AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/5/5-12)
Unit 5 · Topic 5.12
5.12 Exploring Behaviors of Implicit Relations
Curves defined implicitly, like x² + xy + y² = 12, can be analyzed just like functions. Implicit differentiation gives dy/dx in terms of x and y. Horizontal tangents come from a zero numerator, vertical tangents from a zero denominator, and d²y/dx² gives concavity.
Key terms
- implicit differentiation
- dy/dx
- horizontal tangent
- vertical tangent
- second derivative
Horizontal and vertical tangents
Write dy/dx as a single fraction N/D, where N and D are expressions in x and y.
- Horizontal tangent: N = 0 and D ≠ 0. The slope is 0.
- Vertical tangent: D = 0 and N ≠ 0. The slope is undefined because the line is vertical.
- If both N = 0 and D = 0 at a point, the formula can't decide. You need more analysis.
- In every case, the point must also be on the curve. Solve the condition together with the original equation.
Finding the points
Setting N = 0 gives a relationship between x and y, like y = −2x. Substitute that into the original equation and solve to get the actual points. Then check that D isn't 0 at those points.
Points where dy/dx = 0 or dy/dx doesn't exist are called critical points of the relation. They're where a curve can have a top, bottom, leftmost or rightmost point.
Second derivatives and concavity
Differentiate dy/dx again with respect to x. The result can contain x, y and dy/dx. Substitute the expression for dy/dx (or its value at the point) to finish.
At a point with a horizontal tangent, dy/dx = 0, which often simplifies d²y/dx² a lot. Then the Second Derivative Test works just as for functions: d²y/dx² < 0 means the curve has a local maximum there (the top of a hump), and d²y/dx² > 0 means a local minimum.
Tangent lines and approximations
Tangent line equations work as usual: plug the point into dy/dx for the slope and use point-slope form. You can also use the tangent line to approximate nearby points on the curve, as in 4.6, and use the sign of d²y/dx² to tell whether the estimate is too high or too low.
Leftmost, rightmost, highest and lowest points
On a closed curve like an ellipse or a tilted oval, the highest and lowest points have horizontal tangents, and the leftmost and rightmost points have vertical tangents. For x² + xy + y² = 12 (the first example below), the top and bottom of the oval are (−2, 4) and (2, −4), and its far left and far right are (−4, 2) and (4, −2).
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Horizontal and vertical tangents
For the curve x² + xy + y² = 12, find all points with a horizontal tangent and all points with a vertical tangent.
Show the solutionHide the solution
- Step 1: Differentiate: 2x + y + x·dy/dx + 2y·dy/dx = 0, so dy/dx = −(2x + y)/(x + 2y).
- Step 2: Horizontal: 2x + y = 0, so y = −2x. Substitute: x² − 2x² + 4x² = 3x² = 12, so x = ±2. Points (2, −4) and (−2, 4). Check the denominator: x + 2y = 2 − 8 = −6 ≠ 0 and −2 + 8 = 6 ≠ 0.
- Step 3: Vertical: x + 2y = 0, so x = −2y. Substitute: 4y² − 2y² + y² = 3y² = 12, so y = ±2. Points (−4, 2) and (4, −2). Check the numerator: 2x + y = −8 + 2 = −6 ≠ 0 and 8 − 2 = 6 ≠ 0.
Answer: Horizontal tangents at (2, −4) and (−2, 4). Vertical tangents at (−4, 2) and (4, −2).
- Example 2
Concavity at a horizontal tangent
The curve x² + y² − 6x = 16 has dy/dx = (3 − x)/y. Show that it has a horizontal tangent at (3, 5), and decide whether the curve has a local maximum or minimum there.
Show the solutionHide the solution
- Step 1: Check the point: 9 + 25 − 18 = 16. Yes.
- Step 2: At (3, 5), dy/dx = 0/5 = 0, so the tangent is horizontal.
- Step 3: Quotient rule: d²y/dx² = [(−1)·y − (3 − x)·dy/dx]/y².
- Step 4: At (3, 5) with dy/dx = 0: d²y/dx² = (−5 − 0)/25 = −1/5 < 0.
Answer: The curve has a horizontal tangent at (3, 5) and is concave down there, so (3, 5) is a local maximum of the curve. (It's the top of the circle with center (3, 0) and radius 5.)
- Example 3
Trap: when the numerator and denominator are both 0
For the curve y² = x³, dy/dx = 3x²/(2y). Does the curve have a horizontal tangent at the origin?
Show the solutionHide the solution
- Step 1: At (0, 0), the numerator 3x² = 0, but the denominator 2y = 0 too. The formula gives 0/0, so it can't decide.
- Step 2: Look at the curve directly: y = ±x^(3/2) for x ≥ 0. Both halves start at the origin and flatten out there, meeting in a sharp point called a cusp.
- Step 3: So you can't say “horizontal tangent because the numerator is 0.” The derivative formula is undefined at that point.
Answer: You can't conclude it from dy/dx: the formula gives 0/0 at the origin, where the curve has a cusp. To claim a horizontal or vertical tangent from the dy/dx formula alone, you need one part zero and the other nonzero.
Common mistakes
- Setting dy/dx = 0 and stopping without finding the actual points on the curve.
- Forgetting to check that the denominator isn't also 0 at a horizontal-tangent point.
- Leaving dy/dx inside d²y/dx² when a value at the point is asked for.
On the exam
- This is a common free-response question type: given an implicit curve and dy/dx, find points with horizontal or vertical tangents, write a tangent line, then use d²y/dx² to classify a point.
- If the problem gives dy/dx, use it. You don't need to re-derive it unless asked to show it.
Connected topics
Videos
Check yourself
4 questions on 5.12 Exploring Behaviors of Implicit Relations. Pick an answer to see if you got it, and why.
Consider the curve given by x² − xy + y² = 12. It can be shown that dy/dx = (y − 2x)/(2y − x).
Described curve
At which points does the curve have a vertical tangent line?
Let P be the point on the curve with x-coordinate 1 and a positive y-coordinate. What is the slope of the line tangent to the curve at P?
What is the value of d²y/dx² at the point (2, 4)?
Consider the curve y³ − 3y = x. At which points does the curve have a vertical tangent line?
0 of 4 answered