From Trajectories to Integrals — Antiderivatives and the Area of a Circle
How do scientists measure the area of something curved? Today you will see how slicing a circle into infinitely thin strips lets calculus compute areas that geometry alone cannot easily handle. You will also learn how to “undo” a derivative – a skill called antidifferentiation – and discover that inverse trig functions like arcsine pop up in surprising places when you work with circles. Along the way, you will prove the trajectory of a thrown ball is a parabola.
Integration is the second great pillar of calculus — it lets you add up infinitely many tiny pieces to find totals:
- Medicine: doctors calculate the total amount of a drug absorbed by your body over time by integrating the absorption rate — getting the dosage wrong could be dangerous
- Space travel: NASA engineers integrate acceleration data to figure out how far a rocket has traveled and how fast it is going at every moment
- Architecture: architects compute the area of curved surfaces (like domes and arches) by slicing them into thin strips and adding up the pieces — that is integration
- Sports analytics: tracking how far a soccer player runs during a match means integrating their speed over time
- Animation: Pixar uses integration to calculate how light bounces off curved surfaces, making 3D characters look realistic
Topics Covered
- Projectile motion: recovering position from velocity via antiderivatives (\(x = ut\), \(y = \tfrac{1}{2}gt^2\))
- Eliminating the parameter \(t\) to find the trajectory \(y = \dfrac{g}{2u^2}\,x^2\)
- Introduction to integral calculus: assembling infinitesimals
- Area of a circle by vertical slices: \(dA = 2\sqrt{r^2 - x^2}\,dx\)
- Review of the seven families of functions and their derivative–antiderivative pairs
- Derivative of \(\arcsin x\) via implicit differentiation: \(\dfrac{d}{dx}\arcsin x = \dfrac{1}{\sqrt{1-x^2}}\)
- Derivative of \(\ln x\) via the inverse-function rule: \(\dfrac{d}{dx}\ln x = \dfrac{1}{x}\)
- Antiderivative involving arcsine: \(\displaystyle\int \frac{1}{\sqrt{r^2 - x^2}}\,dx = \arcsin\!\left(\frac{x}{r}\right) + C\)
Lecture Video
Key Frames from the Lecture
What You Need to Know First
An antiderivative of a function \(f(x)\) is any function \(F(x)\) whose derivative gives back \(f(x)\):
\[F'(x) = f(x)\]
For example, if \(f(x) = 2x\), then \(F(x) = x^2\) is an antiderivative because \(\frac{d}{dx}(x^2) = 2x\).
But \(F(x) = x^2 + 7\) also works! Any constant added to an antiderivative is still an antiderivative, because constants vanish when you differentiate. That is why we write \(+C\) (an arbitrary constant) at the end.
When you throw a ball horizontally, two things happen at once:
- Horizontally, the ball keeps moving at a constant speed \(u\) (no force pushes or pulls it sideways)
- Vertically, gravity pulls the ball downward with constant acceleration \(g \approx 9.8\;\text{m/s}^2\)
The horizontal and vertical motions are independent. We describe them with separate equations:
\[\dot{x} = u \qquad \text{(constant horizontal velocity)}\] \[\dot{y} = g\,t \qquad \text{(vertical velocity increases with time)}\]
The dot notation \(\dot{x}\) means \(\frac{dx}{dt}\), the rate of change with respect to time.
Often in physics, position is given as two separate equations in time: \(x(t)\) and \(y(t)\). The variable \(t\) is called a parameter.
To find the trajectory (the path the object traces in the \(xy\)-plane), we eliminate \(t\) by solving one equation for \(t\) and substituting into the other. The result is a direct relationship \(y = f(x)\) — the shape of the path, independent of when the object was at each point.
A circle centered at the origin with radius \(r\) is described by the implicit equation:
\[x^2 + y^2 = r^2\]
This is not a function (it fails the vertical line test), but we can solve for \(y\):
\[y = \pm\sqrt{r^2 - x^2}\]
The positive root gives the upper half, and the negative root gives the lower half.
If the derivative of \(x^n\) is \(nx^{n-1}\), then running the rule in reverse gives:
\[\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \qquad (n \neq -1)\]
For example, if you know that \(\frac{d}{dt}(t^2) = 2t\), then the antiderivative of \(2t\) is \(t^2 + C\). Equivalently, the antiderivative of \(t\) is \(\frac{t^2}{2} + C\).
Key Concepts
Projectile Motion: From Velocity to Position
We are given the velocity components of a ball thrown horizontally with initial speed \(u\) in a gravitational field:
\[\dot{x} = u \qquad \dot{y} = g\,t\]
Goal: find the trajectory \(y(x)\) by (1) recovering position from velocity, then (2) eliminating \(t\).
