A collection of my strongest work, worked problems, and reflections from AP Calculus — limits through the Fundamental Theorem, in my own words.
Why I built it, and what to expect in each section.
This portfolio tracks my work across the major units of AP Calculus. For each unit I've included a representative problem, the reasoning I used to solve it, and a short reflection on what clicked (or didn't, at first). The goal isn't just to show correct answers — it's to show how my understanding of the ideas developed over the course.
Where calculus starts: making sense of behavior "as we approach," not just "at."
Evaluate \( \displaystyle \lim_{x \to 3} \frac{x^2 - 9}{x - 3} \).
Direct substitution gives \( \frac{0}{0} \), an indeterminate form, so I factored the numerator: \[ \frac{x^2-9}{x-3} = \frac{(x-3)(x+3)}{x-3} = x+3 \quad (x \ne 3) \] Since the limit only cares about behavior near \(x=3\), not at it, I can cancel the common factor and substitute: \[ \lim_{x \to 3} (x+3) = 6 \]
From the limit definition to the chain rule, product rule, and implicit differentiation.
Find \( f'(x) \) for \( f(x) = x^2 \) using the limit definition.
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} \] \[ = \lim_{h \to 0} \frac{2xh + h^2}{h} = \lim_{h \to 0} (2x + h) = 2x \]
Find \( \frac{dy}{dx} \) if \( x^2 + y^2 = 25 \).
Differentiating both sides with respect to \(x\), treating \(y\) as a function of \(x\): \[ 2x + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y} \]
Related rates, optimization, and curve sketching — where derivatives answer real questions.
A 10–ft ladder leans against a wall. The base slides away at 2 ft/s. How fast is the top sliding down when the base is 6 ft from the wall?
With \(x^2+y^2=100\), differentiate: \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\). At \(x=6\), \(y=8\), and \(\frac{dx}{dt}=2\): \[ 6(2) + 8\frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\tfrac{3}{2} \text{ ft/s} \]
With 100 ft of fencing, maximize the area of a rectangular pen.
\(A = x(50-x)\). Setting \(A'(x)=50-2x=0\) gives \(x=25\), so the maximum area is a \(25 \times 25\) square, \(A = 625\) sq ft.
Approximating area under a curve, then making the approximation exact.
Approximate \( \int_0^4 x^2\,dx \) with a right Riemann sum, \(n=4\), then find the exact value.
With \(\Delta x = 1\), heights at \(x=1,2,3,4\): \(1+4+9+16 = 30\). The exact value: \[ \int_0^4 x^2\,dx = \left[\frac{x^3}{3}\right]_0^4 = \frac{64}{3} \approx 21.33 \]
Connecting derivatives and integrals as inverse operations, then applying it to accumulation and differential equations.
If \( g(x) = \int_0^x t^2+1 \, dt \), find \(g'(x)\).
By the Fundamental Theorem of Calculus (Part 1), differentiating an integral with respect to its upper limit just gives back the integrand: \[ g'(x) = x^2 + 1 \]
A running log of what full free-response practice has taught me about communicating calculus, not just computing it.
Lost points for finding velocity correctly but never stating whether the particle was moving left or right using the sign of \(v(t)\). Lesson: a computed number isn't a complete answer without a sentence connecting it back to the question.
Set up \(\int (f(x)-g(x))\,dx\) with the functions in the wrong order on part of the interval. Now I always sketch the region first and check which curve is on top before writing the integral.
Learned that "the answer is x = 3" isn't a justification — the rubric wants the sign change of \(f'(x)\) stated explicitly. Writing full sentences instead of just symbols has fixed most of my lost justification points.
Looking back at the start of the year versus now.
At the start of the course, calculus felt like a long list of unrelated rules — a limit rule here, a derivative shortcut there. What changed is seeing the connections: limits define derivatives, derivatives describe rates and shapes, and integrals reverse the process to recover accumulated change. The Fundamental Theorem of Calculus is the moment that stitched all of it into one idea instead of four separate units.
Going forward, the thing I want to keep practicing is justification — writing the sentence that explains why an answer is correct, not just producing the number.