Chapter 8 — Implicit Differentiation and Related Rates

8.1 The Problem with Solving for $y$

Every function you have differentiated so far arrived in a tidy, obedient form: $y = f(x)$, with $y$ alone on the left and a recipe in $x$ on the right. You feed the rules from Chapter 7 into that recipe and out comes $dy/dx$. But most of the curves that appear in geometry, physics, and economics do not come pre-solved for $y$. They come as relations — equations that tangle $x$ and $y$ together and refuse to be untangled. Consider three:

• $x^2 + y^2 = 25$ — a circle of radius $5$;
• $x^3 + y^3 = 6xy$ — the folium of Descartes, a looped curve from 1638;
• $\sin(xy) = x + y$ — a transcendental tangle with no name and no hope of simplification.

For the circle we could fight our way to $y = \pm\sqrt{25 - x^2}$, but the $\pm$ is a confession: the relation is really two functions glued together, an upper semicircle and a lower one, and solving forces us to pick a branch before we have even started. For the folium the algebra is a cubic in $y$ — solvable, but monstrous. For $\sin(xy) = x + y$ there is simply no way to isolate $y$ using elementary functions at all.

And yet each of these curves has a perfectly good tangent line at almost every point. A tangent line has a slope. That slope is a derivative. So the derivative must exist — the only thing missing is a method to compute it without first solving for $y$. That method is the subject of the first half of this chapter, and it turns out to be nothing more than the chain rule, used with a little courage.

The Key Insight. When an equation defines $y$ implicitly as a function of $x$, that hidden function is still differentiable. You can compute $\frac{dy}{dx}$ by differentiating both sides of the equation as they stand, treating $y$ as an unknown function $y(x)$ and applying the chain rule every time $y$ appears. You never need to solve for $y$ first. The price is that the answer comes out in terms of both $x$ and $y$ — and that is exactly the right amount of information.

This chapter has two halves that share one engine. The first half — implicit differentiation — differentiates a static relation between $x$ and $y$. The second half — related rates — takes the same trick and runs it through time: when several quantities change together, differentiating their connecting equation with respect to $t$ relates all their rates at once. Both are pure applications of the chain rule. That is the recurring lesson of this chapter, and it ties directly to a theme of the whole book: calculus is the mathematics of change, and the chain rule is how change propagates through a chain of dependencies.

8.2 The Method of Implicit Differentiation

Suppose a relation is written $F(x, y) = 0$ (everything moved to one side). The procedure is three steps:

1. Differentiate both sides with respect to $x$. Treat the equation as an identity that holds for every point on the curve, so the two sides have equal derivatives.
2. Apply the chain rule through $y$. Wherever $y$ appears, it stands for the hidden function $y(x)$, so $\frac{d}{dx}[g(y)] = g'(y)\cdot\frac{dy}{dx}$. Each $y$-term sprouts a factor of $\frac{dy}{dx}$.
3. Solve algebraically for $\frac{dy}{dx}$. Collect all $\frac{dy}{dx}$ terms on one side, factor it out, and divide.

The whole technique lives in step 2. A term like $y^3$ does not differentiate to $3y^2$; it differentiates to $3y^2 \cdot y'$, because $y$ is a function and the chain rule demands its derivative ride along. Watch this carefully on the first example.

Example 1: The circle

Start with $x^2 + y^2 = 25$ and differentiate both sides with respect to $x$:

$$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25).$$

The first term is $2x$. The third term is $0$ (the derivative of a constant). The middle term is the one that matters: $\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}$ by the chain rule. So

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

The slope at any point $(x,y)$ on the circle is $-x/y$. At $(3,4)$ it is $-\tfrac34$; at $(4,3)$ it is $-\tfrac43$; at $(-3,4)$ it is $+\tfrac34$. One formula serves the entire curve, upper and lower branch alike — no $\pm$ choice required, because the point you plug in already knows which branch it is on.

Verifying against the explicit method. On the upper semicircle $y = \sqrt{25 - x^2}$, the chain rule gives directly $\frac{dy}{dx} = \frac{-x}{\sqrt{25 - x^2}}$. At $x = 3$ this is $\frac{-3}{\sqrt{16}} = -\tfrac34$. The two methods agree, exactly as they must — implicit differentiation is not a different theory of calculus, just a more economical bookkeeping.

Geometric Intuition. The formula $\frac{dy}{dx} = -\frac{x}{y}$ has a beautiful reading: the tangent to a circle is perpendicular to the radius. The radius to $(x,y)$ has slope $y/x$; the tangent has slope $-x/y$; their product is $-1$, the signature of perpendicular lines. Implicit differentiation just rediscovered a fact Euclid proved geometrically — but now we have it as a derivative, available on any curve, not just circles.

Check Your Understanding. Use implicit differentiation to find $\frac{dy}{dx}$ for the ellipse $\frac{x^2}{9} + \frac{y^2}{16} = 1$, and evaluate it at $(0, 4)$.

