In related rates problems, two or more quantities change over time and are linked by an equation. We differentiate that equation with respect to time $t$ (using the chain rule) to relate their rates of change.
A 10-foot ladder leans against a wall. The bottom slides away from the wall at 2 ft/sec. How fast is the top sliding down when the bottom is 6 ft from the wall?
Setup: Let $x$ = distance from wall (bottom), $y$ = height on wall. Pythagorean: $x^2 + y^2 = 100$.
Differentiate w.r.t. $t$: $2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0$
Find $y$ when $x = 6$: $y = \sqrt{100 - 36} = 8$
Substitute ($\dfrac{dx}{dt} = 2$, $x=6$, $y=8$): $2(6)(2) + 2(8)\dfrac{dy}{dt} = 0$
$\dfrac{dy}{dt} = -\dfrac{24}{16} = -\dfrac{3}{2}$ ft/sec
The top slides down at $\dfrac{3}{2}$ ft/sec (negative = downward).
A spherical balloon is being inflated so its volume increases at 10 cm³/sec. How fast is the radius growing when $r = 5$ cm? (Recall: $V = \frac{4}{3}\pi r^3$)
A 6-foot person walks away from a 15-foot streetlight at 4 ft/sec. How fast is their shadow lengthening?
Let $x$ = person's distance from the lamp, $s$ = shadow length. By similar triangles: $\dfrac{15}{x+s} = \dfrac{6}{s}$, so $15s = 6(x+s)$, giving $9s = 6x \Rightarrow s = \dfrac{2x}{3}$.
Differentiate: $\dfrac{ds}{dt} = \dfrac{2}{3}\dfrac{dx}{dt} = \dfrac{2}{3}(4) = \dfrac{8}{3}$ ft/sec.
The shadow always lengthens at $\dfrac{8}{3}$ ft/sec regardless of position.
AP Exam Tip: Related rates are a classic FRQ topic. Always state what each variable represents and its units. Identify given rates (with sign: negative if decreasing) and the rate being sought. Show the differentiated equation before substituting values — this earns partial credit.
The sliding ladder: as $x$ (distance from wall) increases, $y$ (height) decreases. Use the slider to see both rates in action.
Figure 4.1 — Related Rates: Ladder Sliding Down Wall ($x^2 + y^2 = 100$)
Near a point $(a, f(a))$, a differentiable function is well approximated by its tangent line. This is the foundation of linear approximation.
The linearization of $f$ at $x = a$ is the tangent line: $$L(x) = f(a) + f'(a)(x - a)$$
For $x$ near $a$: $f(x) \approx L(x)$.
Differential: If $y = f(x)$, then $dy = f'(x)\,dx$. For small change $\Delta x \approx dx$: $\Delta y \approx dy = f'(x)\,dx$.
Let $f(x) = \sqrt{x}$, so $f'(x) = \dfrac{1}{2\sqrt{x}}$. Linearize at $a = 4$:
$L(x) = f(4) + f'(4)(x-4) = 2 + \dfrac{1}{4}(x-4)$
At $x = 4.02$: $L(4.02) = 2 + \dfrac{1}{4}(0.02) = 2 + 0.005 = 2.005$
Actual: $\sqrt{4.02} \approx 2.00499...$. Very close!
Use linear approximation to estimate $\sin(0.1)$ using $f(x) = \sin x$ at $a = 0$.
Linear approximation: the tangent line (red) hugs the curve (blue) near the base point. Drag the slider to change the base point $a$.
Figure 4.2 — Linear Approximation of $f(x) = \sqrt{x}$ at $a = 4$
The Mean Value Theorem (MVT) is one of the most important theoretical results in calculus. It guarantees that somewhere on a smooth curve, the instantaneous rate of change equals the average rate of change.
If $f$ is:
then there exists at least one $c \in (a, b)$ such that:
$$f'(c) = \frac{f(b) - f(a)}{b - a}$$
The slope of the tangent line at $x = c$ equals the slope of the secant line from $a$ to $b$.
If additionally $f(a) = f(b)$, then there exists $c \in (a, b)$ with $f'(c) = 0$. (At least one horizontal tangent exists between equal endpoints.)
