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Before You Begin: The Mathematics Physics Assumes

School physics is a collection of formulas. University physics is a small set of principles, expressed in vectors and calculus, from which the formulas are derived. The transition fails not because the physics gets harder but because the language changes: a statement like ”F=ma\mathbf{F} = m\mathbf{a} where a=r¨\mathbf{a} = \ddot{\mathbf{r}}” assumes you can read vectors, differentiate, and treat an equation between vector quantities as three equations at once.

This page rebuilds that language from first principles; vectors, trigonometry, coordinate geometry, and calculus; defining every term before it is used and proving every rule. A confident reader should take the self-test in §5 and only study the sections whose answers surprise.

Physics quantities come in two kinds. A scalar; mass, energy, temperature - is a number. A vector; displacement, velocity, force; is a number and a direction. The single most important upgrade from school physics is learning to compute with directions: a car turning a corner at constant speed is accelerating, because its velocity vector is changing. Nothing in school algebra can say that sentence; vectors can.

The second upgrade is calculus. Speed is the derivative of position; acceleration is its second derivative. Every equation of motion; the ones school gives you as s=ut+12at2s = ut + \frac{1}{2}at^2; is a differential equation, and the school formulas are what you get by solving it for constant acceleration. University physics derives them; this page gives you the differentiation that derivation needs.

Definition. A vector v\mathbf{v} is a quantity with magnitude and direction, written ∣v∣|\mathbf{v}| for its magnitude. Two vectors are equal when their magnitudes and directions agree.

Components. In a plane, v=vx i^+vy j^\mathbf{v} = v_x\,\hat{\mathbf{i}} + v_y\,\hat{\mathbf{j}} where i^,j^\hat{\mathbf{i}}, \hat{\mathbf{j}} are unit vectors along the axes and

vx=∣v∣cos⁡θ,vy=∣v∣sin⁡θ,∣v∣=vx2+vy2.v_x = |\mathbf{v}|\cos\theta, \qquad v_y = |\mathbf{v}|\sin\theta, \qquad |\mathbf{v}| = \sqrt{v_x^2 + v_y^2}.

These four identities resolve every “resolve the force along the slope” exercise, and they are the definition of component rather than a trick to memorise.

Addition by components. (ux+vx,  uy+vy)(u_x + v_x,\; u_y + v_y); the parallelogram rule from school is component addition, proved by applying each vector’s definition.

Scalar multiplication. c vc\,\mathbf{v} has magnitude ∣c∣∣v∣|c||\mathbf{v}|, direction preserved for c>0c > 0 and reversed for c<0c < 0. Force balance ∑F=0\sum\mathbf{F} = 0 is a statement about component sums, and it is why a book on a table has weight balanced by a normal force of equal magnitude and opposite direction.

The dot product.

u⋅v=uxvx+uyvy=∣u∣ ∣v∣cos⁡θ.\mathbf{u}\cdot\mathbf{v} = u_xv_x + u_yv_y = |\mathbf{u}|\,|\mathbf{v}|\cos\theta.

Proof of the equivalence. Both sides are bilinear in the components, so it suffices to check the two unit vectors along the axes: i^⋅i^=1=1⋅1cos⁡0\hat{\mathbf{i}}\cdot\hat{\mathbf{i}} = 1 = 1\cdot1\cos 0 and i^⋅j^=0=cos⁡90°\hat{\mathbf{i}}\cdot\hat{\mathbf{j}} = 0 = \cos 90°; bilinearity then extends it to every pair. ■\blacksquare

The dot product answers “how much of u\mathbf{u} points along v\mathbf{v}” - which is the question every projection, every work integral, and every component-of-gravity-along-a-slope calculation is asking.

Definition (radian). An angle θ\theta radians subtends arc length θ\theta on the unit circle. All calculus below assumes radians.

The identities, with proofs rather than a list.

  • sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1: the equation of the unit circle.
  • sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta: obtained by rotating the point (cos⁡α,sin⁡α)(\cos\alpha, \sin\alpha) through β\beta.
  • sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\theta and cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2\sin^2\theta: the addition formulas at α=β\alpha = \beta.

The two limits every physics derivation uses:

lim⁡h→0sin⁡hh=1,lim⁡h→01−cos⁡hh2=12.\lim_{h\to 0}\frac{\sin h}{h} = 1, \qquad \lim_{h\to 0}\frac{1 - \cos h}{h^2} = \frac{1}{2}.

They are why ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x = \cos x in radians; and why that derivative is false in degrees. If you have ever differentiated sin⁡\sin in degrees and gotten a wrong trajectory, this is why.

Small-angle physics. sin⁡θ≈θ\sin\theta \approx \theta and cos⁡θ≈1−θ2/2\cos\theta \approx 1 - \theta^2/2 for small θ\theta, which is the linearisation that makes pendulums harmonic and lenses computable; and the second-order term is what makes the correction, so a pendulum’s period grows with amplitude rather than staying constant.

Definition. The gradient of the line through (x1,y1)(x_1,y_1), (x2,y2)(x_2,y_2) is m=y1−y2x1−x2m = \frac{y_1 - y_2}{x_1 - x_2}; two lines are perpendicular iff m1m2=−1m_1 m_2 = -1.

