Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 11 / d / ii

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- Solution ii

Solution

The graph is the upper branch of the hyperbola

d^{2} - (t + 1)^{2} = 1/2

. It is symmetric about

t = - 1

, positive everywhere, and has its lowest point at

(- 1, 1/ 2)

. Its derivatives are

d^{'} (t) = \frac{t + 1}{d ( t )}, d^{''} (t) = \frac{1}{2 d ( t ) ^{3}} > 0,

so it is strictly convex. Since

d (t) - ∣ t + 1∣ = \frac{1/2}{( t + 1 ) ^{2} + 1/2 + ∣ t + 1∣} ⟶ 0,

its two oblique asymptotes are

d = t + 1 (t \to + \infty), d = - t - 1 (t \to - \infty) .

Figure 1.
Distance from a moving point to a skew line
. Original plot of $d (t) = (t + 1)^{2} + 1/2$ , its minimum and its two asymptotes. Details.

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