Исследование свойств неявно заданных зависимостей

Analyzing curves that aren't functions

• Circles, ellipses, and other implicit relations 隐关系 fail the vertical-line test — they aren't $y=f(x)$.
• But the calculus tools still work through implicit differentiation 隐函数求导.
• You can find their tangent slopes, spot horizontal and vertical tangents, and even their concavity.
• This lesson applies Unit 3's implicit method to Unit 5's analysis questions.

Horizontal tangents: numerator zero

• From implicit differentiation you get $\tfrac{dy}{dx}$ as a fraction.
• A horizontal tangent occurs where $\tfrac{dy}{dx}=0$ — set the numerator to zero (with denominator $\neq0$).
• For $x^2+y^2=25$: $\tfrac{dy}{dx}=-\tfrac{x}{y}=0$ when $x=0$, i.e. at $(0,5)$ and $(0,-5)$ (top and bottom).
• These are the highest and lowest points of the curve.

Horizontal tangents where dy/dx = 0; vertical tangents where it is undefined

Vertical tangents: denominator zero

• A vertical tangent occurs where $\tfrac{dy}{dx}$ is undefined — set the denominator to zero (with numerator $\neq0$).
• For $x^2+y^2=25$: $-\tfrac{x}{y}$ is undefined when $y=0$, i.e. at $(5,0)$ and $(-5,0)$ (left and right).
• The tangent line is vertical there — the curve turns straight up and down.
• Numerator zero → horizontal; denominator zero → vertical.

Concavity: the implicit second derivative

• To get concavity, differentiate $\tfrac{dy}{dx}$ again, implicitly, to find $\tfrac{d^2y}{dx^2}$.
• Remember every $y$ still depends on $x$, so the chain rule reappears, and you substitute the known $\tfrac{dy}{dx}$.
• The sign of $\tfrac{d^2y}{dx^2}$ gives concave up/down at a point, just as before.
• It's more algebra, but the same ideas: $f''>0$ up, $f''<0$ down.