Math AA HL IA ideas: differential equations, catenaries and fastest paths

1. Water draining from a bottle using differential equations

Estimate a draining time from a separable differential equation, then test whether changing the outlet produces the predicted effect.

Research question: Can a differential equation model how water height changes as a bottle drains, and what hole size gives a target draining time?

Use a teacher-approved, prepared container vented to the air, with constant cross-sectional area S and a small outlet open to the air. Measure height h above the outlet centre, with initial height h₀. From volume conservation and the ideal Bernoulli outflow relation, derive dh/dt = −k√h. Here k = C_da√(2g)/S, where a is outlet area and C_d represents discharge losses. A tapering bottle needs a changing S(h); the constant-k model does not apply unchanged.

Fit √h against time: its predicted slope is −k/2. Use the initial condition to obtain h(t) = (√h₀ − kt/2)², only for 0 ≤ t ≤ 2√h₀/k. Beyond that time the squared expression would rise again, which is not a physical prediction. For an illustrative h₀ = 0.20 m and k = 0.020 √m/s, the model reaches the outlet level after about 44.7 s and predicts h(10 s) ≈ 0.121 m. These are model calculations, not measured results.

In the formula box, A = √h₀ and B = k/2 are integration constants; S above denotes the tank area. With k > 0, differentiating the height model gives h″ = k²/2 > 0 before the outlet is reached. This checks the direction of the curve as well as its decreasing height.

Mathematics used: Differential equations, separation of variables, initial conditions, parameter estimation and model validation.

Data to collect: Repeated height–time measurements at fixed initial height; estimate uncertainty and compare predicted heights with observations, including a trial not used for fitting. Compare outlet radii only while container area and discharge behaviour remain comparable.

Diagram to include: Container cross-section with h measured above the outlet, outlet radius r and constant tank area S. The predicted height–time curve decreases but is concave upward before it reaches h = 0.

Pilot check: Can video resolve a meaningful height change before the outlet is reached? Test the fitted k on another run before estimating an outlet for a target time; report a range based on uncertainty in the measurements and fitted k.

2. Catenary curve of a hanging chain or cable

Compare a parabola with a catenary on the same chain coordinates, then calculate a length or sag that can be checked independently.

Research question: Does a hanging chain follow a catenary more accurately than a parabola, and how can the model predict cable length and sag?

Use a uniform, flexible chain hanging under its own weight between two supports. Photograph it perpendicular to its plane with a scale in that plane, so perspective does not distort the coordinates. Measure y upwards and set x = 0 at the lowest point for y = a cosh(x/a) + c, with a > 0; otherwise include a horizontal shift. Explain the catenary assumptions before applying the model to a stiff cable or a chain carrying extra loads.

Fit both models to the same coordinates and compare residual patterns as well as squared error. Derive the catenary arc length L = a[sinh(x₂/a) − sinh(x₁/a)] and compare it with a separate length measurement. At small sag, cosh(x/a) ≈ 1 + x²/(2a²), so a parabola can be hard to distinguish within photographic uncertainty. More decimal places in the fitted coefficients do not resolve that limitation.

Mathematics used: Hyperbolic functions, curve fitting, residuals, arc length and model comparison.

Data to collect: Coordinate points from a photographed chain, fitted to parabolic and catenary models.

Diagram to include: Photo grid with chain points, overlaid parabola and catenary, plus a residual comparison.

Pilot check: Photograph one shallow and one deeper sag. If the predicted model difference is smaller than coordinate uncertainty, change the setup or make that indistinguishability the question.

3. Fastest ramp shape for a rolling marble

Compare predicted and measured travel times for three paths sharing endpoints. This tests a defined set of ramps; it does not prove a globally fastest curve.

Research question: Which ramp shape gives the shortest travel time for a marble: straight, circular or cycloid-like?

Keep endpoints, release method and ball fixed, and record repeated runs. Define y as downward displacement for the cycloid below. Derive v = √(2gΔh) for an ideal sliding point mass. For a uniform solid sphere rolling without slipping from rest, use I = 2mr²/5 and translational plus rotational energy to obtain v = √((10/7)gΔh). A hollow ball has a different inertia, and slipping or energy losses require another model.

Approximate time along the ball-centre path with the segment sum below, evaluating velocity at segment midpoints rather than dividing the first segment by the zero starting speed. Halve the segment size and report whether the predicted times and ranking stabilise. Compare that numerical change with variation between trials. Track curvature and ball radius if the centre path differs appreciably from the drawn ramp, and check that the ball maintains contact.

Mathematics used: Parametric equations, energy-based modelling, numerical summation and convergence.

Data to collect: Travel time for three ramp shapes with identical start and end points, repeated and compared with numerical predictions.

Diagram to include: Three ramp curves between the same endpoints, with section lengths and vertical drop marked.

Pilot check: Can you reproduce a release without a push, see any slipping and measure a time difference larger than trial variation? Restrict the conclusion to the tested shapes and assumptions.