Let $f(x) = x^3 - x$ on $[0, 2]$. Find the value(s) of $c$ guaranteed by the MVT.
Average rate: $\dfrac{f(2)-f(0)}{2-0} = \dfrac{(8-2)-0}{2} = \dfrac{6}{2} = 3$
Set $f'(c) = 3$: $f'(x) = 3x^2 - 1 = 3 \Rightarrow 3x^2 = 4 \Rightarrow x^2 = \dfrac{4}{3}$
$c = \dfrac{2}{\sqrt{3}} = \dfrac{2\sqrt{3}}{3} \approx 1.155 \in (0, 2)$ ✓
Verify the MVT applies to $f(x) = x^2$ on $[1, 3]$, then find $c$.
The MVT is often used to prove facts from derivative information. For example: if $f'(x) > 0$ for all $x \in (a,b)$, then $f$ is increasing on $(a,b)$. This follows because for any $x_1 < x_2$ in the interval, the MVT guarantees $f(x_2) - f(x_1) = f'(c)(x_2 - x_1) > 0$.
AP Exam Tip: The MVT is often tested as a justification problem. Given a table of values, you may be asked: "Explain why there must exist a $c$ such that $f'(c) = k$." The key is to first verify the conditions (continuous, differentiable on the interval), then invoke the MVT explicitly by name.
The Mean Value Theorem: the red tangent at $c$ is parallel to the blue secant from $(a, f(a))$ to $(b, f(b))$. Adjust $a$ and $b$ to explore.
Figure 4.3 — Mean Value Theorem for $f(x) = x^3 - x$ on $[0, 2]$
A conical tank (vertex down) has radius 3 m and height 6 m. Water fills at 2 m³/min. How fast is the water level rising when the depth is 4 m? ($V = \frac{1}{3}\pi r^2 h$; by similar triangles, $r = h/2$.)
Two cars leave an intersection simultaneously. Car A travels north at 60 mph, Car B travels east at 80 mph. How fast is the distance between them increasing after 1 hour?
Use linear approximation to estimate $(1.03)^{10}$. Use $f(x) = x^{10}$ at $a = 1$.
A cube's side length is measured as 5 cm with error $\pm 0.02$ cm. Use differentials to estimate the maximum error in the calculated volume.
Verify the MVT conditions for $f(x) = x^2 + 2x$ on $[-1, 2]$ and find all values of $c$.
Use Rolle's Theorem to show $f(x) = x^3 - x$ has a zero of $f'$ in $(-1, 0)$.
A particle's position is $s(t) = t^3 - 6t$ for $t \geq 0$. Find the time(s) when the instantaneous velocity equals the average velocity on $[0, 3]$.
(AP Style) The table shows values of a differentiable function $f$. Explain why there must exist a value $c \in (1, 4)$ with $f'(c) = 2$.
| $x$ | 1 | 4 |
|---|---|---|
| $f(x)$ | 3 | 9 |
If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then $\exists\, c \in (a,b)$ such that $f'(c) = \dfrac{f(b)-f(a)}{b-a}$.
Special case of MVT: if $f(a) = f(b)$, then $\exists\, c$ where $f'(c) = 0$. Used to prove MVT.
If $\lim\frac{f}{g}$ yields $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim\dfrac{f(x)}{g(x)} = \lim\dfrac{f'(x)}{g'(x)}$ (if the latter exists).
$L(x) = f(a) + f'(a)(x-a)$ approximates $f$ near $x = a$ using the tangent line.
$f'(c) = \dfrac{f(b)-f(a)}{b-a}$ — the instantaneous rate equals the average rate at some interior point.
$dy = f'(x)\,dx$ — approximates the change in $f$ for a small change $dx$ in $x$.
$x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$ — iteratively approximates roots of $f(x) = 0$.
$\frac{0}{0}$, $\frac{\infty}{\infty}$, $0\cdot\infty$, $\infty-\infty$, $0^0$, $1^\infty$, $\infty^0$ — apply L'Hôpital or algebraic manipulation.