5.12 - Exploring Behaviors of Implicit Relations

Overview

This lesson takes concepts you already know (increasing/decreasing, concavity, extrema) and applies them to implicitly defined curves. The one big difference: you have to substitute both the

x

and

y

values when evaluating derivatives.

Determining If a Curve Is Increasing or Decreasing

Check the sign of $\frac{dy}{dx}$ at the point.

If $\frac{dy}{dx} > 0$ , the curve is increasing

If $\frac{dy}{dx} < 0$ , the curve is decreasing

Example 1

Determine whether

3x^3 + 3 = \ln(4y^2)

is increasing or decreasing at

(-1, \frac{1}{2})

Step 1: differentiate implicitly with respect to $x$

9x^2 + 0 = \frac{1}{4y^2} \cdot 8y \cdot \frac{dy}{dx}

Step 2: simplify

9x^2 = \frac{8y}{4y^2} \cdot \frac{dy}{dx} \quad \Rightarrow \quad 9x^2 = \frac{2}{y} \cdot \frac{dy}{dx}

Step 3: evaluate at $(-1, \frac{1}{2})$

9(-1)^2 = \frac{2}{\frac{1}{2}} \cdot \frac{dy}{dx} \quad \Rightarrow \quad 9 = 4 \cdot \frac{dy}{dx} \quad \Rightarrow \quad \frac{dy}{dx} = \frac{9}{4}

Step 4: conclude

The curve is increasing at

(-1, \frac{1}{2})

because:

\left(\frac{dy}{dx}\right)_{(-1, \frac{1}{2})} = \frac{9}{4} > 0

Notation for Derivatives at a Point

When evaluating derivatives at specific points in implicit differentiation, use subscript notation:

\left(\frac{dy}{dx}\right)_{(x_0, y_0)}

That just means the derivative evaluated at the point

(x_0, y_0)

Determining Concavity

Check the sign of $\frac{d^2y}{dx^2}$ at the point.

If $\frac{d^2y}{dx^2} > 0$ , the curve is concave up

If $\frac{d^2y}{dx^2} < 0$ , the curve is concave down

Example 2

Determine the concavity of

x^2 = e^y

(-2, \ln 4)

Step 1: first derivative

2x = e^y \cdot \frac{dy}{dx} \quad \Rightarrow \quad \frac{dy}{dx} = \frac{2x}{e^y}

Step 2: second derivative. Use the quotient rule on

\frac{2x}{e^y}

\frac{d^2y}{dx^2} = \frac{2 \cdot e^y - 2x \cdot e^y \cdot \frac{dy}{dx}}{(e^y)^2}

Factor out

e^y

\frac{d^2y}{dx^2} = \frac{e^y\left(2 - 2x \cdot \frac{dy}{dx}\right)}{e^{2y}} = \frac{2 - 2x \cdot \frac{dy}{dx}}{e^y}

Step 3: evaluate at $(-2, \ln 4)$ . First get

\frac{dy}{dx}

there:

\frac{dy}{dx} = \frac{2(-2)}{e^{\ln 4}} = \frac{-4}{4} = -1

Now the second derivative:

\frac{d^2y}{dx^2} = \frac{2 - 2(-2)(-1)}{e^{\ln 4}} = \frac{2 - 4}{4} = -\frac{1}{2}

Step 4: conclude

The curve is concave down at

(-2, \ln 4)

because:

\left(\frac{d^2y}{dx^2}\right)_{(-2, \ln 4)} = -\frac{1}{2} < 0

Important: when finding the second derivative, you have to substitute the expression for $\frac{dy}{dx}$ from the first derivative before evaluating at the point.

Finding Extrema: The Second Derivative Test

To classify a point as a max, min, or neither:

Step 1: verify it’s a critical point. Check whether

\left(\frac{dy}{dx}\right)_{(a,b)} = 0

or does not exist.

Not a critical point → answer is neither

Is a critical point → go to Step 2

Step 2: apply the second derivative test. Evaluate

\frac{d^2y}{dx^2}

at the point.

