Problem 35 - University Calculus: Early Transcendentals

Question

Find all possible functions with the given derivative. a. \(y^{\prime}=-\frac{1}{x^{2}}\) b. \(y^{\prime}=1-\frac{1}{x^{2}}\) c. \(y^{\prime}=5+\frac{1}{x^{2}}\)

Step-by-Step Solution

Answer

a. \(y = -\frac{1}{x} + C\); b. \(y = x + \frac{1}{x} + C\); c. \(y = 5x - \frac{1}{x} + C\).

Step 1: Understanding the Problem

We need to find the antiderivative or integral of a given function. In simple terms, for each part, we are given the derivative of a function and need to find all functions (including constants) that have this derivative.

Step 2: Solving Part (a)

Given: \(y^{\prime}=-\frac{1}{x^{2}}\). We need to find the antiderivative. Integrate \(-\frac{1}{x^{2}}\) with respect to \(x\). The antiderivative of \(x^{-n}\) is \(\frac{x^{-n+1}}{-n+1}\). Here, \(-\frac{1}{x^{2}} = -x^{-2}\), so the integral is:\[ y = \int -x^{-2} \, dx = \int -x^{-2} \, dx = \int -x^{-2} \, dx = x^{-1} + C\] Therefore, \(y = -\frac{1}{x} + C\), where \(C\) is a constant.

Step 3: Solving Part (b)

Given: \(y^{\prime}=1-\frac{1}{x^{2}}\). Integrate the expression with respect to \(x\):\[ y = \int \left(1 - \frac{1}{x^{2}}\right) \, dx = \int 1 \, dx - \int x^{-2} \, dx \]The integral of \(1\) is \(x\), and the integral of \(-x^{-2}\) is \(-x^{-1}\). Therefore:\[ y = x + \frac{1}{x} + C\], where \(C\) is a constant.

Step 4: Solving Part (c)

Given: \(y^{\prime}=5+\frac{1}{x^{2}}\). Integrate the expression with respect to \(x\):\[ y = \int \left(5 + \frac{1}{x^{2}}\right) \, dx = \int 5 \, dx + \int x^{-2} \, dx \]The integral of \(5\) is \(5x\), and the integral of \(x^{-2}\) is \(-x^{-1}\). Thus, we have:\[ y = 5x - \frac{1}{x} + C\], where \(C\) is a constant.

Key Concepts

Integration

Integration is a key concept in calculus, often used to find the original function based on its derivative. This process is the reverse of differentiation.

Integration can be thought of as finding the accumulated area under a curve, commonly represented by the integral sign \( \int \). The function inside the integral is called the integrand, and the result of integrating is known as the antiderivative or integral.

To find the antiderivative, you determine what function could be differentiated to produce the given derivative. One primary rule for finding antiderivatives is: the integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \) and \( C \) represents an arbitrary constant that accounts for any vertical shifts in the function.
- For instance, when finding the antiderivative of \(-\frac{1}{x^2}\), reframe it as an integral of \(-x^{-2}\).

• Calculate: \( \int -x^{-2} \, dx = -\frac{1}{x} + C \).

Calculus

Calculus is the mathematical study of change, focusing on two main operations: differentiation and integration.

While differentiation looks at how functions change and finds the rate of change, integration sums up small pieces to find the total entity, often described as aggregation. These operations are inverses of each other and play fundamental roles in various fields such as physics, engineering, and economics.

In calculus, integration functions as a method to reverse-engineer a derivative to reconstruct the original function, including any constants that were lost during differentiation.
- Working through the given derivatives:

• For \( y' = 1 - \frac{1}{x^2} \), integrate each part separately.
• Integrating gives: \( y = x + \frac{1}{x} + C \).

This showcases calculus’s power to connect rates of change with accumulative measurements, solving problems across diverse disciplines.

Differential Equations

Differential equations involve an equation that relates a function to its derivatives. Solving these equations typically means finding the function which satisfies the given differential condition.

These equations can be simple, such as first-order, or highly complex, involving higher-order derivatives. They are used to model real-world phenomena, such as motion, growth, decay, and many other dynamically changing systems.

When dealing with simple first-order differential equations, the process usually requires integrating the derivative to find the general solution. For example,

• Given \( y' = 5 + \frac{1}{x^2} \), integrate to find: \( y = 5x - \frac{1}{x} + C \).

This specific solution reflects not merely a single function but a family of functions differing by the constant \( C \), illustrating how initial conditions can uniquely determine a solution out of the many possible ones informed by the differential equation.