Implicit Differentiation and a Vertical Tangent
A curve is defined implicitly by x² + xy + y² = 7. A student computes dy/dx and wants to find all points on the curve where the tangent line is horizontal and where it is vertical. A classmate says 'vertical tangents happen where dy/dx is undefined, which is just where the denominator is zero - done.' The student worries this is incomplete because a zero denominator could also signal a cusp or a point not actually on the curve, and because the algebra must be checked against the original equation. She sets up the derivative carefully and plans to verify each candidate point satisfies x² + xy + y² = 7.
1. Use implicit differentiation to find dy/dx as a function of x and y.
Read the worked answer
Differentiate both sides of x² + xy + y² = 7 with respect to x, treating y as a function of x:
2x + (y + x·dy/dx) + 2y·dy/dx = 0,
using the product rule on xy. Group the dy/dx terms:
x·dy/dx + 2y·dy/dx = −2x − y
dy/dx (x + 2y) = −(2x + y)
dy/dx = −(2x + y)/(x + 2y), valid where x + 2y ≠ 0.
2. Find all points on the curve where the tangent is horizontal, and all points where it is vertical. Verify each candidate lies on the curve.
Read the worked answer
Horizontal tangents: dy/dx = 0 requires the numerator 2x + y = 0, i.e. y = −2x. Substitute into x² + xy + y² = 7: x² + x(−2x) + 4x² = x² − 2x² + 4x² = 3x² = 7 → x = ±√(7/3). With y = −2x: points are (√(7/3), −2√(7/3)) and (−√(7/3), 2√(7/3)). Both satisfy the curve equation.
Vertical tangents: need denominator x + 2y = 0, i.e. x = −2y, provided the numerator is nonzero there. Substitute: 4y² + (−2y)y + y² = 4y² − 2y² + y² = 3y² = 7 → y = ±√(7/3), x = ∓2√(7/3). Points: (−2√(7/3), √(7/3)) and (2√(7/3), −√(7/3)). At these, numerator 2x + y = 2(−2y) + y = −3y ≠ 0, so dy/dx is genuinely infinite - true vertical tangents, not cusps.
The classmate's shortcut is incomplete: one must confirm the numerator is nonzero (else the limit of dy/dx could be finite, giving a cusp or smooth point) and confirm the points lie on the curve.