AP® Calculus AB · Unit 5 · Topic 5.12
Exploring Behaviors of Implicit Relations
On a curve defined implicitly, you find peaks and valleys by setting dy/dx = 0 and solving, then confirming with the second derivative, since the curve can loop and fail the vertical line test while still having a clear highest or lowest point.
A relation like x2 + xy + y2 = 12 traces a closed curve, something like a tilted oval. It is not a function of x, because some x-values give two y-values. But it still rises to a top and dips to a bottom, the same way a circle does. Those high and low points are exactly where the tangent line is horizontal, dy/dx = 0, regardless of whether the whole curve passes a vertical line test.
To find dy/dx, differentiate both sides of the equation with respect to x, treating y as a function of x and using the product and chain rules wherever y appears. For x2 + xy + y2 = 12, this gives 2x + y + x(dy/dx) + 2y(dy/dx) = 0, so dy/dx = -(2x+y)/(x+2y). Horizontal tangents need the numerator zero: 2x + y = 0, so y = -2x. Vertical tangents need the denominator zero instead. Substituting y = -2x back into the original equation pins down the actual points: x2 + x(-2x) + (-2x)2 = 12 simplifies to 3x2 = 12, so x = -√4 = -2 or x = 2, giving the points (-2,4) and (2,-4).
Finding where dy/dx = 0 only locates candidates, points where the curve is momentarily flat. This step connects back to the earlier work on critical points for ordinary functions, and forward to the second derivative test: implicit differentiation applied twice gives d2y/dx2, and its sign at a candidate point tells you whether that point is a peak, a valley, or neither.
Key terms
- implicit differentiation
- Differentiating both sides of an equation relating x and y, applying the chain rule to y as an unknown function of x.
- horizontal tangent
- A point on a curve where dy/dx = 0, a candidate for a highest or lowest point.
- vertical tangent
- A point on a curve where dy/dx is undefined because the denominator of the derivative expression equals zero.
- second derivative test
- Using the sign of d2y/dx2 at a candidate point to decide whether it is a maximum, a minimum, or neither.
Watch it explained
- How it worksHow do you find the horizontal and vertical tangents on an implicitly defined curve?
- Worked exampleFind the highest and lowest points on the curve x2 + xy + y2 = 12.
- Common mistakeWhy isn't dy/dx equal to zero enough to guarantee a maximum on an implicit curve?
- Going furtherWhat does the second derivative look like for an implicitly defined relation?
Check yourself
End of topic.