Slopes of curves that are not graphs, a ladder sliding down a wall, and water rising in tanks of every shape
Many quantities are not given as y = f(x) but are tied together by an equation: a point on a circle, the two ends of a ladder, the depth and volume of water in a tank. Differentiate the equation itself, using the chain rule wherever a quantity depends on another, and you get a new equation that ties their rates of change together.
Each curve is the set of points where F(x, y) = 0, and most of them fail the vertical line test. Drag the point around the curve to see its tangent, computed from the slope formula −F_x/F_y, and the gradient arrow standing perpendicular to it. Type your own F to explore.
F_x = 3x² − 3y
F_y = 3y² − 3x
dy/dx = −F_x/F_y = −(−0.06)/(2.795) = 0.02147
Drag anywhere: the point snaps to the nearest part of the curve. The curve is traced by marching squares on a 240 × 240 grid, so a curve that only touches zero without changing sign (such as x² + y² = 0) will not show. At a self-crossing, like the folium’s loop at the origin, both partials vanish and the curve has two tangents, not one.
Near most of its points, a curve F(x, y) = 0 looks like the graph of some function y(x), even when no formula for y(x) is available. Differentiate F(x, y(x)) = 0 with the chain rule:
F_x + F_y · dy/dx = 0, so dy/dx = −F_x / F_y (where F_y ≠ 0).
For the circle x² + y² = 4, differentiating term by term gives 2x + 2y·y′ = 0, so y′ = −x/y. At (√2, √2) the slope is −1; at (2, 0) the formula divides by zero, and the tangent is vertical. The implicit function theorem makes this precise: where F_y ≠ 0, the curve really is the graph of a differentiable y(x) nearby.
Where both partial derivatives vanish, the formula gives 0/0 and the curve may misbehave: the folium of Descartes, x³ + y³ = 3xy, crosses itself at the origin, and the lemniscate does the same. Descartes proposed the folium in 1638 as a challenge to Fermat, whose tangent method handled it with ease.
The pair (F_x, F_y) is the gradient ∇F. Moving along the curve keeps F constant, so the direction of motion has no component along ∇F: the gradient is always perpendicular to the curve. This is the starting point for level curves and steepest ascent in several variables.
The foot of a ladder is pulled away from a wall at a steady speed. Press pull and watch the top: slow at first, then faster and faster as it nears the floor. Turn on the envelope to see the curve the ladder sweeps out.
x² + y² = L² at every instant, so 2x·dx/dt + 2y·dy/dt = 0 and dy/dt = −(x/y)·dx/dt = −(1.4/4.8)·0.5 = −0.1458 m/s. The top is moving more slowly than the foot.
This is the textbook model: the top stays on the wall and x² + y² = L² holds all the way down. A real ladder does not behave like this. Long before the top’s speed becomes enormous, it loses contact with the wall and the ladder falls; for a uniform ladder sliding freely on frictionless surfaces from rest, the top leaves the wall once it has dropped to two-thirds of its starting height.
Let the foot be x metres from the wall and the top y metres up. Both change with time t, but the ladder’s length does not:
x(t)² + y(t)² = L².
Differentiate both sides with respect to t, using the chain rule on each square:
2x·dx/dt + 2y·dy/dt = 0 ⇒ dy/dt = −(x/y)·dx/dt.
For a 5 m ladder with its foot 3 m out, pulled at 0.5 m/s, the top is 4 m up and falls at (3/4)(0.5) = 0.375 m/s. With the foot 4.8 m out the top is 1.4 m up and falls at about 1.71 m/s, and the ratio x/y has no upper bound as y → 0.
The recipe for any related-rates problem is the same: write the equation that links the quantities at every instant, differentiate it with respect to time, then substitute the values at the instant you care about. Substituting before differentiating is the classic mistake; it freezes quantities that are really changing.
Each position of the ladder is a segment from (x, 0) to (0, y). The curve that touches every one of them, their envelope, is the astroid x^(2/3) + y^(2/3) = L^(2/3), parametrised as (L cos³θ, L sin³θ).
Water pours in at a constant rate, but the level does not rise at a constant rate. Pick a vessel, press pour, and compare how fast the surface climbs with the size of the glowing cross-section at the top of the water.
dh/dt = Q/A(h) = 1500 cm³/s ÷ 175.6 cm² = 8.54 cm/s. The vessel widens above the surface, so each centimetre needs more water: the level is slowing down.
The vessels are solids of revolution 40 cm tall; 1 litre is 1000 cm³. The depth is found by inverting the volume V(h) = ∫₀ʰ π r(s)² ds, tabulated numerically. The rate plot is capped where the cone and sphere come to a point, because there A(h) → 0 and dh/dt → ∞ for an instant.
If A(h) is the area of the horizontal cross-section at height h, the volume of water at depth h is V(h) = ∫₀ʰ A(s) ds. Raising the level by a thin layer dh adds about A(h)·dh of water, so dV/dh = A(h) (this is the fundamental theorem of calculus in disguise). By the chain rule,
dV/dt = A(h)·dh/dt ⇒ dh/dt = Q / A(h).
For a cylinder A is constant, so the level rises at a constant rate. For a cone with its point down, r = kh and V = πk²h³/3; with constant inflow, h grows like the cube root of t, very fast at first and then ever more slowly. In a sphere the level slows down until the widest point, then speeds up again.
The reasoning runs the other way too. A graph of h(t), or of dh/dt, determines the shape of the vessel: the wider the vessel at height h, the more slowly the level rises there. This is the idea behind a graduated measuring jug with uneven markings, and behind tapered water clocks: the ancient Egyptian clepsydra at Karnak, from around 1400 BC, has sloping sides so that its water level would fall at a more nearly steady rate.
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