Exploring Behaviors of Implicit Relations: Study Guide (Calculus AB 5.12)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

This page writes derivatives as dy/dx (or y′) and d²y/dx² (or y″). On an implicit curve both are usually expressions in x and y, so you need the coordinates of a point, not just its x-value, to evaluate them.

Applying derivative tests to curves

In Topics 5.4 to 5.7 you used f′ and f″ to describe the graph of y = f(x). Many curves are not written that way. A curve such as y² − 2xy + 2x² = 8 is an implicit relation: x and y are tied together by an equation, and for some x-values there are two y-values.

The good news is that the ideas carry over. Near any point where implicit differentiation gives a value for dy/dx (its denominator is not 0), a small piece of the curve is the graph of a function y(x). On that piece:

• dy/dx > 0 means y is increasing as x increases; dy/dx < 0 means y is decreasing;
• a sign change of dy/dx from + to − is a relative maximum of y; from − to + a relative minimum;
• d²y/dx² > 0 means the piece is concave up; d²y/dx² < 0 means concave down.

The only difference is the algebra. dy/dx and d²y/dx² come from implicit differentiation, so they contain y as well as x.

Critical points of an implicit relation

A critical point of an implicit relation is a point on the curve where dy/dx = 0 or dy/dx does not exist.

Implicit differentiation usually gives dy/dx as a fraction, dy/dx = N/D, where N and D are expressions in x and y. Then:

At a point on the curve	dy/dx is…	The tangent is…
N = 0 and D ≠ 0	0	horizontal
D = 0 and N ≠ 0	undefined (a nonzero number over 0)	vertical
N = 0 and D = 0	0/0, not determined	not decided by the formula; investigate another way

Finding the points. Setting N = 0 gives an equation such as y = 2x. That is a line, not a point. Solve it together with the curve’s equation to find the actual points on the curve. Then check D at each point.

Vertical tangents. These are critical points too, because dy/dx does not exist there. When N ≠ 0, the curve near such a point contains points just above it and just below it, so y has no relative maximum or minimum there; you do not classify them that way. Often the curve folds back at a vertical tangent (it fails the vertical-line test nearby), and then x reaches its greatest or least value there, as in Worked example 1.

When both N and D are 0. The formula gives 0/0, which tells you nothing (just like 0/0 in a limit). For example, on the curve y³ = x², implicit differentiation gives dy/dx = 2x/(3y²), which is 0/0 at the origin. The curve has a sharp point (a cusp) there. Do not call such a point a horizontal or vertical tangent without more evidence.

Increasing, decreasing and concavity

Because dy/dx depends on both x and y, its sign can change across the plane, not just along the x-axis. A useful habit: find where N and D are positive or negative, then combine.

For example, if dy/dx = −3x/y, then dy/dx > 0 exactly when x and y have opposite signs. So on any part of the curve in the second or fourth quadrant, y increases as x increases.

Second derivatives. Differentiate dy/dx again with respect to x. Every y becomes a function of x, so the chain rule produces dy/dx terms. The result is an expression in x, y and dy/dx. Substitute the known dy/dx (or its value at the point) to evaluate it.

At a horizontal tangent, dy/dx = 0. Every term containing dy/dx vanishes, so d²y/dx² is often a very short expression. That makes the second derivative test quick.

Worked example 1: find and classify critical points

Question. Consider the curve y² − 2xy + 2x² = 8.

(a) Show that dy/dx = (y − 2x)/(y − x). (b) Find every point where the tangent is horizontal and every point where it is vertical. (c) Use the second derivative to decide whether y has a relative maximum or a relative minimum at each horizontal tangent.

1. Differentiate (a). Differentiate each term with respect to x. The term −2xy needs the product rule: 2y·y′ − (2y + 2x·y′) + 4x = 0. Collect the y′ terms: y′(2y − 2x) = 2y − 4x. Divide by 2(y − x): dy/dx = (y − 2x)/(y − x).
2. Horizontal tangents (b). N = y − 2x = 0 gives y = 2x. Substitute into the curve: 4x² − 4x² + 2x² = 8, so x² = 4 and x = ±2. The points are (2, 4) and (−2, −4). Check D = y − x: it is 2 at (2, 4) and −2 at (−2, −4), both nonzero. So both are horizontal tangents.
3. Vertical tangents (b). D = y − x = 0 gives y = x. Substitute: x² − 2x² + 2x² = 8, so x² = 8 and x = ±2√2. The points are (2√2, 2√2) and (−2√2, −2√2). Check N = y − 2x: it is −2√2 and 2√2, both nonzero. So both are vertical tangents.
4. Second derivative (c). Use the quotient rule on y′ = (y − 2x)/(y − x), remembering that y depends on x: y″ = [(y′ − 2)(y − x) − (y − 2x)(y′ − 1)] / (y − x)². This is an expression in x, y and y′.
5. Evaluate at the horizontal tangents. There y′ = 0 and y − 2x = 0, so the second term in the numerator is 0 and y″ = −2(y − x)/(y − x)² = −2/(y − x). At (2, 4): y″ = −2/2 = −1 < 0. At (−2, −4): y″ = −2/(−2) = 1 > 0.
6. Conclude. At (2, 4), dy/dx = 0 and d²y/dx² < 0, so y has a relative maximum of 4 there. At (−2, −4), dy/dx = 0 and d²y/dx² > 0, so y has a relative minimum of −4 there.

