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Calculus BC · Unit 5 · 5.12
AP Calculus · Analytical Applications of Differentiation
Exploring Behaviors of Implicit Relations
The derivative contains $y$ — and that is exactly what distinguishes one branch from another.
key idea$\tfrac{dy}{dx}=0\ \text{(numerator)}\Rightarrow\text{horizontal};\quad \text{denominator}=0\Rightarrow\text{vertical}$
Reading the graph
An implicit relation is usually not a function — a vertical line can meet it more than once. Its derivative depends on both $x$ and $y$, and that second variable is what tells the branches apart.
upper branch of $x^{2}+y^{2}=25$
locating special tangents on $x^{2}+y^{2}=25$, where $\tfrac{dy}{dx}=-\tfrac{x}{y}$
| numerator $=0$ | denominator $=0$ | both $=0$ |
|---|
| condition | $x=0$ | $y=0$ | $x=y=0$ |
| points | $(0,\pm5)$ | $(\pm5,0)$ | not on the curve |
| verdict | horizontal tangent | vertical tangent | indeterminate — inspect further |
| To find | Set | Then | Watch for |
|---|
| horizontal tangents | numerator of $\tfrac{dy}{dx}$ $=0$ | substitute back into the relation | solutions not on the curve |
| vertical tangents | denominator $=0$ | substitute back into the relation | same |
| the slope at a point | — | substitute both $x$ and $y$ | using $x$ alone is not enough |
| concavity | — | differentiate $\tfrac{dy}{dx}$ again, then use the relation to simplify | the messy intermediate form |
Every row ends the same way: substitute back into the original relation. A candidate that does not satisfy it is not a point on the curve.
Implicit relationAn equation in $x$ and $y$ that is not solved for $y$ — and often cannot be. It usually fails the vertical line test.
BranchA piece of the relation that is a function. The circle splits into an upper and a lower branch.
Horizontal tangentWhere the numerator of $\tfrac{dy}{dx}$ vanishes and the denominator does not.
Vertical tangentWhere the denominator vanishes and the numerator does not — the slope is undefined.
Worked examples
EX 1
Horizontal and vertical tangents
$x^{2}+y^{2}=25$Find all points on $x^{2}+y^{2}=25$ with horizontal or vertical tangents.
1
Differentiate implicitly
$2x+2y\dfrac{dy}{dx}=0$, so $\dfrac{dy}{dx}=-\dfrac{x}{y}$.
2
Horizontal: numerator zero
$x=0$. Substituting into the relation: $y^{2}=25$, so $y=\pm5$.
3
Vertical: denominator zero
$y=0$. Substituting: $x^{2}=25$, so $x=\pm5$.
4
State the points
Horizontal tangents at $(0,5)$ and $(0,-5)$; vertical at $(5,0)$ and $(-5,0)$.
5
Sanity check
Those are the top, bottom, right and left of the circle — exactly where you would expect flat and vertical tangents.
EX 2
Why the slope needs $y$
two branchesFind the slope of $x^{2}+y^{2}=25$ at $x=3$.
1
Recognise the ambiguity
$x=3$ meets the circle twice: at $y=4$ and $y=-4$.
2
Compute both slopes
At $(3,4)$: $-\tfrac34$. At $(3,-4)$: $-\tfrac{3}{-4}=+\tfrac34$.
3
State the answer
The question is incomplete — the slope depends on which branch. Two answers exist.
4
Explain the structure
Because $\tfrac{dy}{dx}=-\tfrac{x}{y}$ contains $y$, one formula serves both branches and distinguishes them.
5
The general point
An implicit derivative containing $y$ is not a defect. It is what makes a single expression describe a curve that is not a function.
EX 3
Discard the impossible candidate
check the relationFor $x^{2}+xy+y^{2}=3$, where are the horizontal tangents?
1
Differentiate
$2x+y+x\dfrac{dy}{dx}+2y\dfrac{dy}{dx}=0$, so $\dfrac{dy}{dx}=-\dfrac{2x+y}{x+2y}$.
2
Set the numerator to zero
$2x+y=0$, i.e. $y=-2x$.
3
Substitute into the relation
$x^{2}+x(-2x)+(-2x)^{2}=x^{2}-2x^{2}+4x^{2}=3x^{2}=3$, so $x=\pm1$.
4
Find the matching $y$
$x=1\Rightarrow y=-2$; $x=-1\Rightarrow y=2$.
5
Verify the denominator
At $(1,-2)$: $x+2y=1-4=-3\neq0$. ✓ Had it been zero, both parts would vanish and the point would need separate investigation.
EX 4
Second derivative on a relation
concavityFor $x^{2}+y^{2}=25$, determine the concavity of the upper branch.
1
Start from the first derivative
$\dfrac{dy}{dx}=-\dfrac{x}{y}$.
2
Differentiate with the quotient rule
$\dfrac{d^{2}y}{dx^{2}}=-\dfrac{y-x\dfrac{dy}{dx}}{y^{2}}$.
3
Substitute $\tfrac{dy}{dx}$
$=-\dfrac{y+\dfrac{x^{2}}{y}}{y^{2}}=-\dfrac{y^{2}+x^{2}}{y^{3}}$.
4
Use the relation to simplify
$x^{2}+y^{2}=25$, so $\dfrac{d^{2}y}{dx^{2}}=-\dfrac{25}{y^{3}}$.
5
Read the sign
On the upper branch $y\gt 0$, so $y^{3}\gt 0$ and the second derivative is negative — concave down, as the picture shows.
Watch out
Always substitute candidates back into the relation
Setting the numerator to zero produces a line of candidates. Only the points also satisfying the original relation are on the curve.
The slope needs both coordinates
$x$ alone can identify two points with different slopes. Give the full point, or expect two answers.
Numerator zero and denominator zero simultaneously
Both vanishing gives $\tfrac00$ and no conclusion. Such a point needs separate investigation — it is often a self-intersection.
Simplify the second derivative using the relation
$-\dfrac{x^{2}+y^{2}}{y^{3}}$ becomes $-\dfrac{25}{y^{3}}$ only because the relation is substituted back. Without that step the sign is far harder to read.