Problem 88 - Thomas Calculus in SI Units

Question

Solve the initial value problems. $$\frac{d^{3} \theta}{d t^{3}}=0 ; \quad \theta^{\prime \prime}(0)=-2, \quad \theta^{\prime}(0)=-\frac{1}{2}, \quad \theta(0)=\sqrt{2}$$

Step-by-Step Solution

Answer

\( \theta(t) = -t^2 - \frac{1}{2}t + \sqrt{2} \)

Step 1: Integrate the differential equation

The given differential equation is:\[ \frac{d^{3} \theta}{d t^{3}} = 0 \] Integrating this with respect to \( t \), we get: \[ \frac{d^{2} \theta}{d t^{2}} = C_1 \] where \( C_1 \) is a constant of integration.

Step 2: Apply initial condition on second derivative

We use the initial condition \( \theta''(0) = -2 \) to find \( C_1 \). Substituting \( t = 0 \) into \( \frac{d^{2} \theta}{d t^{2}} = C_1 \), we have \( C_1 = -2 \).

Step 3: Integrate to find the first derivative

Integrate \( \frac{d^{2} \theta}{d t^{2}} = -2 \) with respect to \( t \):\[ \frac{d \theta}{d t} = -2t + C_2 \] where \( C_2 \) is another constant of integration.

Step 4: Apply initial condition on first derivative

Use the initial condition \( \theta'(0) = -\frac{1}{2} \). Substitute \( t = 0 \) into \( \frac{d \theta}{d t} = -2t + C_2 \) to find \( C_2 \): \(-\frac{1}{2} = C_2\).

Step 5: Integrate to find the original function

Integrate \( \frac{d \theta}{d t} = -2t - \frac{1}{2} \) with respect to \( t \):\[ \theta(t) = -t^2 - \frac{1}{2}t + C_3 \] where \( C_3 \) is another constant of integration.

Step 6: Apply initial condition on the function

Use the initial condition \( \theta(0) = \sqrt{2} \). Substitute \( t = 0 \) into \( \theta(t) = -t^2 - \frac{1}{2}t + C_3 \) to find \( C_3 \): \( \sqrt{2} = C_3 \).

Step 7: Write the final solution

Substitute all constants back into the function to write the final solution: \[ \theta(t) = -t^2 - \frac{1}{2}t + \sqrt{2} \]

Key Concepts

Differential equations

Differential equations are mathematical equations that involve an unknown function along with its derivatives. In simple terms, they help us describe various real-world phenomena, such as how something changes over time. For example, the given differential equation \( \frac{d^{3} \theta}{d t^{3}} = 0 \) involves a third derivative of the function \( \theta(t) \) with respect to time \( t \). This tells us how fast the change in \( \theta(t) \) is decelerating over time.By solving this equation, we aim to find the function \( \theta(t) \), which shows the behavior of the system. In our problem, since the third derivative is zero, it denotes a system where changes gradually level off after some time.

Constants of integration

When working with differential equations, especially when integrating, we often encounter constants of integration. These constants arise because integration is essentially the reverse operation of differentiation, which means that many functions can have the same derivative.

• After the first integration of \( \frac{d^{3} \theta}{d t^{3}} = 0 \), we obtain \( \frac{d^{2} \theta}{d t^{2}} = C_1 \), where \( C_1 \) is the first constant of integration.
• Integrating again gives us \( \frac{d \theta}{d t} = -2t + C_2 \), introducing another constant \( C_2 \).
• Finally, we integrate once more, leading to \( \theta(t) = -t^2 - \frac{1}{2}t + C_3 \).

Each constant of integration is determined using specific initial conditions provided in the problem, ensuring a unique solution that fits the initial scenario.

Initial conditions

Initial conditions play a crucial role in solving differential equations, particularly in initial value problems. These conditions specify values for the unknown function or its derivatives at a particular point in time, allowing us to find the constants of integration and thus a particular solution.In this exercise, the initial conditions are given as:

• \( \theta''(0) = -2 \)
• \( \theta'(0) = -\frac{1}{2} \)
• \( \theta(0) = \sqrt{2} \)

These values are substituted back into the equations derived during integration to solve for the constants \( C_1 \), \( C_2 \), and \( C_3 \). Applying these initial conditions ensures that the solution satisfies the behavior of the system at \( t = 0 \).

Integration process

The integration process is a method where we calculate antiderivatives to solve differential equations. It is as though we are "backtracking" from the rate of change to the actual quantity.In our problem, the process begins with solving \( \frac{d^{3} \theta}{d t^{3}} = 0 \) through successive integrations:- The first integration results in \( \frac{d^{2} \theta}{d t^{2}} = C_1 \), where \( C_1 \) is later calculated using \( \theta''(0) = -2 \).- The second integration gives \( \frac{d \theta}{d t} = -2t + C_2 \), where \( C_2 \) is determined from \( \theta'(0) = -\frac{1}{2} \).- The third integration leads to \( \theta(t) = -t^2 - \frac{1}{2}t + C_3 \), finalized using \( \theta(0) = \sqrt{2} \).Ultimately, the integration process connects rates of change back to the original quantity, crafting a complete picture of the system's dynamics.