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 / ii

Table of contents
- Solution ii

Solution

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.

 Ancestors (13)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version