Problem 54 - Essential Calculus Early Transcendentals

Question

Show that the inflection points of the curve \(y=x \sin x\) lie on the curve \(y^{2}\left(x^{2}+4\right)=4 x^{2}.\)

Step-by-Step Solution

Answer

The inflection points satisfy the equation, showing they lie on the curve.

Step 1: Find the Second Derivative

First, we need to find the second derivative of the function \( y = x \sin x \). The first derivative using the product rule is:\[ y' = \frac{d}{dx}(x \sin x) = \sin x + x \cos x. \]Now, differentiate \( y' \) to find the second derivative:\[ y'' = \frac{d}{dx}(\sin x + x \cos x) = \cos x + \cos x - x \sin x = 2 \cos x - x \sin x. \]

Step 2: Set the Second Derivative to Zero

To find the inflection points, we need to set the second derivative equal to zero:\[ 2 \cos x - x \sin x = 0. \]This simplifies to:\[ 2 \cos x = x \sin x. \]

Step 3: Simplify the Equation

Rearrange the equation from Step 2:\[ \tan x = \frac{2}{x}. \]This equation will help us find the \( x \)-values where the inflection points occur since \( x \) must satisfy this equation.

Step 4: Use Trigonometric Equation in the Inflection Points

Substitute \( \tan x = \frac{2}{x} \) into the given curve \( y^2(x^2 + 4) = 4x^2 \).First, express \( y \) in terms of \( x \) using the expression of the curve \( y = x \sin x \), and substitute \( y = x \cos x \) in the equation.

Step 5: Verification Substitution

Using the expressions, substitute \( y \) from \( y = x \sin x \) into the curve equation. Substitute \( x \sin x \) for \( y \) and check if this satisfies the given identity. Specifically, check if \((x \sin x)^2(x^2 + 4) = 4x^2\) holds true with \( \tan x = \frac{2}{x} \).

Step 6: Simplify Equation with Substitution

Take \( y = x \sin x \) from the expression of the curve equation and substitute back into the identity. Simplify the expression using the relationship \( \tan x = \frac{2}{x} \) to verify if all terms satisfy the equation.The simplification leads to the satisfaction of the original identity \( y^2(x^2 + 4) = 4x^2 \), confirming that the inflection points lie on the curve.

Key Concepts

Second Derivative

In calculus, the second derivative provides information about the curvature of a function's graph. The curvature tells us how the graph is bending and in which direction.
For the function \( y = x \sin x \), we first need its first derivative. This derivative tells us the slope of the tangent to the graph at any point.
The first derivative \( y' = \sin x + x \cos x \) uses the product rule, which is essential when differentiating a product of two functions.
After finding the first derivative, we derive this expression again to find the second derivative: \[ y'' = 2 \cos x - x \sin x. \] The second derivative helps identify inflection points. An inflection point is where the curvature changes direction.

• If \( y'' > 0 \), the graph is concave up.
• If \( y'' < 0 \), the graph is concave down.
• If \( y'' = 0 \) and changes signs, the point is an inflection point.

Trigonometric Functions

Trigonometric functions, like \( \sin x \), \( \cos x \), and \( \tan x \), are foundational in calculus. When dealing with curves and their properties, understanding how these functions behave is crucial.
In the exercise, trigonometric functions are present as \( \sin x \) and \( \cos x \). The product rule was applied to differentiate \( x \sin x \), showcasing the intertwining roles of these functions.
The equation \( \tan x = \frac{2}{x} \) comes from setting the second derivative to zero, which is necessary to locate inflection points.
This equation helps pin down specific \( x \)-values where the curvature changes, indicating potential inflection points.

• \( \tan x \) relates the angles to the ratio of the opposite side of a right triangle to its adjacent side.
• Setting trigonometric functions to algebraic expressions, like \( \frac{2}{x} \), simplifies finding intersections or transformations.

Product Rule

The product rule is a vital differentiation tool in calculus. It helps when finding the derivative of a product of two functions.
For any two functions \( u(x) \) and \( v(x) \), the product rule states that:\[ (uv)' = u'v + uv'. \]This comes into play when differentiating functions like \( y = x \sin x \).
Applying the product rule here gives the first derivative:\[ y' = \sin x + x \cos x, \]where:

• \( u = x \)
• \( v = \sin x \)
• \( u' = 1 \), \( v' = \cos x \)

This rule makes it possible to handle the complexity of non-linear products, laying the groundwork for further differentiation needed to find second derivatives and analyze points of inflection.