Implicit Differentiation and Critical Points of a Quadratic Curve

Problem Statement

Given the implicit equation:

\[x^2 + x y + y^2 = 12\]

Find:

1. The first derivative $ \frac{dy}{dx} $ using implicit differentiation.
2. Points where $ \frac{dy}{dx} = 0 $ to locate potential extrema.
3. The second derivative $ \frac{d^2 y}{dx^2} $ at these critical points to determine if they are local maxima, minima, or points of inflection.

Solution

Step 1: First Derivative Using Implicit Differentiation

Differentiate both sides with respect to $ x $:

\[\frac{d}{dx} \left( x^2 + x y + y^2 \right) = \frac{d}{dx} (12)\]

Applying the derivative rules term by term:

\[2x + y + x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0\]

Now, isolate $ \frac{dy}{dx} $:

\[(x + 2y) \frac{dy}{dx} = -2x - y\]

Thus,

\[\frac{dy}{dx} = \frac{-2x - y}{x + 2y}\]

Step 2: Finding Critical Points Where $ \frac{dy}{dx} = 0 $

To find the critical points, set $ \frac{dy}{dx} = 0 $:

\[\frac{-2x - y}{x + 2y} = 0\]

This implies:

\[-2x - y = 0 \Rightarrow y = -2x\]

Substitute $ y = -2x $ back into the original equation to solve for $ x $:

\[x^2 + x(-2x) + (-2x)^2 = 12\] \[x^2 - 2x^2 + 4x^2 = 12\] \[3x^2 = 12 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2\]

So, we have two critical points:

• For $ x = 2 $, $ y = -2(2) = -4 $
• For $ x = -2 $, $ y = -2(-2) = 4 $

Step 3: Confirming Validity of Critical Points

For each point, check $ x + 2y \neq 0 $ to ensure that $ \frac{dy}{dx} $ is defined.

• At $ (2, -4) $: $ x + 2y = 2 + 2(-4) = -6 \neq 0 $
• At $ (-2, 4) $: $ x + 2y = -2 + 2(4) = 6 \neq 0 $

Since $ x + 2y \neq 0 $ for both points, they are valid critical points.

Step 4: Second Derivative Calculation

To determine whether these points are maxima, minima, or inflection points, compute the second derivative $ \frac{d^2 y}{dx^2} $:

\[\frac{d^2 y}{dx^2} = \frac{d}{dx} \left( \frac{-2x - y}{x + 2y} \right)\]

Using the quotient rule with $ u = -2x - y $ and $ v = x + 2y $:

\[\frac{d^2 y}{dx^2} = \frac{(u'v - uv')}{v^2}\]

where:

• $ u’ = -2 - \frac{dy}{dx} $
• $ v’ = 1 + 2 \frac{dy}{dx} $

Substituting $ \frac{dy}{dx} = \frac{-2x - y}{x + 2y} $ and simplifying, we obtain:

\[\frac{d^2 y}{dx^2} = -\frac{6(x^2 + xy + y^2)}{(x + 2y)^3}\]

Step 5: Evaluate $ \frac{d^2 y}{dx^2} $ at Critical Points

1. At $ (2, -4) $: \(\frac{d^2 y}{dx^2} = -\frac{6((2)^2 + 2(-4) + (-4)^2)}{(-6)^3} = \frac{1}{3}\) Since $ \frac{d^2 y}{dx^2} > 0 $, this point is a local minimum.

2. At $ (-2, 4) $: \(\frac{d^2 y}{dx^2} = -\frac{6((-2)^2 + (-2)(4) + (4)^2)}{6^3} = -\frac{1}{3}\) Since $ \frac{d^2 y}{dx^2} < 0 $, this point is a local maximum.

Conclusion

The implicit curve defined by $ x^2 + x y + y^2 = 12 $ has:

• A local minimum at $ (2, -4) $
• A local maximum at $ (-2, 4) $

This analysis showcases how to use implicit differentiation and second derivatives to identify local extrema on curves defined by implicit equations.