Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 2 / 13 / a

On the first segment, parameterize

r_{1} (t) = (t, t, 0)^{T}

0 \leq t \leq 1

r_{1}^{'} = (1, 1, 0)^{T}

are

F (r_{1}) \cdot r_{1}^{'} = 2 t, G (r_{1}) \cdot r_{1}^{'} = 2 t cos (t^{2}) .

Their line integrals are one and

sin 1

, respectively.

On the second segment take

r_{2} (t) = (1, 1, t)^{T}

(0, 0, 1)^{T}

. The corresponding integrands are

a - 2 + 2 t

and

c

, whose integrals are

a - 1

and

c

. Adding the segments yields

\int_{C_{1}} F \cdot d r = a, \int_{C_{1}} G \cdot d r = sin 1 + c .

ii

On the curved path,

r (t) = (t, t, t^{2})^{T}

and

r^{'} = (1, 1, 2 t)^{T}

. Direct substitution gives

F (r (t)) \cdot r^{'} (t) = 2 t + (3 a - 6) t^{2} + 4 t^{3},

G (r (t)) \cdot r^{'} (t) = 2 t cos (t^{2}) + 3 c t^{2} .

Thus

\int_{C_{2}} F \cdot d r = 1 + (a - 2) + 1 = a, \int_{C_{2}} G \cdot d r = sin 1 + c .

Both values agree with part (i), for every

a, b, c

. Agreement on these two selected paths alone does not establish a conservative vector field.

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