Step 1 — Find \(x(t)\). Since \(\dot{x} = u\) is constant, the antiderivative is:
\[x(t) = ut + a\]
Using the initial condition \(x(0) = 0\) (the ball starts at the origin), we find \(a = 0\), so:
\[\boxed{x(t) = ut}\]
Step 2 — Find \(y(t)\). Since \(\dot{y} = gt\) is linear in \(t\), we reverse the power rule:
\[y(t) = \frac{1}{2}g\,t^2 + a\]
Again, \(y(0) = 0\) forces \(a = 0\):
\[\boxed{y(t) = \tfrac{1}{2}g\,t^2}\]
Step 3 — Eliminate \(t\). From \(x = ut\) we get \(t = \frac{x}{u}\). Substituting into \(y(t)\):
\[y = \frac{1}{2}g\left(\frac{x}{u}\right)^2\]
By finding position from velocity (antidifferentiation) and then eliminating time, you can prove that a ball thrown horizontally traces out a parabolic path. Gravity controls how steep the curve is, and the throwing speed controls how flat it is.
\[\boxed{y = \frac{g}{2u^2}\,x^2}\]
This is a parabola. Larger \(g\) makes it steeper (the ball drops faster); larger \(u\) makes it flatter (the ball travels farther before falling).
Explore – how \(g\) and \(u\) change the trajectory:
Drag \(g\) to increase gravity (the parabola steepens) or increase \(u\) to throw faster (the parabola flattens). The ball always follows a parabolic path!
From Differential to Integral Calculus
Up to this point we have been doing differential calculus: chopping quantities into infinitesimals and finding rates of change. Now we begin integral calculus: assembling infinitesimals back together to find totals. The key insight is that differentiation and antidifferentiation are inverse operations — just as we recovered position from velocity in the projectile problem above.
Finding the Area of a Circle
How do we prove that the area of a circle of radius \(r\) is \(\pi r^2\)? We slice it into infinitesimally thin vertical strips and add them up.
Setup. Place the circle \(x^2 + y^2 = r^2\) at the origin. Define \(A(x)\) as the area swept from \(-r\) up to position \(x\):
\[A(x) = \text{shaded area from } {-r} \text{ to } x\]
As \(x\) increases by a tiny amount \(dx\), the area increases by a thin rectangular strip:
\[dA = \text{height} \times \text{width} = 2\sqrt{r^2 - x^2}\;\cdot\;dx\]
The factor of \(2\) accounts for both the upper and lower halves of the circle. Therefore, the derivative of the area function is:
To find the area of a circle, imagine sweeping a vertical line across it. Each infinitely thin strip has height \(2\sqrt{r^2 - x^2}\) and width \(dx\). The total area is the sum (integral) of all those strips – and finding that sum means finding an antiderivative.
\[\boxed{A'(x) = 2\sqrt{r^2 - x^2}}\]
Now the problem reduces to finding the antiderivative of \(2\sqrt{r^2 - x^2}\).
Explore – see the vertical slicing of a circle:
Drag \(a\) from \(-3\) to \(3\). The shaded blue region is \(A(a)\). The red strip is the infinitesimal slice \(dA\) — its height is \(2\sqrt{9 - a^2}\). When \(a\) reaches \(3\), the shaded area equals the full circle: \(\pi(3)^2 = 9\pi \approx 28.27\).
Review: Seven Families and Their Derivatives
Before tackling the antiderivative of \(\sqrt{r^2 - x^2}\), we review which functions produce which derivatives. Most families are “closed” — the derivative stays in the same family. The key exceptions are the inverse functions.
| Family | \(f(x)\) | \(f'(x)\) | Stays in family? |
|---|---|---|---|
| Power | \(x^n\) | \(nx^{n-1}\) | Yes |
| Polynomial | \(a_nx^n + \cdots + a_0\) | \(na_nx^{n-1} + \cdots\) | Yes |
| Rational | \(\dfrac{p(x)}{q(x)}\) | \(\dfrac{p'q - pq'}{q^2}\) | Yes |
| Trigonometric | \(\sin x,\;\cos x\) | \(\cos x,\;{-}\sin x\) | Yes |
| Inverse trig | \(\arcsin x\) | \(\dfrac{1}{\sqrt{1 - x^2}}\) | No — gives a radical |
| Exponential | \(a^x\) | \(a^x \ln a\) | Yes |
| Logarithmic | \(\ln x\) | \(\dfrac{1}{x}\) | No — gives a rational function |
The crucial observation: when you see a radical like \(\frac{1}{\sqrt{1-x^2}}\) and need its antiderivative, think inverse trig. When you see \(\frac{1}{x}\), think logarithm. The derivative goes “out” of the family, so the antiderivative goes back “in.”
Deriving the Derivative of \(\arcsin x\)
Let \(y = \arcsin x\). Then \(\sin y = x\). Differentiating both sides with respect to \(x\):
\[\cos y \;\frac{dy}{dx} = 1 \qquad \Longrightarrow \qquad \frac{dy}{dx} = \frac{1}{\cos y}\]
Now use the Pythagorean identity \(\sin^2 y + \cos^2 y = 1\) to rewrite \(\cos y\):
\[\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}\]
Therefore:
The derivative of \(\arcsin x\) produces a radical expression – it “leaves” the inverse trig family. This means that when you see \(\frac{1}{\sqrt{1-x^2}}\) and need its antiderivative, the answer is \(\arcsin x\). This is exactly the link that lets us integrate circular shapes.
\[\boxed{\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}}\]
Explore – see \(\arcsin x\) and its derivative:
Drag \(a\) between \(-1\) and \(1\). The green tangent line to \(\arcsin x\) (blue) has slope \(\frac{1}{\sqrt{1-a^2}}\) (red dashed). Notice the slope blows up near \(x = \pm 1\) — the curve becomes vertical there!
