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 Incoming links: Inverse hyperbolic tangent

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 74 / 2 / Solution 2026-10-03

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Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 2 / 8C / c / Solution Created 2026-09-24 Updated 2026-09-29

The initial point has

H = 1/4

. On the branch leaving

(1/2, 0)

toward negative

y

,

x^{2} = \frac{1}{4} - y^{2} + y^{4} = (\frac{1}{2} - y^{2})^{2}, x = \frac{1}{2} - y^{2} .

Since

\overset{y}{˙} = - x

, the required scalar equation is

\overset{y}{˙} = y^{2} - \frac{1}{2}, y (0) = 0.

Its solution is

y (t) = - \frac{1}{2} tanh (\frac{t}{2}) .

At

(x, y) = (1/4, - 1/2)

, this requires

tanh (t / 2) = 1/ 2

. Using the inverse hyperbolic tangent,

t = 2 artanh \frac{1}{2} = 2 lo g (1 + 2) .

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