Contextual Applications of Differentiation
Unit 4 of AP Calculus AB, worth 10–15% of the exam. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Motion along a line, rates of change in context, related rates, linearisation and approximation, L'Hospital's rule.
How this unit is tested
Start by sorting every problem into one of the five families this unit tests, because each has its own setup ritual. Motion problems give you $s(t)$, $v(t)=s'(t)$, or $a(t)=v'(t)$ and ask about position, speed, or distance traveled — always distinguish velocity (signed) from speed (its absolute value) before answering. Rate-of-change-in-context problems ask you to compute or interpret a derivative with units attached; always attach units and a sentence of interpretation to the number. Related rates and linearization both begin with an equation connecting quantities, but related rates differentiates that equation with respect to time before substituting numbers, while linearization uses the equation of a single tangent line at one fixed point to estimate nearby function values.
For related rates specifically, work in a fixed order every time: draw and label a diagram, write an equation relating all the changing quantities, differentiate implicitly with respect to $t$, THEN substitute the numerical values given at the instant in question, and finally solve for the requested rate and attach units. Skipping the differentiation step or substituting early are the two most common failure points.
For linearization, remember it is just the tangent line: $L(x)=f(a)+f'(a)(x-a)$. To judge whether $L(x)$ over- or under-estimates $f(x)$ near $a$, check the concavity of $f$: concave up means the tangent lies below the curve (underestimate), concave down means the tangent lies above (overestimate).
For L'Hospital's rule, the single most important habit is verifying the indeterminate form (0/0 or ∞/∞) BEFORE differentiating — the rule does not apply to a limit that already evaluates to a number. If you meet a product or difference that looks indeterminate (0·∞ or ∞−∞), rewrite it algebraically as a single fraction first so it becomes 0/0 or ∞/∞, then apply the rule, possibly more than once.
What you have to know
Velocity, speed, and acceleration
If $s(t)$ is position, then $v(t)=s'(t)$ is velocity and $a(t)=v'(t)=s''(t)$ is acceleration. Speed is $|v(t)|$, the magnitude of velocity, and has no sign.
Speed increasing/decreasing test
Speed is increasing at a time $t$ when $v(t)$ and $a(t)$ have the same sign (velocity and acceleration act in the same direction); speed is decreasing when they have opposite signs.
Total distance traveled
Distance traveled on $[t_1,t_2]$ equals $\int_{t_1}^{t_2}|v(t)|\,dt$, computed by finding where $v(t)=0$, evaluating $s(t)$ at those times and the endpoints, and summing the absolute values of the displacements on each sub-interval.
Related rates procedure
Write an equation relating the changing quantities, differentiate both sides implicitly with respect to $t$, substitute the known numerical values only after differentiating, then solve algebraically for the desired rate.
Linearization (tangent line approximation)
Near $x=a$, $f(x)\approx L(x)=f(a)+f'(a)(x-a)$. If $f''(a)>0$ (concave up) the approximation underestimates $f(x)$; if $f''(a)<0$ (concave down) it overestimates $f(x)$.
L'Hospital's Rule
If $\lim_{x\to c} f(x)/g(x)$ has indeterminate form $0/0$ or $\infty/\infty$, and $f,g$ are differentiable near $c$ with $g'(x)\neq 0$, then $\lim_{x\to c} f(x)/g(x)=\lim_{x\to c} f'(x)/g'(x)$, provided the right-hand limit exists or is infinite. This applies as $x\to c$ or $x\to\pm\infty$.
14 practice questions
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A particle moves along a line with position $s(t)=t^3-6t^2+9t$ meters, $t\ge0$ in seconds. At what time(s) is the particle at rest?
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Answer. t = 1 and t = 3 seconds
Velocity is $v(t)=s'(t)=3t^2-12t+9=3(t-1)(t-3)$. The particle is at rest when $v(t)=0$, which happens at t=1 and t=3. -
For the same particle, $v(t)=3t^2-12t+9$ and $a(t)=6t-12$, is the particle's speed increasing or decreasing at t = 1.5 seconds?
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Answer. Speed is increasing
At t=1.5, $v(1.5)=3(2.25)-18+9=-2.25$ and $a(1.5)=6(1.5)-12=-3$. Both are negative, so velocity and acceleration point the same direction, meaning speed is increasing. -
Using $s(t)=t^3-6t^2+9t$ and the fact that v(t)=0 at t=1 and t=3, find the total distance traveled by the particle from t=0 to t=3.
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Answer. 8 meters
Compute s(0)=0, s(1)=1-6+9=4, s(3)=27-54+27=0. Distance = |s(1)-s(0)| + |s(3)-s(1)| = |4-0| + |0-4| = 4+4 = 8 meters, not the net displacement of 0. -
The volume of a sphere is $V=\frac{4}{3}\pi r^3$. If the radius is increasing at 2 cm/min, find the rate of change of the volume when r = 5 cm.
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Answer. $dV/dt = 200\pi \approx 628.3$ cm³/min
Differentiate: $dV/dt = 4\pi r^2 \frac{dr}{dt}$. Substitute r=5 and dr/dt=2 after differentiating: $dV/dt = 4\pi(25)(2) = 200\pi$ cm³/min. -
A spherical balloon is inflated so its volume increases at a constant rate of 100 cm³/s. Find the rate at which the radius is increasing when r = 5 cm.