Deriving the Derivative of \(\ln x\) and \(a^x\)
Natural log. Let \(y = \ln x\), so \(e^y = x\). Differentiating both sides:
\[e^y \frac{dy}{dx} = 1 \qquad \Longrightarrow \qquad \frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}\]
\[\boxed{\frac{d}{dx}\ln x = \frac{1}{x}}\]
General exponential. Let \(y = a^x\), so \(\ln y = x\ln a\). Differentiating: \(\frac{1}{y}\frac{dy}{dx} = \ln a\), giving:
\[\boxed{\frac{d}{dx}\,a^x = a^x \ln a}\]
Antiderivative of \(\dfrac{1}{\sqrt{r^2 - x^2}}\)
Returning to the circle area problem, we need the antiderivative of \(\sqrt{r^2 - x^2}\). We split this into two parts. The first part leads to a term involving \(\frac{1}{\sqrt{r^2 - x^2}}\), which looks similar to the derivative of \(\arcsin x\).
Step 1 — Factor out \(r\). Pull \(r^2\) out of the square root:
\[\frac{1}{\sqrt{r^2 - x^2}} = \frac{1}{r\,\sqrt{1 - (x/r)^2}}\]
Step 2 — Guess the antiderivative. Since \(\frac{d}{dx}\arcsin(x) = \frac{1}{\sqrt{1 - x^2}}\), try \(\arcsin\!\left(\frac{x}{r}\right)\). By the chain rule:
\[\frac{d}{dx}\arcsin\!\left(\frac{x}{r}\right) = \frac{1}{\sqrt{1 - (x/r)^2}} \cdot \frac{1}{r} = \frac{1}{\sqrt{r^2 - x^2}}\]
This matches exactly, so:
\[\boxed{\int \frac{dx}{\sqrt{r^2 - x^2}} = \arcsin\!\left(\frac{x}{r}\right) + C}\]
Step 3 — Scale for the circle area. We need the antiderivative of the full expression \(\frac{r^2}{\sqrt{r^2 - x^2}}\), which appears after rewriting \(\sqrt{r^2 - x^2}\). Multiplying by \(r^2\):
\[\int \frac{r^2}{\sqrt{r^2 - x^2}}\,dx = r^2\,\arcsin\!\left(\frac{x}{r}\right) + C\]
The remaining piece of the circle area (involving \(x\sqrt{r^2 - x^2}\)) has a geometric interpretation as a triangular region — interpreting the arcsine term as a circular sector area reveals what the second term must be.
Explore – the arcsine antiderivative vs. the integrand:
The blue curve is \(\arcsin(x/r)\) (the antiderivative) and the red curve is \(\frac{1}{\sqrt{r^2 - x^2}}\) (the integrand). At every point, the red value equals the slope of the blue curve. Drag \(r\) to change the radius.
Cheat Sheet
| Formula | Meaning |
|---|---|
| \(x(t) = ut\) | Horizontal position under constant velocity \(u\) |
| \(y(t) = \tfrac{1}{2}gt^2\) | Vertical position under constant acceleration \(g\) |
| \(y = \dfrac{g}{2u^2}\,x^2\) | Parabolic trajectory of a horizontal projectile |
| \(dA = 2\sqrt{r^2 - x^2}\;dx\) | Infinitesimal area slice of a circle of radius \(r\) |
| \(\dfrac{d}{dx}\arcsin x = \dfrac{1}{\sqrt{1-x^2}}\) | Derivative of inverse sine |
| \(\dfrac{d}{dx}\ln x = \dfrac{1}{x}\) | Derivative of natural log |
| \(\dfrac{d}{dx}\,a^x = a^x\ln a\) | Derivative of a general exponential |
| \(\displaystyle\int \frac{dx}{\sqrt{r^2 - x^2}} = \arcsin\!\left(\frac{x}{r}\right) + C\) | Key antiderivative for the circle area problem |
The Seven Function Families
\[\frac{d}{dx}(x^n) = nx^{n-1} \qquad\longleftrightarrow\qquad \int x^n\,dx = \frac{x^{n+1}}{n+1}+C\]
\[\frac{d}{dx}(\sin x) = \cos x \qquad \frac{d}{dx}(\cos x) = -\sin x\]
\[\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1-x^2}} \qquad \frac{d}{dx}(\arccos x) = \frac{-1}{\sqrt{1-x^2}}\]
\[\frac{d}{dx}(e^x) = e^x \qquad \frac{d}{dx}(\ln x) = \frac{1}{x}\]
The Inverse Function Derivative Rule
If \(y = f^{-1}(x)\), then:
\[\frac{dy}{dx} = \frac{1}{f'(y)}\]
The derivative of an inverse function is the reciprocal of the derivative of the original function, evaluated at \(y\).