Answer

Differentiate: $\frac{2x}{9} + \frac{2y}{16}\,\frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -\frac{16x}{9y}$. At $(0,4)$ this is $0$ — the tangent at the top of the ellipse is horizontal, exactly as the picture demands.

Example 2: The folium of Descartes

Now a curve you cannot easily solve: $x^3 + y^3 = 6xy$. Differentiate term by term, watching the right side, which needs the product rule and the chain rule:

$$3x^2 + 3y^2\,y' = 6\!\left(y + x\,y'\right).$$

The left side: $x^3 \to 3x^2$ and $y^3 \to 3y^2 y'$. The right side: $6xy$ is a product of $x$ and $y(x)$, so its derivative is $6(1\cdot y + x \cdot y') = 6y + 6x\,y'$. Now gather the $y'$ terms:

$$3y^2 y' - 6x\,y' = 6y - 3x^2 \quad\Longrightarrow\quad (3y^2 - 6x)\,y' = 6y - 3x^2.$$

Divide and simplify by $3$:

$$y' = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}.$$

The point $(3,3)$ lies on the curve, since $27 + 27 = 54 = 6\cdot 9$. There the slope is

$$y'(3,3) = \frac{2(3) - 9}{9 - 2(3)} = \frac{6 - 9}{9 - 6} = \frac{-3}{3} = -1.$$

The tangent at $(3,3)$ has slope $-1$. We computed a tangent to a cubic curve that we never solved and never could solve cleanly — that is the power of the method in one stroke.

Historical Note. René Descartes introduced this folium (Latin folium, "leaf") in 1638 as a challenge to Pierre de Fermat, betting that Fermat's tangent method could not handle it. Fermat found the tangents anyway. The irony is that neither man had the chain rule: implicit differentiation as we now teach it crystallized only after Newton and Leibniz, decades later. The technique you just used in four lines was, in 1638, a research-grade open problem.

Common Pitfall. Many students differentiate $y^3$ as $3y^2$ and stop, forgetting the trailing $y'$. The tell-tale symptom is an answer for $\frac{dy}{dx}$ that contains no $y$ at all, or an equation you cannot solve for $\frac{dy}{dx}$ because the factor is missing. Remember: every term containing $y$ produces a $\frac{dy}{dx}$ when you differentiate with respect to $x$. If $y'$ does not appear in your differentiated equation, you forgot the chain rule somewhere.

8.3 Higher-Order Implicit Derivatives

To find $y''$, simply differentiate the expression for $y'$ a second time — still treating $y$ as a function of $x$, and substituting the known $y'$ wherever it reappears.

Return to the circle $x^2 + y^2 = 25$, where $y' = -x/y$. Apply the quotient rule:

$$y'' = \frac{d}{dx}\!\left(-\frac{x}{y}\right) = -\frac{(1)\,y - x\,y'}{y^2} = -\frac{y - x\,y'}{y^2}.$$

Now substitute $y' = -x/y$:

$$y'' = -\frac{y - x\!\left(-\frac{x}{y}\right)}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{y^2 + x^2}{y^3} = -\frac{25}{y^3},$$

using $x^2 + y^2 = 25$ at the last step. At $(3,4)$, $y'' = -\frac{25}{64} < 0$. The negative sign confirms what the eye already knows: the top of the circle is concave down. Higher derivatives obtained implicitly are not just algebra — they read off concavity from a curve you never solved.

Check Your Understanding. For $x^2 + y^2 = 25$, what is the sign of $y''$ on the lower semicircle, where $y < 0$? What does that say about concavity?

Answer

$y'' = -25/y^3$. For $y < 0$, $y^3 < 0$, so $y'' = -25/(\text{negative}) > 0$. The lower semicircle is concave up — it cups upward, exactly as you would draw it.

8.4 Tangent Lines and Vertical Tangents

Once you have $y'$ as a function of $x$ and $y$, the tangent line at a point $(x_0, y_0)$ on the curve is the usual point-slope line:

$$y - y_0 = y'(x_0, y_0)\,(x - x_0).$$

Implicit differentiation also exposes something the explicit form hides: vertical tangents. A tangent is vertical where the slope blows up — that is, where the denominator of $y'$ vanishes while the numerator does not.

For the circle, $y' = -x/y$ has denominator $y$, so vertical tangents occur where $y = 0$: the points $(\pm 5, 0)$, the left and right edges of the circle. There the curve is momentarily vertical, and indeed $y = \pm\sqrt{25 - x^2}$ has an infinite slope there — the explicit formula misbehaves at exactly the points the implicit formula handles cleanly.

For the folium, $y' = \frac{2y - x^2}{y^2 - 2x}$ has vertical tangents where $y^2 = 2x$ (and $2y \ne x^2$), tracing out the rightmost extent of the famous loop. Locating such features by hand is one of the everyday uses of implicit differentiation in curve sketching, which we take up in Chapter 9.