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Answer. $dr/dt = 1/\pi \approx 0.318$ cm/s
From $V=\frac{4}{3}\pi r^3$, differentiate to get $dV/dt=4\pi r^2\, dr/dt$. Substitute dV/dt=100 and r=5: $100=4\pi(25)\,dr/dt=100\pi\,dr/dt$, so $dr/dt=1/\pi$ cm/s. -
Water drains from a conical tank (vertex down) where the radius is always half the height, $r=h/2$, at a rate $dV/dt=-3$ ft³/min. Find $dh/dt$ when h = 4 ft.
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Answer. $dh/dt = -3/(4\pi) \approx -0.239$ ft/min
Substitute r=h/2 into $V=\frac{1}{3}\pi r^2 h$ to get $V=\frac{\pi}{12}h^3$. Differentiate: $dV/dt=\frac{\pi}{4}h^2\,dh/dt$. Substitute h=4 and dV/dt=-3: $-3=4\pi\,dh/dt$, so $dh/dt=-3/(4\pi)$ ft/min. -
Car A travels north from an intersection at 40 mph; Car B travels east from the same intersection at 30 mph, both starting at t=0. Find the rate at which the distance between them is increasing at t = 1 hour.
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Answer. 50 mph
Let x=30t (east), y=40t (north), $z^2=x^2+y^2$. At t=1: x=30, y=40, z=50. Differentiate: $z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$, giving $50\,dz/dt = 30(30)+40(40)=2500$, so $dz/dt=50$ mph. -
Use linearization to estimate $\sqrt{4.1}$ using $f(x)=\sqrt{x}$ centered at a = 4.
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Answer. 2.025
$f(4)=2$, $f'(x)=\frac{1}{2\sqrt{x}}$ so $f'(4)=1/4$. The linearization is $L(x)=2+\frac{1}{4}(x-4)$, and $L(4.1)=2+0.25(0.1)=2.025$. -
Does the linear approximation of $\sqrt{4.1}$ from the previous card overestimate or underestimate the true value?
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Answer. Overestimates
$f''(x)=-\frac{1}{4x^{3/2}}<0$ for x>0, so $f(x)=\sqrt{x}$ is concave down near a=4. A concave-down function lies below its tangent line, so the linearization gives a value larger than the true $\sqrt{4.1}$. -
Evaluate $\lim_{x\to 0}\dfrac{e^x-1-x}{x^2}$ using L'Hospital's rule.
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Answer. 1/2
Direct substitution gives 0/0. Differentiate top and bottom: $\lim \frac{e^x-1}{2x}$, still 0/0 at x=0, so apply again: $\lim \frac{e^x}{2}=\frac{1}{2}$. -
Evaluate $\lim_{x\to\infty}\dfrac{3x^2+5x}{2x^2-7}$ using L'Hospital's rule.
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Answer. 3/2
This is ∞/∞. Differentiate: $\lim \frac{6x+5}{4x}$, still ∞/∞, so differentiate again: $\lim \frac{6}{4}=\frac{3}{2}$. -
Evaluate $\lim_{x\to 0^+} x\ln x$.
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Answer. 0
This is the indeterminate form $0\cdot(-\infty)$, not directly usable with L'Hospital's rule. Rewrite as $\lim_{x\to0^+}\frac{\ln x}{1/x}$, an $\infty/\infty$ form (actually $-\infty/\infty$), then differentiate: $\lim \frac{1/x}{-1/x^2}=\lim(-x)=0$. -
A tank's volume $V(t)$ in liters is measured at selected times (t in minutes): V(0)=100, V(2)=150, V(5)=270. Estimate the average rate of change of volume on [2,5] and state its units and meaning.
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Answer. 40 liters per minute; the volume increased on average by 40 L/min over that interval
Average rate of change is $\frac{V(5)-V(2)}{5-2}=\frac{270-150}{3}=\frac{120}{3}=40$ liters per minute, which estimates how fast the tank was filling on average between t=2 and t=5. -
In a related rates problem, why must you differentiate the relating equation with respect to time BEFORE substituting the specific numerical values given at the instant of interest?
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Answer. Because the variables are functions of time, not constants; substituting numbers first freezes them and produces incorrect (often zero) rates
If you plug in the known instantaneous values before differentiating, you are differentiating a constant equation, which forces every derivative term to vanish. Differentiating the general relationship first preserves the $dx/dt$, $dy/dt$ terms so the actual rates can be solved for after substitution.
What people get wrong
- Substituting the numerical values into the related-rates equation before differentiating. Differentiate the general equation with respect to t first, and only plug in the specific instant's numbers afterward.
- Treating speed and velocity as the same thing on a motion problem. Speed is $|v(t)|$; a particle can have decreasing speed while its velocity is becoming more negative. Always check the sign of both v and a.
- Computing total distance as $\int v(t)\,dt$ (which gives net displacement) instead of $\int|v(t)|\,dt$. Find where v(t)=0 first, then add the absolute values of displacement on each piece.
- Applying L'Hospital's rule to a limit that isn't actually 0/0 or ∞/∞. Plug in the limiting value first to confirm the indeterminate form; if it isn't indeterminate, just evaluate directly.
- Assuming a linear approximation is exact rather than an estimate, or guessing over/underestimate without checking concavity. Compute $f''(a)$'s sign to justify the direction of the error.
- Giving a rate-of-change answer as a bare number with no units or sentence of interpretation, which loses points on free-response context questions.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of AP Calculus AB. Free, and no account needed to start.