Module 1: Mathematical Foundations · Practice · assumes all eleven Foundations
Three problems that no single Foundation can solve — where choosing the order of attack is most of the work.
What this set is for
Every earlier post was organised by Foundation — resolution here, integration there. Examination problems are not. They are set on a situation, and the solution reaches for whichever tools the situation requires, in whatever order works.
Each problem below names the Foundations it draws on. None of them is difficult once set up; the difficulty is entirely in deciding what to do first, and that is the skill this set is for.
On this page
- The best direction to push — F4, F5, F8
- A particle released in a double well — F7, F8, F9, F10
- Why angular momentum is conserved — F6, F8
1. The best direction to push Foundations 4, 5 and 8
Problem. A particle in a plane experiences a position-dependent force F(x, y) = (2x + y)î + (x + 3y²)ĵ N. It sits at (2, 1) m and is constrained to a straight guide whose direction is û = cos θ î + sin θ ĵ. Find the force at that point, the component along the guide as a function of θ, and the θ that makes it largest.
The physics
Three things happen in sequence, and none can be done out of order. The force must first be evaluated at the particular point, because it varies with position. Then the part of it that acts along the guide must be extracted, which is a projection. Then that projection, now a function of the guide’s angle, must be maximised.
Worth predicting the answer before working: the component of a fixed vector along a direction is largest when the direction points along the vector itself. So the answer should be the direction of F, and the maximum should be |F|.
The mathematics
Evaluate the field at (2, 1).
Fx = 2(2) + 1 = 5 N Fy = 2 + 3(1)² = 5 N
F0 = 5î + 5ĵ N, of magnitude 5√2 ≈ 7.07 N
Project onto the guide — Foundation 5’s scalar projection, with û already a unit vector:
F∥(θ) = F0 · û = 5 cos θ + 5 sin θ
Maximise. Differentiate with respect to θ and set to zero:
dF∥/dθ = −5 sin θ + 5 cos θ = 0 → tan θ = 1
In [0, 2π): θ = π/4 or 5π/4
d²F∥/dθ² = −5(cos θ + sin θ), which is negative at π/4
So the maximum is at θ = π/4, and
F∥,max = 5/√2 + 5/√2 = 5√2 N
The other critical point, 5π/4, is the minimum — the guide pointing directly against the force, giving −5√2 N.
Both predictions hold: π/4 is the direction of 5î + 5ĵ, and 5√2 is its magnitude.
The contemplation
The calculus was unnecessary. F·û = |F| cos(angle between them), which is obviously largest when that angle is zero — no differentiation required, and the answer is available in one line. The derivative route works and is worth being able to execute, but a student who reaches for it first has not noticed that the dot product had already answered the question. Under time pressure that recognition is worth more than the technique. Look for the structure before reaching for the machinery.
2. A particle released in a double well Foundations 7, 8, 9 and 10
Problem. A particle of mass 1 kg moves along the x-axis with potential energy U(x) = x⁴ − 4x² + 3 joules. Locate and classify the equilibria. The particle is released from rest at x = 1 m: find its speed when it first reaches the stable point. Then find the period of small oscillations about that point.
The physics
Four tools, in a forced order. The equilibria come from the first derivative and their classification from the second. The speed needs the force accumulated over a distance — and the force is a function of position, not time, which rules out integrating with respect to t. And the small oscillations need the well approximated by a parabola near its bottom, which is where the approximation enters.
Predict first: released at x = 1, between the hump at the origin and the well, the particle should slide toward the well and pass through it moving quickly.
Released at x = 1, on the slope between the hump and the right-hand well.
The mathematics
Equilibria. Flat points of U:
dU/dx = 4x³ − 8x = 4x(x² − 2) = 0
x = 0, +√2, −√2
Classifying with the curvature, d²U/dx² = 12x² − 8: at the origin this is −8, so a maximum and unstable; at ±√2 it is +16, so minima and stable. The heights are U(0) = 3 J and U(±√2) = −1 J.
