 Solution

ID: past-exam-of-the-computer-science-course-of-the-university-of-oxford/2015/geometric-modelling/2/g/solution

The join has matching positions and first derivatives:

F (1) = E (0) = (0, 0), F^{'} (t) = (4 - 4 t, 2 - 4 t), E^{'} (t) = (4 t, - 2 + 4 t),

F^{'} (1) = E^{'} (0) = (0, - 2) \neq = (0, 0) .

Because both pieces have unit parameter intervals, these are also the one-sided derivatives of

S

u = 1

. Thus

S

has first-order parametric continuity, and hence first-order geometric continuity, with no corner or cusp. Its nonzero tangent vector points vertically downward.

F^{''} (1) = (- 4, - 4), E^{''} (0) = (4, 4) .

So this is

C^{1}

but not

C^{2}

κ = (x^{'} y^{''} - y^{'} x^{''}) / (x^{'2} + y^{'2})^{3/2}

has one-sided values

κ_{F} (1) = \frac{- 8}{8} = - 1, κ_{E} (0) = \frac{8}{8} = 1.

The unsigned curvatures agree, but the bending switches sides; there is no second-order geometric continuity through a regular orientation-preserving reparametrization. The join is a reversal of bending with a curvature jump, rather than a smooth zero-curvature inflection.

For the “gradient” compute

d y / d x = y^{'} / x^{'}

away from the join:

\frac{F _{y}^{'} ( t )}{F _{x}^{'} ( t )} = \frac{2 - 4 t}{4 - 4 t} ⟶ - \infty (t \to 1^{-}), \frac{E _{y}^{'} ( t )}{E _{x}^{'} ( t )} = \frac{- 2 + 4 t}{4 t} ⟶ - \infty (t \to 0^{+}) .

Hence

no finite d y / d x at the join; the tangent line is x = 0.

The zero horizontal derivative is not a stationary point of the parametric curve, since its vertical derivative is

- 2

. It means that the locally regular curve has a vertical tangent line and cannot have a finite graph derivative as

y (x)

there.

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