Check. On the upper part of the curve, y = x + √(8 − x²). At x = 1.5, dy/dx ≈ 0.37 > 0; at x = 2.5, dy/dx ≈ −0.89 < 0. The sign change from + to − agrees with a maximum at x = 2. Figure 1 shows the same thing.

What about the vertical tangents? They are critical points because dy/dx does not exist. They are not maxima or minima of y. On this curve they are where x is greatest (2√2) and least (−2√2).

Worked example 2: a justification with the first derivative test

Question. The curve y³ + 3y = 6x − x² + 5 passes through (3, 2).

(a) Find dy/dx. (b) Show that the curve has a horizontal tangent at (3, 2) and no vertical tangents. (c) Justify that y has a relative maximum at x = 3, and say whether it is the absolute maximum of y on the curve. (d) Confirm with d²y/dx².

1. Check the point. 2³ + 3(2) = 14 and 6(3) − 9 + 5 = 14. The point is on the curve.
2. Differentiate (a). 3y²·y′ + 3y′ = 6 − 2x, so y′(3y² + 3) = 6 − 2x and dy/dx = (6 − 2x)/(3y² + 3) = 2(3 − x)/(3(y² + 1)).
3. Tangents (b). At (3, 2): N = 6 − 6 = 0 and D = 3(4) + 3 = 15 ≠ 0, so the tangent is horizontal. The denominator 3(y² + 1) is at least 3 for every y, so it is never 0: there are no vertical tangents.
4. Sign of dy/dx (c). The denominator is always positive, so dy/dx has the same sign as 3 − x. For x < 3, dy/dx > 0, so y is increasing; for x > 3, dy/dx < 0, so y is decreasing. dy/dx changes from positive to negative at x = 3, so y has a relative maximum at x = 3.
5. Absolute? The only point where dy/dx = 0 is at x = 3 (and dy/dx always exists). y increases for all x < 3 and decreases for all x > 3. So y = 2 is the absolute maximum value of y on the curve.
6. Second derivative (d). Differentiate y′(3y² + 3) = 6 − 2x with the product rule: y″(3y² + 3) + y′·6y·y′ = −2. At (3, 2), y′ = 0, so 15y″ = −2 and y″ = −2/15 < 0. The curve is concave down there, which agrees with a maximum.

Why this works. The left side y³ + 3y always increases as y increases, so each x gives exactly one y. Here the curve really is the graph of a function, even though you cannot easily solve for y. The derivative tests apply everywhere.

Answer sentence. “y has a relative maximum at x = 3 because dy/dx = 2(3 − x)/(3(y² + 1)) changes from positive to negative there; the denominator is always positive. Since this is the only critical point, y = 2 is the absolute maximum.”

Common misconceptions

• Stopping at a line. “y = 2x” is the set where dy/dx could be 0. Solve it with the curve’s equation to get points.
• Forgetting to check the other part of the fraction. N = 0 gives a horizontal tangent only if D ≠ 0 at that point.
• Calling 0/0 a horizontal tangent. When N and D are both 0, the formula says nothing. Investigate another way.
• Using only x to evaluate dy/dx. On most implicit curves one x has two points. You need both coordinates.
• Treating y as a constant when finding d²y/dx². y depends on x, so each y gives a y′ term by the chain rule.
• Ignoring vertical tangents. dy/dx not existing also makes a critical point.
• Writing “it has a maximum”. Name what has the maximum: “y has a relative maximum of 4 at x = 2”.
• Giving the sign of dy/dx without the reason. Show why the sign is what it is, for example “the denominator 3(y² + 1) is always positive”.

Where this leads

This topic completes Unit 5. You have now used derivatives to describe functions given by formulas, graphs, tables, contexts and implicit equations. If you need a refresher, see Topic 3.2, Implicit Differentiation and Topic 5.11, Solving Optimization Problems. Unit 6 begins with Topic 6.1, Exploring Accumulations of Change, where you turn from rates of change to the amounts that build up from them. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.