Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 3 / 4A (Vector Calculus)

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- Solution 4A

Solution

Use

x > 0

as the parameter of the smooth curve:

r (x) = (x, lo g x, 0) .

Its velocity vector and speed are

r^{'} (x) = (1, \frac{1}{x}, 0), ∣ r^{'} (x) ∣ = \frac{x ^{2} + 1}{x} .

The unit tangent vector is therefore

t (x) = \frac{( x , 1 , 0 )}{x ^{2} + 1} .

The parameter-independent formula for the curvature of a space curve gives

κ (x) = \frac{∣ r ^{'} ( x ) \times r ^{''} ( x ) ∣}{∣ r ^{'} ( x ) ∣ ^{3}} = \frac{x}{( x ^{2} + 1 ) ^{3/2}} .

Differentiating,

κ^{'} (x) = \frac{1 - 2 x ^{2}}{( x ^{2} + 1 ) ^{5/2}} .

Hence

κ

increases for

0 < x < 1/ 2

and decreases for

x > 1/ 2

. Its global maximum occurs at

x = 1/ 2

and equals

κ_{m a x} = \frac{2}{3 3} .

The corresponding point is

(\frac{1}{2}, - \frac{1}{2} lo g 2, 0) .

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