Math Major Sidebar — When does a relation actually define a function? (The Implicit Function Theorem). We have been blithely assuming that $F(x,y) = 0$ secretly defines a differentiable function $y(x)$ near any point of interest. That is not always true — and the formal third tier of rigor is the theorem that says exactly when it is. The Implicit Function Theorem states: if $F$ has continuous partial derivatives near a point $(x_0, y_0)$ on the curve $F = 0$, and if $\frac{\partial F}{\partial y}(x_0, y_0) \ne 0$, then near that point the relation does define a unique differentiable function $y = g(x)$, and its derivative is $$\frac{dy}{dx} = -\frac{\partial F / \partial x}{\partial F / \partial y}.$$ Check it on the circle, $F = x^2 + y^2 - 25$: here $\frac{\partial F}{\partial x} = 2x$ and $\frac{\partial F}{\partial y} = 2y$, so the formula gives $\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}$ — precisely our §8.2 result. The condition $\frac{\partial F}{\partial y} \ne 0$ is the rigorous reason vertical tangents are special: they are exactly the points where $\frac{\partial F}{\partial y} = 2y = 0$, the theorem's hypothesis fails, and no single-valued $y(x)$ exists. We prove this theorem in Chapter 30; for now, know that implicit differentiation rests on a genuine existence guarantee, not wishful thinking.

8.5 Inverse Functions via Implicit Differentiation

Implicit differentiation is the cleanest route to the derivative of an inverse function. If $y = f^{-1}(x)$, then by definition $x = f(y)$. Differentiate that equation implicitly with respect to $x$:

$$1 = f'(y)\cdot \frac{dy}{dx} \quad\Longrightarrow\quad \frac{dy}{dx} = \frac{1}{f'(y)}.$$

In closed form, $(f^{-1})'(x) = \dfrac{1}{f'\!\big(f^{-1}(x)\big)}$. This single identity generates every inverse-trigonometric and inverse-hyperbolic derivative. Watch it produce three of the most important formulas in calculus.

Example: the derivative of $\arctan$

Let $y = \arctan x$, which means $x = \tan y$ for $y \in (-\tfrac{\pi}{2}, \tfrac{\pi}{2})$. Differentiate implicitly:

$$1 = \sec^2 y \cdot y' \quad\Longrightarrow\quad y' = \frac{1}{\sec^2 y} = \cos^2 y.$$

Now convert back to $x$. From $\tan y = x$, a right triangle with opposite $x$ and adjacent $1$ has hypotenuse $\sqrt{1 + x^2}$, so $\cos y = \frac{1}{\sqrt{1+x^2}}$ and $\cos^2 y = \frac{1}{1+x^2}$. Therefore

$$\frac{d}{dx}\arctan x = \frac{1}{1 + x^2}.$$

This is a foundational formula — it is the antiderivative behind the appearance of $\pi$ in Chapter 14's integral $\int_0^1 \frac{dx}{1+x^2} = \frac{\pi}{4}$.

Example: the derivative of $\arcsin$

Let $y = \arcsin x$, so $x = \sin y$ with $y \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$. Then $1 = \cos y \cdot y'$. On that interval $\cos y \ge 0$, so $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$, giving

$$\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}.$$

Example: rediscovering $(\ln x)' = 1/x$

Let $y = \ln x$, so $x = e^y$. Differentiating implicitly, $1 = e^y \cdot y' = x\, y'$, hence $y' = 1/x$. This is how the derivative of the natural logarithm is derived from the derivative of $e^x$ — the two are inverse functions, and implicit differentiation flips one into the other.

Computational Note. Implicit differentiation of $x = f(y)$ is the standard derivation route for all inverse derivatives, but there is one subtlety the algebra hides: you must pick the correct sign when you take a square root, and that choice is dictated by the range of the inverse function. For $\arcsin$, the range $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ forces $\cos y \ge 0$, so the positive root is correct. Choose the wrong branch and you get a derivative with the wrong sign. The geometry of the inverse function's domain is doing real work here.

8.6 Logarithmic Differentiation

Implicit differentiation powers a labor-saving trick called logarithmic differentiation. When $y$ is a tangle of products, quotients, and powers, take the natural log of both sides first — logarithms turn products into sums and powers into multipliers — and only then differentiate implicitly.

Starting from $y = f(x)$, write $\ln y = \ln f(x)$ and differentiate. The left side, by the chain rule, is $\frac{y'}{y}$:

$$\frac{y'}{y} = \frac{d}{dx}\big[\ln f(x)\big] \quad\Longrightarrow\quad y' = y \cdot \frac{d}{dx}\big[\ln f(x)\big].$$

Example: the genuinely hard derivative $y = x^x$

The function $x^x$ has a variable in both the base and the exponent, so neither the power rule nor the exponential rule applies. Logarithms rescue it. Take $\ln y = x\ln x$ and differentiate the right side with the product rule:

$$\frac{y'}{y} = \ln x + x\cdot\frac{1}{x} = \ln x + 1 \quad\Longrightarrow\quad y' = x^x(\ln x + 1).$$

There is no honest way to get this without logarithmic differentiation; the trick is not a convenience but a necessity.

