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You already know how to differentiate $x^2 + y^2 = 25$ implicitly with respect to $x$. The result involves $\frac{dy}{dx}$. In related rates, we do something similar but differentiate with respect to time $t$ instead.
Here's the key insight: when $x$ and $y$ are both functions of time, differentiating $x^2 + y^2 = 25$ with respect to $t$ gives us $\frac{dx}{dt}$ and $\frac{dy}{dt}$—the rates we're looking for.
| Property | Value |
|---|---|
| Chapter | 2 - Derivatives |
| Section | 2.8 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
When differentiating any expression with respect to $t$, every variable that depends on $t$ requires the chain rule.
$$\frac{d}{dt}[\text{expression in } x] = \frac{d}{dx}[\text{expression}] \cdot \frac{dx}{dt}$$
For any variable $u$ that depends on $t$:
$$\frac{d}{dt}[u^n] = n \cdot u^{n-1} \cdot \frac{du}{dt}$$
Examples: | Expression | Derivative with respect to $t$ | |------------|-------------------------------| | $x^2$ | $2x \cdot \frac{dx}{dt}$ | | $r^3$ | $3r^2 \cdot \frac{dr}{dt}$ | | $\sqrt{y} = y^{1/2}$ | $\frac{1}{2}y^{-1/2} \cdot \frac{dy}{dt} = \frac{1}{2\sqrt{y}}\frac{dy}{dt}$ | | $\frac{1}{x} = x^{-1}$ | $-x^{-2} \cdot \frac{dx}{dt} = -\frac{1}{x^2}\frac{dx}{dt}$ |
When two time-dependent quantities are multiplied, use the product rule:
$$\frac{d}{dt}[xy] = \frac{dx}{dt} \cdot y + x \cdot \frac{dy}{dt}$$
Example: If area $A = \ell w$ where both length and width change: $$\frac{dA}{dt} = \frac{d\ell}{dt} \cdot w + \ell \cdot \frac{dw}{dt}$$
$$\frac{d}{dt}[\sin \theta] = \cos \theta \cdot \frac{d\theta}{dt}$$
$$\frac{d}{dt}[\tan \theta] = \sec^2 \theta \cdot \frac{d\theta}{dt}$$
Critical distinction:
The $4\pi$ is constant; the $r^2$ involves a variable.
If $A = \pi r^2$ (area of a circle), find $\frac{dA}{dt}$ in terms of $r$ and $\frac{dr}{dt}$.
For a right triangle where both legs $x$ and $y$ change with time, and the hypotenuse $z$ satisfies $x^2 + y^2 = z^2$, find $\frac{dz}{dt}$ in terms of $x$, $y$, $z$, $\frac{dx}{dt}$, and $\frac{dy}{dt}$.
A cone has volume $V = \frac{1}{3}\pi r^2 h$. If both the radius $r$ and height $h$ change with time, differentiate to find $\frac{dV}{dt}$.
In a right triangle, $\tan \theta = \frac{y}{x}$ where both $x$ and $y$ are functions of time, and $\theta$ is the angle opposite side $y$.
Find $\frac{d\theta}{dt}$ in terms of $x$, $y$, $\frac{dx}{dt}$, and $\frac{dy}{dt}$.
Two sides of a triangle have fixed lengths $a = 5$ and $b = 7$. The angle $\theta$ between them is increasing. The third side $c$ satisfies the Law of Cosines: $$c^2 = a^2 + b^2 - 2ab\cos\theta$$
(a) Differentiate to find $\frac{dc}{dt}$ in terms of $c$, $\theta$, and $\frac{d\theta}{dt}$.
(b) Verify that your formula gives $\frac{dc}{dt} > 0$ when $0 < \theta < \pi$ and $\frac{d\theta}{dt} > 0$. Explain why this makes geometric sense.
| Formula | Derivative with respect to $t$ |
|---|---|
| $A = s^2$ (square) | $\frac{dA}{dt} = 2s\frac{ds}{dt}$ |
| $A = \pi r^2$ (circle) | $\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$ |
| $V = s^3$ (cube) | $\frac{dV}{dt} = 3s^2\frac{ds}{dt}$ |
| $V = \frac{4}{3}\pi r^3$ (sphere) | $\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$ |
| $V = \pi r^2 h$ (cylinder, fixed $r$) | $\frac{dV}{dt} = \pi r^2\frac{dh}{dt}$ |
| $x^2 + y^2 = z^2$ | $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}$ |
When differentiating $V = \frac{4}{3}\pi r^3$ with respect to $t$, a student writes:
$$\frac{dV}{dt} = 4\pi r^2$$
What is wrong with this answer?
(A) The coefficient should be $\frac{4}{3}$, not $4$ (B) The exponent should be $3$, not $2$ (C) The factor $\frac{dr}{dt}$ is missing (D) The formula for sphere volume is wrong
For the rectangle area $A = \ell w$, which correctly gives $\frac{dA}{dt}$?
(A) $\frac{dA}{dt} = \frac{d\ell}{dt} \cdot \frac{dw}{dt}$
(B) $\frac{dA}{dt} = \ell \frac{dw}{dt} + w \frac{d\ell}{dt}$
(C) $\frac{dA}{dt} = 2(\ell w) \cdot \frac{d(\ell w)}{dt}$
(D) $\frac{dA}{dt} = \ell \cdot w \cdot \frac{dt}{dt}$
| Error | Example | Correct Form |
|---|---|---|
| Forgetting chain rule | $\frac{d}{dt}[r^2] = 2r$ | $\frac{d}{dt}[r^2] = 2r\frac{dr}{dt}$ |
| Wrong product rule | $\frac{d}{dt}[\ell w] = \frac{d\ell}{dt} \cdot \frac{dw}{dt}$ | $\frac{d}{dt}[\ell w] = \frac{d\ell}{dt} \cdot w + \ell \cdot \frac{dw}{dt}$ |
| Differentiating constants | $\frac{d}{dt}[5] = \frac{d5}{dt}$ | $\frac{d}{dt}[5] = 0$ |
| Missing parentheses | $\frac{d}{dt}[2x] = 2\frac{dx}{dt}$ ✓ | But $\frac{d}{dt}[(2x)^2] = 2(2x) \cdot 2\frac{dx}{dt} = 8x\frac{dx}{dt}$ |
The Gears Analogy:
Think of differentiating with respect to time like a system of gears. When $r$ changes with time, it's a gear turning. The formula $r^2$ is like a gear ratio—it transforms the motion. But the chain rule says: to find how fast the output changes, you multiply the gear ratio by how fast the input is turning.
$$\frac{d(r^2)}{dt} = \underbrace{2r}_{\text{gear ratio}} \times \underbrace{\frac{dr}{dt}}_{\text{input speed}}$$
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| Related Rates Setup | Ch2 §8 Skills | Related Rates Problem Solving |
Last updated: 2026-01-22