Problem 30 - Thomas Calculus

Question

Find all possible functions with the given derivative. a. \(y^{\prime}=\frac{1}{2 \sqrt{x}} \quad\) b. \(y^{\prime}=\frac{1}{\sqrt{x}} \quad\) c. \(y^{\prime}=4 x-\frac{1}{\sqrt{x}}\)

Step-by-Step Solution

Answer

a: \(y = \sqrt{x} + C\); b: \(y = 2\sqrt{x} + C\); c: \(y = 2x^2 - 2\sqrt{x} + C\).

Step 1: Understanding the Problem

The problem asks us to find the original functions given the derivatives. This requires integrating the given derivatives with respect to variable \(x\).

Step 1: Integrate the First Derivative

For part (a), the derivative is \(y' = \frac{1}{2\sqrt{x}}\). To find \(y\), we integrate \(\frac{1}{2\sqrt{x}}\) with respect to \(x\). The integral of \(\frac{1}{2\sqrt{x}}\) is \(\int \frac{1}{2\sqrt{x}} \, dx = \sqrt{x} + C\), where \(C\) is an integration constant.

Step 2: Integrate the Second Derivative

For part (b), the derivative is \(y' = \frac{1}{\sqrt{x}}\). The integral is \(\int \frac{1}{\sqrt{x}} \, dx = 2\sqrt{x} + C\). This gives us the function \(y = 2\sqrt{x} + C\).

Step 3: Integrating a More Complex Derivative

For part (c), the derivative is \(y' = 4x - \frac{1}{\sqrt{x}}\). We need to integrate each term separately: \(\int 4x \, dx = 2x^2\) and \(\int -\frac{1}{\sqrt{x}} \, dx = -2\sqrt{x}\). Combining these results gives \(y = 2x^2 - 2\sqrt{x} + C\), where \(C\) is the integration constant.

Key Concepts

Integration Techniques

When we talk about finding the original function from a given derivative, we are diving into the world of integration. Integration is the process of reversing differentiation, and there are various techniques to achieve this. For simple functions, direct integration is often feasible.

• **Power Rule**: This is useful when integrating terms like \(x^n\). When differentiating, the exponent \(n\) comes down as a coefficient and the exponent decreases by one. Integration does the reverse, where \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\), provided \(n eq -1\).
• **Substitution**: Sometimes the function isn't in a form that's easy to integrate directly. Substitution can "reformulate" the integral by substituting a part of the integrand with a new variable.
• **Partial Fractions**: For rational functions, decomposing them into simpler fractions can simplify integration.

In our current exercise, to integrate \(\frac{1}{\sqrt{x}}\), recognize it as \(x^{-1/2}\), and apply the power rule. The integration becomes straightforward as \(\int x^{-1/2} \, dx = 2\sqrt{x} + C\).
Understanding these techniques is crucial for solving a wide range of integration problems with ease.

Integration Constant

When we perform an indefinite integral, a very important component crops up — the integration constant, denoted by \(C\). This constant is essential because it represents the family of all possible functions that we can obtain when integrating a derivative.
Here is why it's relevant:

• **Arbitrary Nature**: Since differentiation of a constant yields zero, any constant added to our function has no effect during the differentiation process. This means when we integrate, \(C\) could have been any number when the derivative was taken.
• **Initial Conditions**: The specific value of \(C\) can be determined if additional information such as initial conditions or boundary values are available. These conditions "fix" the constant, providing one precise function out of the infinite possibilities.
• **Family of Functions**: Without an initial condition, the solution remains a general form expressing the idea that many possible original functions could exist.

As seen in the exercise, each solution contains a \(C\) reflecting the set of all potential antiderivatives.

Differential Equations

Differential equations involve equations that relate a function with its derivatives. Solving these equations often entails integration, as it lets us reconstruct the original function from its rate of change.
Differential equations can be:

• **Ordinary Differential Equations (ODEs)**: These are equations involving derivatives of a function of a single variable. Our exercise deals with ordinary derivatives like \(y' = \frac{1}{2\sqrt{x}}\).
• **Partial Differential Equations (PDEs)**: These involve partial derivatives, which are the derivatives of functions of multiple variables.

Solving a differential equation involves finding a function that satisfies the given relation. For the given derivatives in the exercise, integrating term by term of the function produces a solution, which potentially could be a part of what solves more complex systems of equations.
Understanding differential equations greatly enhances one's ability to model real-world phenomena where change and rates are naturally occurring, across sciences and engineering.