Example: taming a product-quotient-power monster

$$y = \frac{x^2\sqrt{x+1}}{(2x+3)^3}.$$

Differentiating this directly would require the quotient rule wrapped around a product rule wrapped around two chain rules. Instead, take logs and let them flatten the structure:

$$\ln y = 2\ln x + \tfrac12\ln(x+1) - 3\ln(2x+3).$$

Differentiate each simple term:

$$\frac{y'}{y} = \frac{2}{x} + \frac{1}{2(x+1)} - \frac{6}{2x+3}.$$

Multiply through by $y$ to finish. The hard work — three nested differentiation rules — collapsed into three easy derivatives of logarithms. This is the everyday payoff of logarithmic differentiation.

8.7 Related Rates: Change Running Through Time

We now turn the same machinery on a moving target. In a related rates problem, two or more quantities change as time passes, and they are bound together by an equation. You are told the rate of one and asked for the rate of another. The connecting equation does the relating; differentiating it with respect to time converts a relationship between quantities into a relationship between their rates.

This is where the recurring theme of the book becomes vivid: calculus is the mathematics of change, and related rates is its most direct expression. A geometric truth like "the ladder has fixed length" or "the tank is a cone" holds at every instant. Differentiate that truth with respect to $t$ and you learn how the moving pieces must move in concert.

The engine is again the chain rule. If a volume $V$ depends on a height $h$, and $h$ in turn depends on time $t$, then

$$\frac{dV}{dt} = \frac{dV}{dh}\cdot\frac{dh}{dt}.$$

Every variable in a related-rates equation is secretly a function of $t$, so every term picks up a time-derivative factor when differentiated — exactly as every $y$-term picked up a $y'$ in §8.2. Implicit differentiation and related rates are the same act aimed at different variables.

The Key Insight. A related-rates problem is implicit differentiation with respect to time. Write the equation that relates the quantities at every instant, differentiate it with respect to $t$ (so each variable contributes its own rate via the chain rule), and only then substitute the numbers for the instant you care about. The geometry holds always; the numbers hold only now.

The seven-step method

Every related-rates problem yields to the same disciplined procedure. We will follow it explicitly in each worked example below.

1. Read and identify. What is changing? What rate is given? What rate is asked?
2. Draw a diagram and label the changing quantities with letters, not numbers.
3. Record the given and wanted rates in symbols (e.g., given $\frac{dx}{dt} = 1$, want $\frac{dy}{dt}$).
4. Find the relating equation from the geometry or physics — Pythagoras, similar triangles, a volume formula, a trig ratio.
5. Differentiate both sides with respect to $t$. Apply the chain rule to every variable.
6. Substitute the known values for the chosen instant and the known rates.
7. Solve for the unknown rate, then check units and sign.

Common Pitfall. The single most destructive error in related rates is substituting numbers before differentiating (skipping the discipline of steps 4–6). If you set $x = 6$ before differentiating, then $x$ has become a constant, $\frac{dx}{dt} = 0$, and you have thrown away the very motion the problem is about. Differentiate while every quantity is still a variable; freeze the numbers only at step 6.

8.8 Worked Example: The Sliding Ladder

A $10$-foot ladder leans against a vertical wall. The base is pulled away from the wall at $1$ ft/s. How fast is the top sliding down the wall when the base is $6$ ft from the wall?

Steps 1–3 (identify, draw, label). Let $x$ be the distance from the wall to the base and $y$ the height of the top up the wall. Picture a right triangle whose hypotenuse is the fixed ladder. We are given $\frac{dx}{dt} = 1$ ft/s and want $\frac{dy}{dt}$ at the instant $x = 6$.

Step 4 (relating equation). The ladder length is constant, so Pythagoras holds at every instant:

$$x^2 + y^2 = 10^2 = 100.$$

Step 5 (differentiate with respect to $t$). Both $x$ and $y$ are functions of time:

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

Step 6 (substitute the instant). When $x = 6$, the relation $36 + y^2 = 100$ gives $y = 8$. With $\frac{dx}{dt} = 1$:

$$\frac{dy}{dt} = -\frac{6}{8}\,(1) = -\frac{3}{4}\ \text{ft/s}.$$

Step 7 (interpret). The negative sign means the top is descending: the ladder's top slides down at $\tfrac34$ ft/s when the base is $6$ ft out. The units check ($\text{ft} \div \text{ft} \times \text{ft/s} = \text{ft/s}$), and the sign matches the physics — pull the base out, the top falls.

