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Chapter 4: Problem 88

Solve the initial value problems in Exercises \(71-90\) $$\frac{d^{3} \theta}{d t^{3}}=0 ; \quad \theta^{\prime \prime}(0)=-2, \quad \theta^{\prime}(0)=-\frac{1}{2}, \quad \theta(0)=\sqrt{2}$$

Short Answer

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

Step by step solution

01

Recognize the Problem Type

The given problem is a third-order differential equation where the third derivative of the function, \( \theta(t) \), is zero. This indicates that the function \( \theta(t) \) is a polynomial of degree 2 or less.
02

Integrate to Find \( \theta''(t) \)

First, integrate the equation \( \frac{d^3 \theta}{dt^3} = 0 \) with respect to \( t \). This will give us \( \frac{d^2 \theta}{dt^2} = C_1 \), where \( C_1 \) is an integration constant.
03

Apply Initial Condition for \( \theta''(0) \)

We know \( \theta''(0) = -2 \). Thus, substitute into \( \theta''(t) = C_1 \): \( C_1 = -2 \). So, \( \theta''(t) = -2 \).
04

Integrate to Find \( \theta'(t) \)

Integrate \( \theta''(t) = -2 \) with respect to \( t \). This gives \( \theta'(t) = -2t + C_2 \), where \( C_2 \) is another integration constant.
05

Apply Initial Condition for \( \theta'(0) \)

Using the initial condition \( \theta'(0) = -\frac{1}{2} \), we substitute into \( \theta'(t) = -2t + C_2 \). Solving for \( C_2 \), we find that \( C_2 = -\frac{1}{2} \). Therefore, \( \theta'(t) = -2t - \frac{1}{2} \).
06

Integrate to Find \( \theta(t) \)

Integrate \( \theta'(t) = -2t - \frac{1}{2} \) with respect to \( t \). This results in \( \theta(t) = -t^2 - \frac{1}{2}t + C_3 \).
07

Apply Initial Condition for \( \theta(0) \)

With the initial condition \( \theta(0) = \sqrt{2} \), substitute into \( \theta(t) = -t^2 - \frac{1}{2}t + C_3 \). Solving for \( C_3 \), we find \( C_3 = \sqrt{2} \). Thus, \( \theta(t) = -t^2 - \frac{1}{2}t + \sqrt{2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Third-Order Differential Equation
A third-order differential equation involves terms up to the third derivative of a function. In our exercise, the equation \( \frac{d^{3} \theta}{d t^{3}}=0 \) clearly states this. A third-order derivative being zero simplifies things greatly, as it suggests the original function could be a polynomial of degree at most 2. Why?
  • If the third derivative is zero, integrating once gives a constant second derivative.
  • Integrating the constant second derivative results in a linear first derivative.
  • A subsequent integration of a linear first derivative yields a quadratic function.
Thus, our function \( \theta(t) \) has a form that doesn't go beyond second-degree, matching up with the expectations of the given differential equation.
Integration Constants
In differential equations, integration constants arise when integrating expressions. Our initial differential equation gets simplified as we integrate it step by step. Each integration introduces a constant that remains indeterminate unless initial conditions are applied. For example:
  • After the first integration of \( \frac{d^{3} \theta}{d t^{3}}=0 \), we derive \( \theta''(t) = C_1 \).
  • A further integration results in \( \theta'(t) = -2t + C_2 \).
  • Finally, integrating again gives \( \theta(t) = -t^2 - \frac{1}{2}t + C_3 \).
These constants \( C_1, C_2, \) and \( C_3 \) need specific initial conditions to be determined. Using initial conditions helps convert this general solution into a specific solution.
Polynomial Solution
Since our third derivative is zero, the function \( \theta(t) \) is a polynomial. By integrating stepwise, starting from the zero third derivative, we end up with a polynomial. The structure of the resulting polynomial matches that of a normal quadratic expression:
  • The function form is \( \theta(t) = -t^2 - \frac{1}{2}t + C_3 \).
  • The coefficients of \( t^2 \) and \( t \) were determined through integration and the application of the initial conditions.
Polynomials make solutions predictable, approachable, and, in many cases, simpler to manipulate. They also demonstrate neat connections between calculus operations and algebraic forms.
Initial Conditions
Initial conditions are crucial in solving differential equations, especially when finding specific solutions. For our problem, initial conditions were given at \( t=0 \):
  • \( \theta''(0) = -2 \)
  • \( \theta'(0) = -\frac{1}{2} \)
  • \( \theta(0) = \sqrt{2} \)
These help solve for the integration constants so that the differential equation's general solution becomes a precise function conforming to the specified conditions. By applying these values during the integration constant solving steps, each constant becomes known, leading us to the concrete answer \( \theta(t) = -t^2 - \frac{1}{2}t + \sqrt{2} \). This method ensures the polynomial derived respects the problem's specific constraints.

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Most popular questions from this chapter

A right circular cone is circumscribed in a sphere of radius \(1 .\) Determine the height \(h\) and radius \(r\) of the cone of maximum volume.

You are to construct an open rectangular box with a square base and a volume of 48 \(\mathrm{ft}^{3} .\) If material for the bottom costs \(\ 6 / \mathrm{ft}^{2}\) and material for the sides costs \(\ 4 / \mathrm{ft}^{2},\) what dimensions will result in the least expensive box? What is the minimum cost?

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