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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge
/
2014
/
Paper 2
/
9
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Year of the natural sciences course of the University of Cambridge
Year 1 of the natural sciences course of the University of Cambridge
Mathematics course of the year 1 of the natural sciences course of the University of Cambridge
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge
2014
Paper 2
2026-10-18
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Table of contents
a
9
Solution
a
b
9
Solution
b
a
0
0
0
9
Solution
0
0
0
a
Only the
x
component of the
vector field
is nonzero, so
F
⋅
d
r
=
(
x
2
+
y
2
)
d
x
. On the vertical segment,
d
x
=
0
throughout. Hence
∫
C
F
⋅
d
r
=
0
.
(1)
b
0
0
0
9
Solution
0
0
0
b
On the horizontal segment
y
=
0
, while
x
decreases from
1
to
−
1
. Keeping this orientation gives
∫
C
F
⋅
d
r
=
∫
1
−
1
x
2
d
x
=
−
3
2
.
(1)
The minus
sign
follows from traversing opposite to the positive
x
direction.
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(11)
Paper 2
2014
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge
Mathematics course of the year 1 of the natural sciences course of the University of Cambridge
Year 1 of the natural sciences course of the University of Cambridge
Year of the natural sciences course of the University of Cambridge
Natural sciences course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
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