Warning. Push this example to the limit and something dramatic happens. The formula $\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}$ shows that as the base nears the wall's full reach ($x \to 10$, so $y \to 0$), the downward speed $\frac{x}{y}\frac{dx}{dt} \to \infty$. The top would have to fall infinitely fast. This is not a paradox of calculus; it is a sign that the idealized model (rigid ladder, base pulled at constant speed all the way to the end) stops describing reality — a real ladder leaves the wall first. Related rates faithfully reports when a model breaks, if you read the answer honestly.

8.9 Worked Example: The Conical Tank

Water flows into an inverted cone (vertex down) at $2$ m³/min. The cone has height $10$ m and top radius $4$ m. How fast is the water level rising when the water is $5$ m deep?

Steps 1–4 (set up). Let $h$ be the water depth and $r$ the surface radius of the water, and let $V$ be the water's volume. We are given $\frac{dV}{dt} = 2$ m³/min and want $\frac{dh}{dt}$ when $h = 5$. The volume of a cone is

$$V = \frac{1}{3}\pi r^2 h.$$

This relation has two changing lengths, $r$ and $h$. We must eliminate one before differentiating, or we will be stuck with two unknown rates. Similar triangles do it: the water cone is geometrically similar to the full tank, so $\frac{r}{h} = \frac{4}{10} = \frac{2}{5}$, giving $r = \frac{2h}{5}$. Substitute:

$$V = \frac{1}{3}\pi\left(\frac{2h}{5}\right)^2 h = \frac{1}{3}\pi\cdot\frac{4h^2}{25}\cdot h = \frac{4\pi}{75}\,h^3.$$

Step 5 (differentiate). Now $V$ depends on the single variable $h$:

$$\frac{dV}{dt} = \frac{4\pi}{75}\cdot 3h^2\cdot\frac{dh}{dt} = \frac{4\pi h^2}{25}\,\frac{dh}{dt}.$$

Step 6 (substitute the instant). At $h = 5$ with $\frac{dV}{dt} = 2$:

$$2 = \frac{4\pi (5)^2}{25}\,\frac{dh}{dt} = \frac{4\pi \cdot 25}{25}\,\frac{dh}{dt} = 4\pi\,\frac{dh}{dt}.$$

Step 7 (solve and interpret). Therefore $\frac{dh}{dt} = \frac{1}{2\pi} \approx 0.159$ m/min — about $16$ cm per minute. Notice the structure of the answer: because $\frac{dh}{dt} = \frac{25}{4\pi h^2}\frac{dV}{dt}$, the level rises slower as the water gets deeper, since the cross-section widens with $h$. The same inflow spreads over a larger surface, so the level creeps up more slowly. The mathematics encodes the everyday observation that a funnel fills fast at the tip and slowly at the brim.

Real-World Application — Chemical and civil engineering. Tank fill- and drain-rate calculations are precisely this problem, and engineers solve them constantly: a process tank's level sensor reads $h$, but the operator cares about volume $V$, and the conversion rate $\frac{dV}{dt} \leftrightarrow \frac{dh}{dt}$ depends entirely on the tank's geometry. The same calculus sizes the spillway of a dam (how fast does the reservoir rise during a flood?) and calibrates the float gauge in a conical hopper. Get the similar-triangles substitution wrong and a tank overflows.

Check Your Understanding. In the same conical tank, is $\frac{dh}{dt}$ larger when $h = 2$ m or when $h = 8$ m (same inflow $2$ m³/min)?

Answer

$\frac{dh}{dt} = \frac{25}{4\pi h^2}\cdot 2 = \frac{25}{2\pi h^2}$, which decreases as $h$ grows. So the level rises faster at $h = 2$ (smaller surface) than at $h = 8$ (larger surface). At $h=2$: $\frac{dh}{dt} = \frac{25}{8\pi} \approx 0.99$ m/min; at $h=8$: $\frac{25}{128\pi}\approx 0.062$ m/min — a sixteenfold difference, since the surface area scales as $h^2$.

8.10 Worked Example: The Approaching (Separating) Cars

Two cars leave the same intersection at the same moment. Car A drives east at $60$ mph; Car B drives north at $80$ mph. How fast is the straight-line distance between them changing $30$ minutes later?

Steps 1–4 (set up). Let $x$ be Car A's eastward distance, $y$ Car B's northward distance, and $D$ the distance between the cars. The roads are perpendicular, so

$$D^2 = x^2 + y^2.$$

We square the distance deliberately — differentiating $D^2$ avoids the square root and the messier algebra it brings. We are given $\frac{dx}{dt} = 60$ and $\frac{dy}{dt} = 80$ mph and want $\frac{dD}{dt}$ at $t = 0.5$ h.

