#Critical Points of Implicit Relations

#Table of Contents

1. Introduction to Critical Points
2. Identifying Critical Points in Implicit Equations
3. Horizontal and Vertical Tangents
4. Worked Example
5. Practice Questions
6. Glossary
7. Summary and Key Takeaways

#Introduction to Critical Points

Critical points are crucial in determining the behavior of functions. They are points on the graph of a function where the derivative is zero or undefined. These points help in identifying local maxima, local minima, and points of inflection.

#Identifying Critical Points in Implicit Equations

Implicit equations can have critical points, which are defined similarly to those in explicit functions.

#Key Concepts

The application of first and second derivatives to classify the nature of points on a graph can be extended to implicit functions.

#Classifying Critical Points

Here is a summary of how the first and second derivatives are used to classify critical points:

#Horizontal and Vertical Tangents

To determine points where the tangent to an implicitly-defined curve is horizontal or vertical, we use the following conditions:

#Horizontal Tangents

The tangent line is horizontal at a point on the curve where: $$\frac{dy}{dx} = 0$$

#Vertical Tangents

The tangent line is vertical at a point on the curve where: $$\frac{dx}{dy} = 0$$

This is equivalent to a point where $$\frac{dy}{dx}$$ has a denominator equal to zero and a numerator not equal to zero. Recall: $$\frac{dx}{dy} = \frac{1}{\left(\frac{dy}{dx}\right)}$$

#Worked Example

#Example Problem

Consider the function $y = f(x)$ whose curve is given by the equation: $$2y^2 - 6 = y \sin 2x \quad \text{for} \quad y > 0$$

#(a) Finding Horizontal Tangents

To find where the tangent is horizontal, we need to solve for where the derivative is zero.

Differentiate both sides of the equation with respect to $x$: $$\frac{d}{dx}(2y^2 - 6) = \frac{d}{dx}(y \sin 2x)$$ Using the product rule and chain rule: $$4y \frac{dy}{dx} = \frac{dy}{dx} \sin 2x + y \cdot 2 \cos 2x$$

Rearrange to solve for $\frac{dy}{dx}$: $$4y \frac{dy}{dx} - \frac{dy}{dx} \sin 2x = 2y \cos 2x$$ $$\frac{dy}{dx}(4y - \sin 2x) = 2y \cos 2x$$ $$\frac{dy}{dx} = \frac{2y \cos 2x}{4y - \sin 2x}$$

Set this equal to zero to find the critical point: $$\frac{2y \cos 2x}{4y - \sin 2x} = 0$$ $$2y \cos 2x = 0$$

Ensure the denominator is not zero: $$4y - \sin 2x \neq 0$$

Solve $2y \cos 2x = 0$: $$\cos 2x = 0$$ $$2x = \frac{\pi}{2}, \frac{3\pi}{2}, \ldots$$ $$x = \frac{\pi}{4}, \frac{3\pi}{4}, \ldots$$

Given $0 \le x \le \frac{\pi}{2}$: $$x = \frac{\pi}{4}$$

Find $y$ using the original equation: $$2y^2 - 6 = y \sin \left(2 \cdot \frac{\pi}{4}\right)$$ $$2y^2 - 6 = y \sin \left(\frac{\pi}{2}\right)$$ $$2y^2 - 6 = y$$ $$2y^2 - y - 6 = 0$$ $$(2y + 3)(y - 2) = 0$$

Since $y > 0$, $y = 2$.

Verify the denominator condition: $$4(2) - \sin \left(2 \cdot \frac{\pi}{4}\right) = 8 - 1 = 7 \neq 0$$

The point where the tangent to the curve is horizontal is $\left(\frac{\pi}{4}, 2\right)$.

#(b) Classifying the Critical Point

To determine if the point $\left(\frac{\pi}{4}, 2\right)$ is a relative minimum, maximum, or neither, we examine the second derivative:

Write the first derivative: $$\frac{dy}{dx} = \frac{2y \cos 2x}{4y - \sin 2x}$$

Differentiate both sides with respect to $x$: $$\frac{d^2 y}{dx^2} = \frac{d}{dx} \left( \frac{2y \cos 2x}{4y - \sin 2x} \right)$$

Using the quotient rule: $$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$$ with $u = 2y \cos 2x$ and $v = 4y - \sin 2x$

Differentiating $v$: $$v = 4y - \sin 2x$$ $$v' = 4 \frac{dy}{dx} - 2 \cos 2x$$

Differentiating $u$ using the product rule: $$u = 2y \cos 2x$$ $$p = 2y, \quad q = \cos 2x$$ $$p' = 2 \frac{dy}{dx}, \quad q' = -2 \sin 2x$$ $$u' = p'q + p q'$$ $$u' = 2 \frac{dy}{dx} \cos 2x - 4y \sin 2x$$

Applying the quotient rule: $$\frac{d^2 y}{dx^2} = \frac{\left(2 \cos 2x \frac{dy}{dx} - 4y \sin 2x\right)(4y - \sin 2x) - (2y \cos 2x)(4 \frac{dy}{dx} - 2 \cos 2x)}{(4y - \sin 2x)^2}$$

Since $\frac{dy}{dx} = 0$ at the point: $$\frac{d^2 y}{dx^2} = \frac{\left(-8 \sin \left(2 \cdot \frac{\pi}{4}\right)\right)(8 - 1)}{(8 - 1)^2}$$ $$\frac{d^2 y}{dx^2} = \frac{-8}{49} = -\frac{56}{49}$$

The first derivative is zero, and the second derivative is negative at $\left(\frac{\pi}{4}, 2\right)$. Therefore, it is a relative maximum.

#Practice Questions

Practice Question

1. Given an implicitly defined curve $x^2 + y^2 = 1$:
   • Find the points where the tangent is horizontal.
   • Find the points where the tangent is vertical.

Practice Question

2. Consider the implicitly defined curve $x^3 + y^3 = 6xy$:
   • Determine the critical points on the curve.
   • Classify each critical point as a relative minimum, maximum, or neither.

#Glossary

• Critical Point: A point where the derivative of a function is zero or undefined.
• Horizontal Tangent: A tangent line that is horizontal, occurring where $\frac{dy}{dx} = 0$.
• Vertical Tangent: A tangent line that is vertical, occurring where $\frac{dx}{dy} = 0$.
• First Derivative: The rate of change of a function, indicating the slope of the tangent line.
• Second Derivative: The rate of change of the first derivative, indicating the concavity of the function.
• Point of Inflection: A point where the concavity of the function changes.

#Summary and Key Takeaways

• Implicit equations can have critical points where the derivative is zero or undefined.
• Horizontal tangents occur where $\frac{dy}{dx} = 0$.
• Vertical tangents occur where $\frac{dx}{dy} = 0$.
• The first and second derivatives help classify critical points into local minima, maxima, and points of inflection.
• The second derivative must change sign at a point for it to be a point of inflection.

#Key Takeaways

1. Critical points in implicit equations are found similarly to explicit functions.
2. Use the first derivative to find horizontal and vertical tangents.
3. The second derivative is essential for classifying the nature of critical points.
4. Practice differentiating implicit functions and applying the quotient and product rules.