Problem 36 - University Calculus: Early Transcendentals

Question

Find all possible functions with the given derivative. a. \(y^{\prime}=\frac{1}{2 \sqrt{x}}\) 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. \(y = 2x^2 - 2\sqrt{x} + C\) where \(C\) is a constant.

Step 1: Understand the Derivative

Identify the derivative given in each part. For part (a), the derivative is \(y' = \frac{1}{2 \sqrt{x}}\). For part (b), the derivative is \(y' = \frac{1}{\sqrt{x}}\). For part (c), the derivative is \(y' = 4x - \frac{1}{\sqrt{x}}\).

Step 2: Integrate the Derivative for Part (a)

To find the function \(y\) for part (a), integrate \(y' = \frac{1}{2 \sqrt{x}}\). The integral is \(y = \int \frac{1}{2 \sqrt{x}} \, dx = \int \frac{1}{2} x^{-1/2} \, dx\). Integrating, we get \(y = \frac{1}{2} \cdot \frac{2}{1} x^{1/2} + C = \sqrt{x} + C\), where \(C\) is a constant.

Step 3: Integrate the Derivative for Part (b)

Integrate \(y' = \frac{1}{\sqrt{x}}\) to find \(y\). This can be rewritten as \(y' = x^{-1/2}\). The integral is \(y = \int x^{-1/2} \, dx = 2x^{1/2} + C\), where \(C\) is a constant.

Step 4: Integrate the Derivative for Part (c)

For part (c), integrate \(y' = 4x - \frac{1}{\sqrt{x}}\). The integral can be split as \(y = \int (4x) \, dx - \int \frac{1}{\sqrt{x}} \, dx\). Integrating separately, we get \(y = 2x^2 - 2x^{1/2} + C\), where \(C\) is a constant.

Key Concepts

Calculus

Calculus is a fundamental branch of mathematics. It provides tools to solve problems related to change and motion. The two basic concepts in calculus are differentiation and integration.

• Limits: The concept of a limit is used to define both derivatives and integrals. It helps in understanding the behavior of functions as they approach particular points or extent to infinity.
• Continuity: A function is continuous if small changes in the input lead to small changes in the output.
• Derivatives: Derivatives represent the rate of change of a function. It's the slope of the function at a particular point.
• Integrals: Integrals are used to find areas under curves and are the "reverse" of derivatives.

Once these basic concepts are understood, calculus can be applied to various fields such as physics, engineering, economics, and more. It allows us to describe real-world phenomena in a quantitative way.

Differentiation

Differentiation is a key concept in calculus. It deals with finding the derivative of a function.
The derivative measures how a function changes as the input changes. Think of it as the function's "instantaneous rate of change."

• Basic Rule: For a function \( f(x) = x^n \), the derivative is \( f'(x) = n \cdot x^{n-1} \).
• Chain Rule: Useful for differentiating composite functions, it states \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \).
• Product Rule: Used when differentiating a product of two functions, it gives \( (uv)' = u'v + uv' \).
• Quotient Rule: For the division of two functions, the derivative is \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \).

Differentiation is widely used in solving problems that involve finding the maxima or minima of functions, and also in understanding how quantities change with time.

Indefinite Integrals

Indefinite integrals, also known as antiderivatives, are reversals of differentiation. Finding an indefinite integral involves calculating the original function from its derivative.

• General Form: The indefinite integral of a function \( f(x) \) is represented by \( \int f(x) \, dx \).
• Constants when integrating: Always add a constant \( C \) to your result because integration reverses differentiation. The derivative of a constant is zero.
• Basic Integral Rules: For example, the integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} + C \), for \( n eq -1 \).

Each integral represents a family of functions differing by a constant. Indefinite integrals are used in calculating areas under curves and solving differential equations.

Constant of Integration

The constant of integration is a crucial concept when dealing with indefinite integrals. When integrating a function, you should always add a constant to the result. This constant is often noted as \( C \).
Why is this constant important? Differentiation removes constants. If you had two functions, like \( f(x) = x^2 + 3 \) and \( g(x) = x^2 - 5 \), their derivatives are both \( 2x \). Integrating involves reversing this process. Therefore, when you integrate \( 2x \), you get \( x^2 + C \). Without specifying \( C \), you wouldn’t know the exact function.

• Real-world Impact: In physics, calculating velocity from acceleration involves using a constant of integration to account for initial velocity.
• Mathematical Impact: In solving differential equations, the constant allows representing all possible solutions.

The constant of integration ensures the completeness and accuracy of the mathematical model derived through integration.