Implicit & Related Rates

01Why implicit differentiation exists

Not every curve hands you $y$ solved in terms of $x$. The circle $x^2 + y^2 = 25$ defines $y$ only implicitly — and solving gives two messy branches, $y = \pm\sqrt{25 - x^2}$. Curves like $x^3 + y^3 = 6xy$ can't be solved for $y$ at all. Implicit differentiation sidesteps the solving entirely: differentiate both sides of the equation as-is, treating $y$ as an unnamed function of $x$, and extract $\frac{dy}{dx}$ algebraically at the end.

02The key move: y is a function of x

Everything reduces to one chain-rule observation: since $y$ depends on $x$, differentiating any expression in $y$ produces a trailing $\frac{dy}{dx}$:

$$\frac{d}{dx}\,y^2 = 2y\,\frac{dy}{dx} \qquad \frac{d}{dx}\,\sin y = \cos y\,\frac{dy}{dx} \qquad \frac{d}{dx}\,x y = y + x\frac{dy}{dx}$$

(That last one is the product rule — $xy$ is a product of two functions of $x$.) Compare $\frac{d}{dx}x^2 = 2x$ with $\frac{d}{dx}y^2 = 2y\,y'$: the only difference is the tag-along $y'$. Forgetting that tag-along is the implicit differentiation error.

03The procedure, worked

Find $\frac{dy}{dx}$ for $x^2 + y^2 = 25$. Differentiate both sides term by term:

$$2x + 2y\,\frac{dy}{dx} = 0 \quad\Longrightarrow\quad \frac{dy}{dx} = -\frac{x}{y}$$

The answer contains both $x$ and $y$ — normal and fine, since a given $x$ on a circle corresponds to two points with different slopes. At $(3, 4)$ the slope is $-\tfrac{3}{4}$; at $(3, -4)$ it's $+\tfrac{3}{4}$. General recipe: (1) differentiate both sides, tagging every $y$-derivative with $y'$; (2) collect all $y'$ terms on one side; (3) factor out $y'$ and divide. It's always linear in $y'$, so step 3 always works.

04Tangent lines to implicit curves

Once $\frac{dy}{dx}$ is in hand, tangent lines work exactly as before. For the curve $x^3 + y^3 = 6xy$ (the folium of Descartes) at $(3, 3)$: differentiating gives $3x^2 + 3y^2 y' = 6y + 6xy'$, so $y' = \dfrac{6y - 3x^2}{3y^2 - 6x} = \dfrac{2y - x^2}{y^2 - 2x}$. At $(3,3)$: $y' = \frac{6 - 9}{9 - 6} = -1$, and the tangent is $y = 3 - (x - 3) = -x + 6$. Verify the point satisfies the original equation before trusting anything — graders check, and so should you.

05From implicit to related rates

Now replace $x$ with time. In a related rates problem, several quantities depend on $t$ and are linked by an equation; differentiating that equation with respect to $t$ links their rates. The chain rule move is identical — every variable picks up a $\frac{d}{dt}$ tag:

$$A = \pi r^2 \quad\Longrightarrow\quad \frac{dA}{dt} = 2\pi r\,\frac{dr}{dt}$$

If a ripple's radius grows at $\frac{dr}{dt} = 2$ ft/s, then when $r = 10$ ft the area grows at $\frac{dA}{dt} = 2\pi(10)(2) = 40\pi$ ft²/s. One equation about quantities became one equation about speeds.

06The five-step method

1. Draw and label. Assign symbols to everything that changes; write given rates as derivatives with signs ($\frac{dx}{dt} = -3$ if $x$ shrinks).
2. Relate. Find one equation connecting the variables — Pythagorean theorem, similar triangles, area/volume formulas are the usual suspects.
3. Differentiate with respect to $t$ — before plugging in any numbers.
4. Substitute the snapshot values and known rates.
5. Solve and sanity-check the sign and units.

The cardinal sin is substituting numbers before differentiating: a value like $x = 6$ that holds only at one instant becomes a constant with derivative zero, silently deleting a term.

07The classic: sliding ladder

A 10-ft ladder leans against a wall; the base slides away at 1 ft/s. How fast is the top sliding down when the base is 6 ft out? Let $x$ = base distance, $y$ = height: $x^2 + y^2 = 100$. Differentiate: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$. At the snapshot $x = 6$, so $y = 8$, and $\frac{dx}{dt} = 1$:

$$2(6)(1) + 2(8)\frac{dy}{dt} = 0 \quad\Longrightarrow\quad \frac{dy}{dt} = -\frac{3}{4} \text{ ft/s}$$

The negative sign says the top moves down — the model reporting physics back to you. Note $y = 8$ came from the constraint equation, not the problem statement; deriving missing snapshot values is part of step 4.

08Cones, shadows, and similar triangles

Two more patterns cover most homework. Draining cones: water in a cone has $V = \frac{1}{3}\pi r^2 h$ with two changing variables — use similar triangles ($\frac{r}{h}$ = constant from the cone's dimensions) to eliminate $r$ before differentiating, leaving $V$ in terms of $h$ alone. Shadows and distances: a person walking from a lamppost creates similar triangles relating shadow length to distance; two cars driving apart hide a Pythagorean relation. In every case the geometry supplies step 2's equation — related rates is 20% calculus, 80% setting up the right picture.