Step 5 (differentiate).

$$2D\,\frac{dD}{dt} = 2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt}.$$

Step 6 (substitute the instant). At $t = 0.5$ h, $x = 60(0.5) = 30$ and $y = 80(0.5) = 40$, so $D = \sqrt{30^2 + 40^2} = \sqrt{2500} = 50$ mi. Then

$$2(50)\,\frac{dD}{dt} = 2(30)(60) + 2(40)(80) = 3600 + 6400 = 10000.$$

Step 7 (solve). $\frac{dD}{dt} = \frac{10000}{100} = 100$ mph. The cars separate at $100$ mph. The clean number is no accident: because the velocity vectors are perpendicular, the separation speed is the magnitude of the combined velocity, $\sqrt{60^2 + 80^2} = 100$ — a $6$–$8$–$10$ right triangle scaled by ten. Related rates and vector geometry agree.

8.11 Worked Example: The Walking Shadow

A $6$-foot-tall person walks away from an $18$-foot-tall streetlamp at $4$ ft/s. (a) How fast does the length of the shadow grow? (b) How fast does the tip of the shadow move?

Steps 1–4 (set up). Let $x$ be the distance from the lamp's base to the person and $s$ the length of the shadow (from the person's feet to the shadow's tip). The lamp, the person, and the ground form two similar triangles — the large triangle from the lamp top to the shadow tip, and the small triangle from the person's head to the shadow tip:

$$\frac{s}{6} = \frac{x + s}{18}.$$

Cross-multiplying: $18 s = 6(x + s) = 6x + 6s$, so $12 s = 6 x$, giving the clean relation $s = \tfrac12 x$. (The lamp is exactly three times the person's height, which is why the numbers land so neatly.)

Step 5–7, part (a). Differentiate $s = \tfrac12 x$ with respect to $t$:

$$\frac{ds}{dt} = \frac{1}{2}\,\frac{dx}{dt} = \frac{1}{2}(4) = 2\ \text{ft/s}.$$

The shadow lengthens at $2$ ft/s. Notice this rate is constant — it does not depend on where the person is, because $s$ is a fixed fraction of $x$.

Part (b). The shadow's tip sits at distance $x + s$ from the lamp. Since $x + s = x + \tfrac12 x = \tfrac32 x$,

$$\frac{d}{dt}(x + s) = \frac{3}{2}\,\frac{dx}{dt} = \frac{3}{2}(4) = 6\ \text{ft/s}.$$

The tip races along at $6$ ft/s — faster than the $4$ ft/s person who casts it.

Geometric Intuition. Why does the shadow's tip outrun the person? The tip's motion is the sum of two motions: the person walking forward ($4$ ft/s) and the shadow stretching out behind them ($2$ ft/s). The geometry of the light cone amplifies the person's pace into the tip's pace. Stand under a tall lamp at night and watch your own shadow tip skate ahead of you — you are watching a related rate in real time.

8.12 Worked Example: The Filling Trough

A trough is $10$ ft long. Its cross-section is an isosceles triangle, point down, $2$ ft deep with a $2$-ft top. Water flows in at $1$ ft³/min. How fast is the water level rising when the water is $1$ ft deep?

Steps 1–4 (set up). Let $h$ be the water depth and $V$ the water volume. At depth $h$, the water's cross-section is a smaller similar triangle. The full triangle is $2$ ft deep and $2$ ft across, so its width equals its depth; by similarity, at depth $h$ the surface width is also $b = h$. The cross-sectional area is

$$A = \frac{1}{2}\,b\,h = \frac{1}{2}\,h\cdot h = \frac{h^2}{2},$$

and since the trough has constant length $10$,

$$V = 10\,A = 5h^2.$$

Step 5 (differentiate). $\dfrac{dV}{dt} = 10h\,\dfrac{dh}{dt}.$

Steps 6–7 (substitute and solve). At $h = 1$ with $\frac{dV}{dt} = 1$:

$$1 = 10(1)\,\frac{dh}{dt} \quad\Longrightarrow\quad \frac{dh}{dt} = 0.1\ \text{ft/min} = 1.2\ \text{in/min}.$$

The level rises at one-tenth of a foot per minute. As with the cone, the rise slows as the water deepens (since $\frac{dh}{dt} = \frac{1}{10h}\frac{dV}{dt}$), because a wider surface at greater depth spreads the same inflow more thinly.

8.13 Worked Example: A Changing Angle

A bird watcher stands $50$ m from the base of a tall tree and watches a bird climb straight up the trunk at $3$ m/s. How fast is the watcher's angle of elevation $\theta$ increasing when the bird is $50$ m up?

Steps 1–4 (set up). Let $h$ be the bird's height and $\theta$ the angle of elevation. The horizontal distance is fixed at $50$ m, so

$$\tan\theta = \frac{h}{50}.$$

We are given $\frac{dh}{dt} = 3$ m/s and want $\frac{d\theta}{dt}$ when $h = 50$.

