For a curve defined by an equation F(x,y) = 0, you find dy/dx through implicit differentiation. Critical points occur where dy/dx = 0 (horizontal tangent) or where dy/dx is undefined (vertical tangent or cusp). To find the second derivative d2y/dx2 for an implicit relation, differentiate dy/dx implicitly again; the result is typically expressed in terms of x, y, and dy/dx. Use the sign of d2y/dx2 to determine concavity at a point, substituting the coordinates and the value of dy/dx at that point.
- Horizontal tangent on implicit curve: Occurs where dy/dx = 0; find by setting the numerator of dy/dx equal to zero (while the denominator is nonzero).
- Vertical tangent on implicit curve: Occurs where dy/dx is undefined; find by setting the denominator of dy/dx equal to zero (while the numerator is nonzero).
- Second derivative of implicit relation: d2y/dx2 is found by differentiating dy/dx implicitly; the result involves x, y, and dy/dx, so substitute all three at the point of interest.
- Implicit function theorem: Guarantees that an implicit relation can be treated locally as a function y = f(x) near a point where dy/dx is defined and the partial derivative with respect to y is nonzero.
For x^2 + y^2 = 25, find all points with horizontal tangents, find all points with vertical tangents, and determine the concavity at (3,4) using the second derivative.
| Feature | Condition on dy/dx | How to find it |
|---|
| Horizontal tangent | dy/dx = 0 | Set numerator of dy/dx = 0, check denominator is nonzero |
| Vertical tangent | dy/dx undefined | Set denominator of dy/dx = 0, check numerator is nonzero |
| Concavity | Sign of d2y/dx2 | Differentiate dy/dx implicitly; substitute x, y, and dy/dx at the point |