Mathematics

Problem

Consider an implicitly defined curve in the

xy

–plane whose first derivative

\dfrac{dy}{dx}

is finite except possibly at isolated points. Which of the following conditions must hold at a critical point

(x_0, y_0)

of the curve? a)

\displaystyle \frac{dy}{dx}(x_0,y_0)=0

and

(x_0,y_0)

satisfies the curve’s equation b) Either

\displaystyle \frac{dy}{dx}(x_0,y_0)=0

\dfrac{dy}{dx}

is undefined at

(x_0,y_0)

, and

(x_0,y_0)

satisfies the curve’s equation c) Either

\displaystyle \frac{dy}{dx}(x_0,y_0)=0

\dfrac{dy}{dx}

is undefined at

(x_0,y_0)

, regardless of whether

(x_0,y_0)

lies on the curve d)

\displaystyle \frac{dy}{dx}

is undefined at

(x_0,y_0)

and

(x_0,y_0)

satisfies the curve’s equation

Intuitive explanation

A critical point on a curve is where the tangent line becomes vertical—meaning $\frac{dy}{dx}$ is undefined or infinite. The key insight is that implicit differentiation lets you find $\frac{dy}{dx}$ without solving for $y$ explicitly, and a vertical tangent occurs when the denominator of that derivative expression equals zero while the numerator doesn't. At such a point, the curve has a well-defined vertical direction, even though $\frac{dy}{dx}$ itself blows up.

Intuitive explanation

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