Exploring Behaviors of Implicit Relations
Introduced| English |
|---|
| implicit relations/ɪmˈplɪsɪt rɪˈleɪʃnz/ |
| implicit differentiation/ɪmˈplɪsɪt ˌdɪfəˌrenʃɪˈeɪʃn/ |
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.
Tangents on a circle
y = √(25 − x²) (upper half)
On $x^2+y^2=25$ the top and bottom have horizontal tangents; the left and right have vertical tangents.
A horizontal tangent on an implicit curve occurs where $\tfrac{dy}{dx}$...
Horizontal ⇔ slope $0$ ⇔ numerator of $\tfrac{dy}{dx}$ is zero.
For $x^2+y^2=25$ ($\tfrac{dy}{dx}=-\tfrac{x}{y}$), select all points with a horizontal tangent.
$\tfrac{dy}{dx}=0$ when $x=0$: the top and bottom points.
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.
A vertical tangent occurs where the ____ of $\tfrac{dy}{dx}$ is zero.
Denominator zero ⇒ slope undefined ⇒ vertical tangent.
Setting the denominator of $\tfrac{dy}{dx}$ to zero finds tangents that are...
Denominator zero ⇒ vertical tangent (numerator zero ⇒ horizontal).
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.
A candidate point must satisfy the original equation (lie on the curve) to count.
Check the point is actually on the curve, not just the slope condition.
To find concavity on an implicit curve you...
The implicit second derivative $\tfrac{d^2y}{dx^2}$ gives concavity by its sign.
For horizontal tangents set the numerator of $\tfrac{dy}{dx}$ to zero; for vertical tangents set the denominator to zero — don't swap them. And a point only qualifies if it actually lies on the curve: check it satisfies the original equation, not just the derivative condition.
Find the horizontal and vertical tangents of $x^2+y^2=25$.
- $2x+2y\tfrac{dy}{dx}=0\Rightarrow\tfrac{dy}{dx}=-\tfrac{x}{y}$.
- Horizontal ($\tfrac{dy}{dx}=0$): $x=0\Rightarrow(0,5),(0,-5)$.
- Vertical ($\tfrac{dy}{dx}$ undefined): $y=0\Rightarrow(5,0),(-5,0)$.
Analyze an implicit relation with implicit differentiation: $\tfrac{dy}{dx}=0$ (numerator zero) gives horizontal tangents; $\tfrac{dy}{dx}$ undefined (denominator zero) gives vertical tangents. Differentiate again implicitly for $\tfrac{d^2y}{dx^2}$ and read concavity from its sign. Always confirm points lie on the curve.