Proof of perpendicularity. Direction vectors (1,m1)(1, m_1) and (1,m2)(1, m_2) are perpendicular iff their dot product vanishes: 1+m1m2=01 + m_1 m_2 = 0. ■\blacksquare

Straight-line graphs as linear models. y=mx+cy = mx + c is the statement that the response changes at a constant rate mm per unit of the input. Every “equation of the form y=mx+cy = mx + c” exercise in school is a linear model, and reading mm and cc off a graph is reading a derivative and an initial condition.

Areas and the quadratic. Projectile range R=v2sin⁡2θgR = \frac{v^2\sin 2\theta}{g} is maximised at θ=45°\theta = 45° because sin⁡2θ\sin 2\theta peaks at 11 when 2θ=90°2\theta = 90°; a trigonometry fact doing the work of a differentiation, and a preview of how the university course will use both.

Definition. f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h\to 0}\frac{f(x+h) - f(x)}{h} when the limit exists; geometrically the gradient of the tangent.

The table, with the two proofs that matter.

f(x)f(x)f′(x)f'(x)
xrx^rrxr−1rx^{r-1}
sin⁡x\sin xcos⁡x\cos x
cos⁡x\cos x−sin⁡x-\sin x
exe^xexe^x

Proof for xrx^r from first principles. Expand (x+h)r(x+h)^r by the binomial theorem, subtract xrx^r, divide by hh; every surviving term carries one factor hh, so the limit is rxr−1r x^{r-1}. ■\blacksquare

Proof for sin⁡\sin. From the addition formula, sin⁡(x+h)−sin⁡xh=sin⁡x cos⁡h−1h+cos⁡x sin⁡hh→cos⁡x\frac{\sin(x+h)-\sin x}{h} = \sin x\,\frac{\cos h - 1}{h} + \cos x\,\frac{\sin h}{h} \to \cos x, using the two limits of §2. ■\blacksquare

Rules. Product: (fg)′=f′g+fg′(fg)' = f'g + fg'. Chain: (g∘f)′=g′(f) f′(g \circ f)' = g'(f)\,f'. The chain rule is the one that makes physics work: if xx is a function of tt and v=dxdtv = \frac{dx}{dt}, then dvdt=dvdx⋅dxdt=v dvdx\frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt} = v\,\frac{dv}{dx}, which converts every acceleration-into-position problem into a first-order separable equation.

Integration. ∫abf dx\int_a^b f\,dx is the signed area under the curve, and the fundamental theorem of calculus says it is computed by antiderivatives: ∫abf=F(b)−F(a)\int_a^b f = F(b) - F(a). The work–energy theorem W=∫F⋅dr=Δ(12mv2)W = \int \mathbf{F}\cdot d\mathbf{r} = \Delta(\tfrac12 mv^2) is that theorem applied to Newton’s second law.

Differential equations, in one move. The statement

d2xdt2=−ω2x\frac{d^2 x}{dt^2} = -\omega^2 x

is the simple harmonic motion equation, and x=Acos⁡(ωt+ϕ)x = A\cos(\omega t + \phi) is its solution; verifiable by differentiating twice. Verification is a legitimate solution method at this level, and it is how the school formulas are recovered from the principles.

  1. A force of 50 N50\,\mathrm{N} acts at 37°37° above the horizontal. Give its horizontal and vertical components. (§1)
  2. Two forces F1=(3,4)\mathbf{F}_1 = (3, 4) and F2=(−4,3)\mathbf{F}_2 = (-4, 3) act on a body. Find the resultant’s magnitude. (§1)
  3. What is cos⁡2θ\cos 2\theta in terms of sin⁡θ\sin\theta only? Prove it. (§2)
  4. Why is ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x = \cos x false in degrees? (§2)
  5. Show that the lines y=3x+1y = 3x + 1 and x+3y=7x + 3y = 7 are perpendicular. (§3)
  6. Differentiate f(x)=x2sin⁡xf(x) = x^2\sin x. (§4)
  7. Verify that x=Acos⁡(ωt)x = A\cos(\omega t) solves x¨=−ω2x\ddot{x} = -\omega^2 x. (§4)
  8. A ball is thrown at 20 m/s20\,\mathrm{m/s} at 30°30°. Find its range using the double-angle identity, then state the angle that maximises it. (§2, §3)

Answers. (1) (50cos⁡37°,50sin⁡37°)≈(39.9,30.1) N(50\cos 37°, 50\sin 37°) \approx (39.9, 30.1)\,\mathrm{N}. (2) (3−4, 4+3)=(−1,7)(3-4,\, 4+3) = (-1, 7), magnitude 50≈7.07\sqrt{50} \approx 7.07. (3) cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2\sin^2\theta, from cos⁡(θ+θ)\cos(\theta+\theta) with the addition formula. (4) Because sin⁡hh→π180\frac{\sin h}{h} \to \frac{\pi}{180} when hh is in degrees, so the derivative picks up that factor. (5) Gradients 33 and −13-\tfrac13; product −1-1. (6) 2xsin⁡x+x2cos⁡x2x\sin x + x^2\cos x by the product rule. (7) x˙=−Aωsin⁡(ωt)\dot{x} = -A\omega\sin(\omega t), x¨=−Aω2cos⁡(ωt)=−ω2x\ddot{x} = -A\omega^2\cos(\omega t) = -\omega^2 x. (8) R=400sin⁡60°9.81≈35.3 mR = \frac{400\sin 60°}{9.81} \approx 35.3\,\mathrm{m}; maximised at θ=45°\theta = 45°.