$\frac{d^2y}{dx^2} > 0$ → minimum (concave up)

$\frac{d^2y}{dx^2} < 0$ → maximum (concave down)

$\frac{d^2y}{dx^2} = 0$ → inconclusive

Example 3

Does

y^3 - y = 2x^2

have a maximum, minimum, or neither at

(0, 1)

Step 1: first derivative

3y^2 \cdot \frac{dy}{dx} - \frac{dy}{dx} = 4x

Factor:

\frac{dy}{dx}(3y^2 - 1) = 4x \quad \Rightarrow \quad \frac{dy}{dx} = \frac{4x}{3y^2 - 1}

Verify it’s a critical point at

(0, 1)

\left(\frac{dy}{dx}\right)_{(0,1)} = \frac{4(0)}{3(1)^2 - 1} = \frac{0}{2} = 0 \quad ✓

The first derivative is zero, so it’s a critical point. On to Step 2.

Step 2: second derivative. Quotient rule on

\frac{4x}{3y^2 - 1}

\frac{d^2y}{dx^2} = \frac{4(3y^2 - 1) - 4x \cdot 6y \cdot \frac{dy}{dx}}{(3y^2 - 1)^2}

Evaluate at

(0, 1)

, where

\frac{dy}{dx} = 0

\frac{d^2y}{dx^2} = \frac{4(3 - 1) - 4(0)(6)(1)(0)}{(3 - 1)^2} = \frac{8 - 0}{4} = 2

Step 3: conclude

The curve has a minimum at

(0, 1)

because:

$\left(\frac{dy}{dx}\right)_{(0,1)} = 0$

$\left(\frac{d^2y}{dx^2}\right)_{(0,1)} = 2 > 0$

The positive second derivative means the curve is concave up at the critical point, which makes it a minimum.

5.12 - Exploring Behaviors of Implicit Relations

Finding Critical Points of an Implicit Relation

In Example 3 the point was given. When it isn’t, you have to find the critical points yourself. A critical point of an implicit relation is a point on the curve where

\frac{dy}{dx} = 0

\frac{dy}{dx}

does not exist. Since

\frac{dy}{dx}

usually has both

x

and

y

in it, you need full points

(x, y)

, not just

x

-values.

Where $\frac{dy}{dx} = 0$ : set the numerator equal to zero, then substitute that into the original equation to find the actual points on the curve. The tangent line is horizontal there.

Where $\frac{dy}{dx}$ doesn’t exist: set the denominator equal to zero, then substitute that into the original equation too. The tangent line is usually vertical there.

Example 4

Find all critical points of

x^2 + xy + y^2 = 27

. Then classify the ones where

\frac{dy}{dx} = 0

Step 1: differentiate implicitly with respect to

x

2x + \left(x\frac{dy}{dx} + y\right) + 2y\frac{dy}{dx} = 0

\frac{dy}{dx}(x + 2y) = -(2x + y) \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{2x + y}{x + 2y}

Step 2: find where

\frac{dy}{dx} = 0

The numerator is zero when

2x + y = 0

, so

y = -2x

. Substitute into the original equation:

x^2 + x(-2x) + (-2x)^2 = 27 \quad \Rightarrow \quad 3x^2 = 27 \quad \Rightarrow \quad x = \pm 3

Since

y = -2x

, the points are

(3, -6)

and

(-3, 6)

. The denominator

x + 2y

-9

and

9

there, not zero, so

\frac{dy}{dx} = 0

at both.

Step 3: find where

\frac{dy}{dx}

doesn’t exist

The denominator is zero when

x + 2y = 0

, so

x = -2y

. Substitute:

(-2y)^2 + (-2y)y + y^2 = 27 \quad \Rightarrow \quad 3y^2 = 27 \quad \Rightarrow \quad y = \pm 3

Since

x = -2y

, the points are

(-6, 3)

and

(6, -3)

. The numerator isn’t zero there, so

\frac{dy}{dx}

doesn’t exist, and the tangent lines are vertical.