Step 5 (differentiate). Differentiating with respect to $t$, the left side needs the chain rule for $\tan\theta$:

$$\sec^2\theta\,\frac{d\theta}{dt} = \frac{1}{50}\,\frac{dh}{dt}.$$

Step 6 (substitute the instant). When $h = 50$, the triangle has equal legs ($50$ and $50$), so $\theta = \tfrac{\pi}{4}$ and $\sec^2\theta = (\sqrt{2})^2 = 2$. Then

$$2\,\frac{d\theta}{dt} = \frac{1}{50}(3) = \frac{3}{50}.$$

Step 7 (solve). $\frac{d\theta}{dt} = \frac{3}{100} = 0.03$ rad/s. The angle opens at $0.03$ radians per second (about $1.7°$/s) at that instant. This is the calculus behind tracking: a camera following a rising rocket, or radar locking onto a climbing aircraft, must rotate at exactly this rate to keep the target centered.

Real-World Application — Physics and tracking systems. Angle-rate problems govern any device that must aim at a moving object: a telescope tracking a satellite, an anti-aircraft gun, a solar panel following the sun, an automated camera on a sports field. The required angular velocity $\frac{d\theta}{dt}$ depends on both the target's speed and its current position — which is why tracking mounts must constantly recompute their slew rate. The relation $\sec^2\theta\,\frac{d\theta}{dt} = \frac{1}{d}\frac{dh}{dt}$ is, in miniature, the mathematics inside a tracking servo loop.

8.14 Problems with More Than Two Rates

Some related-rates problems set several quantities in motion at once. The method does not change — write the relations, differentiate each with respect to $t$ — but you now juggle more rates.

Example: the expanding balloon. A spherical balloon is inflated so its radius grows at $2$ cm/s. How fast are its volume and surface area increasing when $r = 10$ cm?

For volume, $V = \frac{4}{3}\pi r^3$, so

$$\frac{dV}{dt} = 4\pi r^2\,\frac{dr}{dt} = 4\pi(10)^2(2) = 800\pi\ \text{cm}^3/\text{s} \approx 2513\ \text{cm}^3/\text{s}.$$

For surface area, $S = 4\pi r^2$, so

$$\frac{dS}{dt} = 8\pi r\,\frac{dr}{dt} = 8\pi(10)(2) = 160\pi\ \text{cm}^2/\text{s} \approx 503\ \text{cm}^2/\text{s}.$$

Note that $\frac{dV}{dt} = \frac{S}{1}\cdot\frac{dr}{dt}$ exactly: the volume grows at (surface area) × (radial speed), because each instant adds a thin shell of thickness $\frac{dr}{dt}\,dt$ over the whole surface. That identity, $dV = S\,dr$, is itself a glimpse of integration — accumulating shells — which Chapter 18 turns into the shell method.

8.15 Implicit Differentiation in Economics

Related rates and implicit differentiation are not confined to geometry. In economics they appear wherever a constraint binds variables together.

Consider a Cobb–Douglas production function, in which output $Q$ depends on labor $L$ and capital $K$ as $Q = A\,L^{\alpha}K^{\beta}$. Suppose a firm holds output fixed at a target level $Q_0$ (an isoquant) and asks: if it sheds capital, how must it add labor to keep output constant? The constraint $A\,L^{\alpha}K^{\beta} = Q_0$ relates $L$ and $K$ implicitly. Differentiating with respect to $L$ (treating $K = K(L)$ along the isoquant) gives

$$A\alpha L^{\alpha-1}K^{\beta} + A\beta L^{\alpha}K^{\beta-1}\frac{dK}{dL} = 0 \quad\Longrightarrow\quad \frac{dK}{dL} = -\frac{\alpha K}{\beta L}.$$

That slope is (minus) the marginal rate of technical substitution — the rate at which capital can be traded for labor without changing output. It is implicit differentiation of a constraint, exactly like the circle's $-x/y$, and it is a workhorse of microeconomic theory. We will meet this idea again, formalized as the gradient and Lagrange multipliers, in Chapter 31.

Common Pitfall. When a quantity is held constant, its rate is zero, but it is not absent from the equation. In the isoquant above, $Q$ is constant so $\frac{dQ}{dt} = 0$ — yet $Q_0$ still appears as the value the constraint must equal. Students sometimes drop a constant entirely and lose the equation. A held-constant quantity contributes a zero derivative, not a zero value.

