Problem 53 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(53-58\). $$ \frac{d s}{d t}=12 t\left(3 t^{2}-1\right)^{3}, \quad s(1)=3 $$

Step-by-Step Solution

Answer

The solution is \( s(t) = \frac{1}{2}(3t^2-1)^4 - 5 \).

Step 1: Identify the Problem Structure

This is an initial value problem involving a differential equation. The function to solve for is \( s(t) \), given \( \frac{ds}{dt} = 12t(3t^2-1)^3 \) with an initial condition \( s(1) = 3 \). The goal is to find the function \( s(t) \) that satisfies both the differential equation and the initial condition.

Step 2: Integrate the Differential Equation

To find \( s(t) \), integrate the given derivative with respect to \( t \). \[\int \frac{ds}{dt} \, dt = \int 12t(3t^2-1)^3 \, dt.\]Focus on the right-hand side. Use substitution: let \( u = 3t^2 - 1 \), then \( du = 6t \, dt \), so \( t \, dt = \frac{1}{6} du \). Substitute and integrate:\[\int 12t(3t^2-1)^3 \, dt = 2 \int u^3 \, du = 2 \left(\frac{u^4}{4}\right) + C = \frac{1}{2}u^4 + C.\]Re-substitute \( u = 3t^2 - 1 \), so \( \frac{1}{2}(3t^2-1)^4 + C \) is the antiderivative.

Step 3: Apply the Initial Condition

Use the initial condition \( s(1) = 3 \) to find the constant \( C \). Substitute \( t = 1 \) and \( s = 3 \) into the expression obtained from the integration:\[3 = \frac{1}{2}(3(1)^2-1)^4 + C = \frac{1}{2}(2)^4 + C = \frac{1}{2} \times 16 + C = 8 + C.\]Solve for \( C \):\[C = 3 - 8 = -5.\]

Step 4: Write the Final Solution

Substitute \( C = -5 \) back into the expression for \( s(t) \):\[s(t) = \frac{1}{2}(3t^2-1)^4 - 5.\]This function \( s(t) \) satisfies both the differential equation and the initial condition.

Key Concepts

Differential Equation

A differential equation involves an unknown function and its derivatives. These equations express how a function changes and are fundamental in representing various natural processes. In this particular exercise, the differential equation is given as \( \frac{ds}{dt} = 12t(3t^2-1)^3 \). It describes the rate of change of the function \( s(t) \) with respect to time \( t \). Understanding this change is key to finding the actual function \( s(t) \) that solves the equation. By solving differential equations, we gain insight into the behaviour of dynamic systems and can predict future outcomes based on current states.
Differential equations come in many types and forms. Some of the most common are:

• Ordinary Differential Equations (ODEs) - equations with one independent variable, like time.
• Partial Differential Equations (PDEs) - involve multiple independent variables.

In this exercise, we are dealing with an ordinary differential equation as the variables involved are \( s \) and \( t \). It’s crucial to set up the differential equation correctly before proceeding further with methods like integration.

Integration

Integration is the process of finding a function from its derivative, essentially the reverse process of differentiation. To solve the given differential equation, we need to integrate the expression to find \( s(t) \). This involves evaluating the integral of the expression \( \int 12t(3t^2-1)^3 \, dt \). Integration allows us to sum up small pieces to find a whole value, which in this context helps us find \( s(t) \) knowing how it changes.
There are different methods of integration:

• Indefinite Integrals - where the result includes a constant of integration \( C \).
• Definite Integrals - give a specific number, adjusting for limits of integration.

When integrating the differential equation, we used indefinite integration since the limits were not specified beyond \( t \). Here, integration takes the form of finding an antiderivative, allowing us to determine \( s(t) \). Only after integration, we can apply initial conditions to find both specific constants and solutions.

Substitution Method

The substitution method is a powerful technique to simplify complex integrals. It transforms the original variable into a new variable that makes the integral easier to evaluate. For the integral \( \int 12t(3t^2-1)^3 \, dt \), we make the substitution \( u = 3t^2 - 1 \). This simplifies the problem by reducing complex expressions to simpler ones.

Here's how substitution works in this exercise:

• Define \( u = 3t^2 - 1 \).
• Differentiate \( u \) to find \( du = 6t \, dt \).
• Solve for \( t \, dt \), giving \( t \, dt = \frac{1}{6} du \).

The integral then becomes \( 2 \int u^3 \, du \), which is easier to integrate. Once integrating in terms of \( u \) is complete, substitute back \( u = 3t^2 - 1 \) to express the solution in terms of the original variable \( t \). This substitution method is often used in calculus to handle integrals involving composite functions, trigonometric expressions, and other complex algebraic forms.

Initial Condition

An initial condition allows us to find a specific solution to a differential equation by providing a starting point for that solution. In the context of differential equations, they give us a specific piece of information that helps determine the constant value in our solution after integration. In this exercise, the initial condition is given as \( s(1) = 3 \).

The steps involving initial conditions are straightforward:

• Use the condition \( s(1) = 3 \) to substitute \( t = 1 \) in the integrated result.
• Solve for the constant \( C \) by setting the integrated equation equal to the initial condition value.

This problem shows how initial conditions transform a general solution into a specific one, valid for the scenario described. By incorporating initial conditions, we anchor the integration result to a specific point on the curve \( s(t) \), ensuring the solution is unique and directly applicable.