Step 4: second derivative. Use the quotient rule on

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

\frac{d^2y}{dx^2} = -\frac{\left(2 + \frac{dy}{dx}\right)(x + 2y) - (2x + y)\left(1 + 2\frac{dy}{dx}\right)}{(x + 2y)^2}

This second derivative is a relation in

x

y

, and

\frac{dy}{dx}

, which is normal with implicit differentiation. At both horizontal-tangent points,

\frac{dy}{dx} = 0

and

2x + y = 0

, so it simplifies to:

\frac{d^2y}{dx^2} = -\frac{2(x + 2y) - 0}{(x + 2y)^2} = -\frac{2}{x + 2y}

Step 5: conclude

(3, -6)

\left(\frac{d^2y}{dx^2}\right)_{(3, -6)} = -\frac{2}{-9} = \frac{2}{9} > 0

, so the curve has a minimum there.

(-3, 6)

\left(\frac{d^2y}{dx^2}\right)_{(-3, 6)} = -\frac{2}{9} < 0

, so the curve has a maximum there.

The curve is a tilted ellipse. All four points are critical points:

(3, -6)

and

(-3, 6)

have horizontal tangents (the lowest and highest points), and

(-6, 3)

and

(6, -3)

have vertical tangents (the farthest left and right).

Why the Second Derivative Test Works Here

The logic is identical to explicit functions:

First Derivative	Second Derivative	Shape	Result
$\frac{dy}{dx} = 0$	$\frac{d^2y}{dx^2} > 0$	Concave up ∪	Minimum
$\frac{dy}{dx} = 0$	$\frac{d^2y}{dx^2} < 0$	Concave down ∩	Maximum
$\frac{dy}{dx} = 0$	$\frac{d^2y}{dx^2} = 0$	Indeterminate	Inconclusive

Note: the first derivative test is a pain with implicit relations, because testing values on either side of a critical point means you’d need both the

x

and

y

coordinates of nearby points, which isn’t practical. The second derivative test only needs one point.

Summary of Justification Statements

Increasing: the curve is increasing at

(a, b)

because

\left(\frac{dy}{dx}\right)_{(a,b)} > 0

Decreasing: the curve is decreasing at

(a, b)

because

\left(\frac{dy}{dx}\right)_{(a,b)} < 0

Concave up: the curve is concave up at

(a, b)

because

\left(\frac{d^2y}{dx^2}\right)_{(a,b)} > 0

Concave down: the curve is concave down at

(a, b)

because

\left(\frac{d^2y}{dx^2}\right)_{(a,b)} < 0

Minimum: there’s a minimum at

(a, b)

because

\left(\frac{dy}{dx}\right)_{(a,b)} = 0

and

\left(\frac{d^2y}{dx^2}\right)_{(a,b)} > 0

Maximum: there’s a maximum at

(a, b)

because

\left(\frac{dy}{dx}\right)_{(a,b)} = 0

and

\left(\frac{d^2y}{dx^2}\right)_{(a,b)} < 0

Neither: neither a max nor a min at

(a, b)

because

\left(\frac{dy}{dx}\right)_{(a,b)} \neq 0

, so it isn’t a critical point

Problem-Solving Strategy

Differentiate implicitly with respect to $x$

Simplify the expression for $\frac{dy}{dx}$ (solve for it if you’ll need the second derivative)

Substitute the point $(x_0, y_0)$ , and remember you need both coordinates

Evaluate the numerical result

Interpret the sign (positive, negative, or zero)

Write the conclusion with justification

Key Takeaways

Nothing new conceptually: same ideas as before, just applied to implicit curves

Both coordinates needed: substitute both $x$ and $y$ when evaluating

Subscript notation: $\left(\frac{dy}{dx}\right)_{(a,b)}$ means the derivative evaluated at $(a, b)$

Finding critical points: set the numerator of $\frac{dy}{dx}$ equal to zero (horizontal tangents) or the denominator equal to zero (vertical tangents), then substitute back into the original equation to get the points

Increasing: $\frac{dy}{dx} > 0$ at the point

Decreasing: $\frac{dy}{dx} < 0$ at the point

Concave up: $\frac{d^2y}{dx^2} > 0$ at the point

Concave down: $\frac{d^2y}{dx^2} < 0$ at the point

Second derivative: substitute the $\frac{dy}{dx}$ expression before evaluating

Extrema require two checks: first derivative equals 0, then check the sign of the second derivative

Use the second derivative test: the first derivative test is impractical for implicit relations

Always justify: state the derivative value and its sign as your reason

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