Why This Unit Exists
Calculus did not begin as a collection of classroom procedures. It grew out of physical questions: How fast is something changing right now? How much has accumulated over time? How do small changes add up to a complete motion, area, or impulse?
A derivative measures an instantaneous rate of change. An integral adds up many small pieces. A vector carries both magnitude and direction. These are not just math rules; they are the language of mechanics.
We will clean up the core tools students need before serious AP Physics C work: derivative rules, integral rules, u-substitution, vector notation, dot products, and cross products.
1. Derivatives: Slope, Rate, and Instantaneous Change
You first learned slope as rise over run:
The derivative asks what happens when the two points used for slope get infinitely close together.
Derivative notation
These are common ways to say the same idea:
\(f'(x)\)\(\frac{d}{dx}f(x)\)\(\frac{dy}{dx}\)\(\dot{x}\) for time derivatives
Power rule
Examples:
\[f(x)=x^2\]
\[f'(x)=2x\]
\[f(x)=3x^2+2x+4\]
\[f'(x)=6x+2\]
\[f(x)=x^6\]
\[f'(x)=6x^5\]
Physical meaning: if \(x(t)\) is position, then \(\frac{dx}{dt}=v(t)\). If \(v(t)\) is velocity, then \(\frac{dv}{dt}=a(t)\). If \(\theta(t)\) is angle, then \(\frac{d\theta}{dt}=\omega(t)\).
2. Integrals: Area, Accumulation, and Adding Tiny Pieces
An integral adds up small pieces. A rectangle has area \(A=hw\). If the height changes from one small piece to the next, we use many thin rectangles of width \(dx\) and height \(f(x)\).
Power rule for integration
The constant \(C\) appears for indefinite integrals because many functions can have the same derivative.
Indefinite integral
Finds a family of antiderivatives.
\[\int x^3\,dx=\frac{x^4}{4}+C\]
Definite integral
Finds a numerical accumulation over bounds.
\[\int_{-2}^{2}x^3\,dx=\left.\frac{x^4}{4}\right|_{-2}^{2}=0\]
Physical meaning: \(\int a(t)\,dt\) gives change in velocity, \(\int v(t)\,dt\) gives change in position, and \(\int F(t)\,dt\) gives impulse, or change in momentum.
3. Brief Focus: U-Substitution
U-substitution reverses the chain rule. It is useful when one expression is nested inside another and its derivative is also present.
Let \(u=g(x)\). Then \(du=g'(x)dx\), so the integral becomes simpler.
Example
Evaluate \(\int 2x(x^2+3)^4\,dx\).
Let \(u=x^2+3\). Then \(du=2x\,dx\).
\[\int 2x(x^2+3)^4\,dx=\int u^4\,du=\frac{u^5}{5}+C=\frac{(x^2+3)^5}{5}+C\]
Common AP Physics C use: u-substitution often appears when force, acceleration, or electric field depends on position through a square root, power, or expression like \((R^2+x^2)\).
4. Vectors in \(\hat{i},\hat{j},\hat{k}\) Notation
A scalar has magnitude only. A vector has magnitude and direction.
Scalars
distance, speed, mass, time, energy, temperature
Vectors
displacement, velocity, acceleration, force, momentum, impulse, torque
The unit vectors define the coordinate directions:
A general 3D vector is written as
Its magnitude comes from the 3D Pythagorean theorem:
Example
For \(\vec{a}=2\hat{i}+3\hat{j}-\hat{k}\),
\[|\vec{a}|=\sqrt{2^2+3^2+(-1)^2}=\sqrt{14}\]
Adding vectors
Add like components:
5. Dot Product: Projection and Work
The dot product multiplies like components and adds the results.
It also has a geometric definition:
Physical meaning: the dot product measures how much one vector points in the direction of another. That is why work is \(W=\vec{F}\cdot\Delta\vec{x}=F\Delta x\cos\theta\).
Example
Let \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\) and \(\vec{b}=-\hat{i}-\hat{j}+\hat{k}\).
\[\vec{a}\cdot\vec{b}=(1)(-1)+(1)(-1)+(1)(1)=-1\]
6. Cross Product: Perpendicular Direction and Rotation
The cross product produces a vector perpendicular to the two original vectors.
The direction comes from the right-hand rule. In physics, this appears in torque, angular momentum, and magnetic force.
Formal determinant method
Visual diagonal method
Rewrite the first two columns to the right, multiply down-right diagonals as positive, and down-left diagonals as negative.
Example
Let \(\vec{a}=2\hat{i}-\hat{j}+3\hat{k}\) and \(\vec{b}=-\hat{i}+4\hat{j}+\hat{k}\).
\[\vec{a}\times\vec{b}=(-1\cdot1-3\cdot4)\hat{i}-(2\cdot1-3\cdot(-1))\hat{j}+(2\cdot4-(-1)(-1))\hat{k}\]
\[\vec{a}\times\vec{b}=-13\hat{i}-5\hat{j}+7\hat{k}\]
7. Physics Connections You Should Recognize Immediately
| Math operation | Physics meaning | Example |
|---|---|---|
| Derivative | Instantaneous rate of change | \(v=\frac{dx}{dt}\), \(a=\frac{dv}{dt}\) |
| Integral | Accumulation of small changes | \(\Delta x=\int v\,dt\), \(\Delta v=\int a\,dt\) |
| Area under \(F(t)\) | Impulse | \(J=\int F\,dt=\Delta p\) |
| Dot product | Parallel component / projection | \(W=\vec{F}\cdot\Delta\vec{x}\) |
| Cross product | Rotational effect / perpendicular vector | \(\vec{\tau}=\vec{r}\times\vec{F}\) |
8. Practice Problems with Hidden Solutions
Derivative as velocity
A particle has position \(x(t)=3t^2+2t+4\). Find \(v(t)\) and \(v(2)\).
Show solution
\[v(t)=\frac{dx}{dt}=6t+2\]
\[v(2)=14\]
Integral as displacement
A particle has velocity \(v(t)=t^2+4t\). Find the displacement from \(t=0\) to \(t=3\).
Show solution
\[\Delta x=\int_0^3(t^2+4t)dt=\left.\left(\frac{t^3}{3}+2t^2\right)\right|_0^3=9+18=27\]
U-substitution
Evaluate \(\int 6x(3x^2+1)^5dx\).
Show solution
Let \(u=3x^2+1\), so \(du=6x dx\).
\[\int 6x(3x^2+1)^5dx=\int u^5du=\frac{u^6}{6}+C=\frac{(3x^2+1)^6}{6}+C\]
Vector magnitude
Find the magnitude of \(\vec{a}=4\hat{i}+0\hat{j}+3\hat{k}\).
Show solution
\[|\vec{a}|=\sqrt{4^2+0^2+3^2}=5\]
Dot product and work
A force \(\vec{F}=3\hat{i}+4\hat{j}\,\text{N}\) moves an object through \(\Delta\vec{x}=2\hat{i}+1\hat{j}\,\text{m}\). Find the work.
Show solution
\[W=\vec{F}\cdot\Delta\vec{x}=(3)(2)+(4)(1)=10\text{ J}\]
Cross product and torque
A force \(\vec{F}=0\hat{i}+5\hat{j}+0\hat{k}\) N is applied at \(\vec{r}=2\hat{i}+0\hat{j}+0\hat{k}\) m. Find \(\vec{\tau}=\vec{r}\times\vec{F}\).
Show solution
\[\vec{\tau}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&0&0\\0&5&0\end{vmatrix}=10\hat{k}\text{ N m}\]