2}{1} + 3 \times 1 + C_2 \]Solve for \(C_2\): \[ 1 = 2 + 3 + C_2 \Rightarrow C_2 = 1 - 5 = -4 \] Thus, the solution for \(r(t)\) is: \[ r(t) = \frac{2}{t} + 3t - 4 \]

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve derivatives, which are expressions representing the rates of change of a function. They are used to model a wide variety of real-world phenomena, such as mechanics, heat, and biological processes.

• Types: Differential equations can be ordinary or partial. The former involves one independent variable, while the latter involves multiple.
• Order: The order of a differential equation is determined by the highest derivative present in the equation.

Consider the equation from the exercise: \(\frac{d^{2} r}{d t^{2}}=\frac{2}{t^{3}}\). This is a second-order ordinary differential equation because it involves the second derivative of \(r\) with respect to \(t\), and there is only one independent variable. By integrating such equations, we work towards finding the original function \(r(t)\), which describes how \(r\) changes with \(t\). Understanding differential equations is crucial, as they provide the basis for predicting future states of dynamic systems.

Integration

Integration is the mathematical process of finding the integral of a function, which is essentially the inverse operation of differentiation. When we integrate, we are looking to build a function from its derivative. This process allows us to find the original function that gives rise to the given rate of change.

• Definite vs. Indefinite: A definite integral provides an actual number and represents the area under a curve between two limits. An indefinite integral, like those used in this exercise, represents a family of functions and includes an arbitrary constant.
• Finding Integrals: To solve the problem, we integrated \(\frac{2}{t^3}\) to get the first derivative \(\frac{dr}{dt}\), and then integrated that result to find \(r(t)\).

Each step involved calculating the integral of a part of the equation. For instance, the function \(-\frac{2}{t^2}\) resulted from integrating \(-\frac{2}{t^3}\), and integrating the constant 3 yielded \(3t\). The solutions are part of constructing a bigger picture that brings out the relationship between the variables involved. Integration helps us recover the original function from its rate of change, providing deeper insights into the behavior of dynamic systems.

Constant of Integration

In the process of integration, we invariably come across the concept of the "constant of integration." This constant, typically represented as \(C\), arises because the derivative of a constant is zero, meaning that any function differing only by a constant will have the same derivative.

• Essential Addition: When solving indefinite integrals, always include a constant of integration, \(C\), since the original value of the function could be shifted by any constant amount.
• Solution Uniqueness: The constant of integration is crucial for determining a unique solution to a differential equation, especially in conjunction with initial conditions.

In the exercise, when integrating \(-\frac{2}{t^2} + 3\), we added a constant \(C_2\) to form the function \(r(t) = \frac{2}{t} + 3t + C_2\). This constant reflects infinite possible vertical shifts of the function, which can be narrowed down using given initial conditions. Such constants ensure that our solutions to differential equations reflect specific situations described by additional information provided, like initial values.

Initial Conditions

Initial conditions are specific values given in a problem that allow you to find the constants of integration in the solution of a differential equation. They effectively pinpoint a specific member of the family of solutions you obtain from indefinite integration.

• Purpose: Initial conditions create definite and practical solutions from general ones, ensuring the function satisfies certain known values at particular points.
• Applying Initial Conditions: In each instance where a derivative is integrated, the resulting expression contains a constant. Applying an initial condition helps solve for this constant.

For instance, we applied the condition \(\left.\frac{d r}{d t}\right|_{t=1}=1\) to find \(C_1\), which resulted in the first solution \(\frac{dr}{dt} = -\frac{2}{t^2} + 3\). Another condition, \(r(1)=1\), was applied to solve for \(C_2\), allowing us to conclude \(r(t) = \frac{2}{t} + 3t - 4\). These conditions allow us to tailor our general solution, aligning it with known characteristics of the system being modeled, thereby capturing its unique behavior.