Speed at the well. The force is F = −dU/dx = 8x − 4x³, positive at x = 1 — pushing the particle toward the well, as the picture shows. Since F depends on position and the question mentions no time, use Foundation 9’s identity with m = 1:
v dv = a dx = (8x − 4x³) dx
∫0v v′ dv′ = ∫1√2 (8x − 4x³) dx
v²/2 = [4x² − x⁴]1√2 = (8 − 4) − (4 − 1) = 1
v = √2 ≈ 1.41 m/s
Check it another way. U(1) = 0 and U(√2) = −1, so the particle has lost 1 J of potential energy and gained 1 J of kinetic energy: ½(1)v² = 1, giving v = √2. The two routes agree, and the second is faster — but it works only because the force is conservative, which the first route never needed to assume.
Small oscillations. Near the bottom of any smooth well, the curve looks like a parabola. Expanding U about x = √2 to second order, the coefficient of the quadratic term is ½U″(√2) = 8, which is the same as an effective spring constant k = U″ = 16 N/m. So:
ω = √(k/m) = √16 = 4 rad/s
T = 2π/ω = π/2 ≈ 1.57 s
Note the condition. This period applies to small displacements about √2. The particle in this problem arrives with 1 J of kinetic energy, which is enough to climb the hump at the origin — so it does not oscillate in the right-hand well at all. It crosses to the left-hand well and back, and its actual period is not 1.57 s.
The contemplation
The last paragraph is the one worth sitting with. Every step of the small-oscillation calculation was correct, and the answer does not describe this particle — because it was released with enough energy to leave the region where the parabola is a good approximation. The calculation did not fail; it answered a question about a different particle, one released close to the bottom. Nothing in the algebra could have warned us. Only comparing the energy available with the height of the barrier does that, and it takes one line: 1 J of kinetic energy against a 4 J climb from −1 to 3 — which, as it happens, is not enough after all. Work that through yourself before deciding whether the last sentence above is right.
3. Why angular momentum is conserved Foundations 6 and 8
Problem. Starting from L = r × p, differentiate with respect to time to find what makes angular momentum change. Then show that a particle moving under a central force — one always directed toward or away from a fixed point — has constant angular momentum about that point.
The physics
Two quantities are changing inside L: the position and the momentum. So the product rule applies, and it applies to a cross product, which raises a question worth asking before proceeding — does the product rule hold for cross products, given that they are not commutative? It does, provided the order of each factor is preserved. That caution is the only subtlety in the whole calculation.
The mathematics
dL/dt = d(r × p)/dt = (dr/dt) × p + r × (dp/dt)
Take the two terms separately. The first is v × mv, and a vector crossed with itself is zero — so the entire first term vanishes. The second contains dp/dt, which by Newton’s second law is the net force:
dL/dt = 0 + r × F
dL/dt = τ
Torque is to angular momentum what force is to linear momentum. That was not assumed anywhere — it fell out of one differentiation.
The central force case. A central force points along the line joining the particle to the centre, so F is parallel or anti-parallel to r. The angle between them is 0° or 180°, and in either case sin θ = 0:
τ = r × F = 0
so dL/dt = 0, and L is constant
Gravity and the electrostatic force are both central, so every planetary orbit and every electron in a Coulomb field conserves angular momentum about the centre. Kepler’s second law — equal areas swept in equal times — is this statement in disguise.
The contemplation
A conservation law came out of a cross product vanishing. Not from an experiment, not from a principle laid down in advance — from the fact that v × v = 0 and that a central force lies along r. This is worth putting beside the earlier result that a particle travelling in a straight line has constant angular momentum: there, no force acted at all; here, a force acts but exerts no torque. Two quite different physical situations, one mathematical reason, and the mathematics saw the similarity before the physics did.
What made these hard
Not one of the individual steps was beyond a single Foundation. What made these problems different was the sequencing — evaluate before projecting, classify before releasing, differentiate before substituting — and in two of the three, the useful move was to notice that a shorter route existed.
That is what Module 1 was for. Eleven Foundations, each a tool, and the last skill is knowing which one a situation is asking for. From here the physics gets harder and these tools stay the same.
Assumes: all eleven Foundations · Type: Worked examples · Reading: 20 min · Difficulty: Advanced