Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that the indicated function is a solution of the given differential equation, that is, substitute the indicated function for y to see that it produces an equality. $$\frac{d y}{d x}+\frac{x}{y}=0 ; y=\sqrt{1-x^{2}}$$

Step-by-Step Solution

Answer

The function does not satisfy the differential equation as expected.

Step 1: Differentiate the Function

Differentiate the given function \( y = \sqrt{1-x^2} \) with respect to \( x \). Use the chain rule for differentiation:\[ \frac{d}{dx}(1-x^2)^{-1/2} = -\frac{1}{2}(1-x^2)^{-3/2} (-2x) = \frac{x}{\sqrt{1-x^2}} \] Thus, \( \frac{dy}{dx} = \frac{x}{\sqrt{1-x^2}}. \)

Step 2: Substitute into the Differential Equation

Substitute \( \frac{dy}{dx} = \frac{x}{\sqrt{1-x^2}} \) and \( y = \sqrt{1-x^2} \) into the given differential equation:\[ \frac{dy}{dx} + \frac{x}{y} = \frac{x}{\sqrt{1-x^2}} + \frac{x}{\sqrt{1-x^2}} = \frac{x + x}{\sqrt{1-x^2}} = \frac{2x}{\sqrt{1-x^2}} \] The original equation is \( \frac{dy}{dx} + \frac{x}{y} = 0 \). Therefore, these equalities confirm that \( y = \sqrt{1-x^2} \) does not satisfy the differential equation as expected since both terms do not cancel each other.

Key Concepts

Chain Rule in Differentiation

The chain rule is a fundamental tool in calculus, essential for differentiating composite functions. When you have a function composed of one function inside another, the chain rule helps determine the derivative.
For example, if you have a function of the form \( y = g(f(x)) \), to find the derivative \( \frac{dy}{dx} \), you would use:

• Find the derivative of the outer function \( g \) with respect to \( f \), denoted as \( g' \).
• Then, multiply by the derivative of the inner function \( f \) with respect to \( x \), denoted as \( f'(x) \).

Essentially, the chain rule is expressed as:\[\frac{dy}{dx} = g'(f(x)) \cdot f'(x)\]In the solution for differentiating \( y = \sqrt{1-x^2} \), treated as \( (1-x^2)^{1/2} \), the chain rule was applied:

• The outer function \( g(z) = z^{1/2} \)'s derivative is \( \frac{1}{2}z^{-1/2} \).
• The inner function is \( f(x) = 1-x^2 \), with \( f'(x) = -2x \).

Together, applying these yields the derivative \( \frac{dy}{dx} = \frac{x}{\sqrt{1-x^2}} \). This demonstrates how the chain rule makes handling nested or composite functions straightforward.

Basics of Differentiation

Differentiation is a key concept in calculus, focusing on how to find the rate at which a function is changing at any point. It provides the slope of the tangent line to a function at a given point, indicating how the function behaves around that specific area.
When differentiating basic functions, certain rules and techniques are very handy:

• Power Rule: If \( y = x^n \), then \( \frac{dy}{dx} = nx^{n-1} \).
• Constant Rule: The derivative of any constant is zero.
• Sum/Difference Rule: The derivative of a sum or difference of functions is the sum or difference of their derivatives.

These rules simplify finding derivatives and allow for quick calculation. With more complex functions, like \( y = \sqrt{1-x^2} \), you may need to apply a combination of these rules along with the chain rule. Differentiation helps not only to understand how a function evolves but also to solve differential equations by finding these derivative values.

Understanding Mathematical Proof

A mathematical proof is a logical argument demonstrating the truth of a given statement. Proofs are crucial in mathematics as they establish results as facts based on previously verified truths and logical reasoning.
To understand the proof associated with proving that a function solves a differential equation, follow these stages:

• Start with the given differential equation and the proposed solution.
• Substitute the proposed solution into the equation.
• Simplify and manipulate both sides of the equation using algebraic techniques and differentiation as needed.

In our exercise, we initially differentiated the proposed solution \( y = \sqrt{1-x^2} \) to get \( \frac{dy}{dx} = \frac{x}{\sqrt{1-x^2}} \). This result was substituted back into the differential equation \( \frac{dy}{dx} + \frac{x}{y} = 0 \), expecting them to cancel to zero if the function was an actual solution.
The verification process highlighted discrepancies, indicating the need for further validation or review of substitutions, reinforcing how proofs are vital for identifying the accuracy and correctness of mathematical assertions.