8.16 Computation: Implicit Differentiation in Python

The three-tier pattern of this book — pose it, solve it by hand, then verify by machine — applies cleanly here. SymPy can differentiate an implicit relation and confirm the folium result we found in §8.2.

import sympy as sp

# Implicitly differentiate the folium of Descartes: x^3 + y^3 = 6xy
x = sp.symbols('x')
y = sp.Function('y')(x)          # declare y as an unknown function of x

relation = sp.Eq(x**3 + y**3, 6*x*y)
# Differentiate both sides w.r.t. x, then solve for y'
yprime = sp.solve(sp.diff(relation.lhs - relation.rhs, x), sp.Derivative(y, x))[0]
print(sp.simplify(yprime))
# Output: (-x**2 + 2*y(x)) / (y(x)**2 - 2*x)   ==  (2y - x^2)/(y^2 - 2x)

# Evaluate the slope at the point (3, 3) on the curve
slope_at_33 = yprime.subs({x: 3, y: 3})
print(slope_at_33)
# Output: -1

The symbolic engine reproduces our hand result $y' = \frac{2y - x^2}{y^2 - 2x}$ and the slope $-1$ at $(3,3)$. This is the recurring division of labor in the book: hand computation builds the understanding, machine computation delivers the power. You should be able to derive the formula yourself; SymPy then lets you apply it to relations far too tangled to differentiate by hand, and catches your algebra slips along the way.

Computational Note. Declaring y = sp.Function('y')(x) is the crucial step — it tells SymPy that $y$ depends on $x$, so that sp.diff applies the chain rule through $y$ and produces the Derivative(y, x) terms. If you instead declared y = sp.symbols('y') as an ordinary independent variable, SymPy would treat $\frac{dy}{dx} = 0$ and hand back a useless $2x = 0$. The single line that makes $y$ a function of $x$ is exactly the human insight of §8.2, encoded for the machine.

8.17 Where These Techniques Lead

Implicit differentiation and related rates are not a sideshow; they are load-bearing tools that reappear throughout the rest of calculus and its applications:

• Curve analysis (Chapter 9) uses implicit $y'$ and $y''$ to locate tangents, vertical tangents, and concavity on curves that are not graphs of functions.
• Optimization (Chapter 10) frequently differentiates a constraint implicitly to eliminate a variable, the precursor of Lagrange multipliers.
• Differential equations (Chapter 19) often produce solutions defined only implicitly — a relation $F(x,y) = C$ that no algebra can solve for $y$, yet whose slope field $y'$ is exactly what implicit differentiation provides.
• Multivariable calculus (Chapters 30–31) generalizes the whole story: the Implicit Function Theorem guarantees when a relation defines a function locally, and the gradient is the natural multivariable successor to the $-x/y$ slope.
• Physics and thermodynamics relate state variables $(P, V, T)$ through equations of state such as $PV = nRT$; differentiating these implicitly is daily work in those fields.

Each of these is the same chain-rule idea — change propagating through a chain of dependencies — pushed into a new setting. That continuity is the point.

Add to Your Modeling Portfolio. Add a related-rates or implicit-constraint relationship to your model — one equation that ties two of your quantities together and lets a known rate determine an unknown one. Biology: if tumor volume is $V = \frac{4}{3}\pi r^3$ and you measure radial growth $\frac{dr}{dt}$ from imaging, derive the volume growth rate $\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$ — the quantity oncologists actually track. Economics: write your production constraint $Q(L,K) = Q_0$ as an isoquant and compute the marginal rate of technical substitution $\frac{dK}{dL}$ by implicit differentiation. Physics: take the ideal gas law $PV = nRT$ and, holding temperature fixed, relate $\frac{dP}{dt}$ to $\frac{dV}{dt}$ during a compression. Data Science: express a constraint surface (e.g. a budget or a normalization $\sum p_i = 1$) implicitly and differentiate it to find how one coordinate must move when another is adjusted.

Looking Ahead

You arrived at this chapter able to differentiate any function handed to you in the form $y = f(x)$. You leave it able to differentiate relations that are not solved for $y$, to find tangents on curves no algebra can untangle, to derive every inverse-function derivative, and to follow a chain of quantities changing together through time. All of it rested on one idea — the chain rule, applied through a hidden dependency.

Chapter 9 turns derivatives loose on the shape of functions: where they rise and fall, where they bend, where they peak. Several of those analyses lean on the implicit $y'$ and $y''$ you computed here. Chapter 10 then uses derivatives to find the best possible value of a quantity — the most for the least — and constrained optimization will repeatedly call on the implicit differentiation of constraints you practiced in §8.15. The technique you just learned is not finished; it is foundation.

Reflection

There is a quiet lesson in this chapter worth carrying forward. We did not invent a new kind of derivative. Implicit differentiation and related rates are the chain rule of Chapter 7, applied with the willingness to differentiate an equation we cannot solve and to let the answer involve more than one variable. The mathematics rewards that willingness: a circle's tangent falls out as $-x/y$, a ladder's fall as $-\tfrac{x}{y}\frac{dx}{dt}$, an economy's substitution rate as $-\frac{\alpha K}{\beta L}$ — all the same gesture, aimed at different worlds. Calculus is the mathematics of change, and change, it turns out, almost never arrives pre-solved for $y$. Now you can differentiate it anyway.