Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 2 / 5 / a

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii

i

Solution

Parametrize the straight line by

r (t) = (t, t, t)

,

0 \leq t \leq 1

, with

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

. Substituting into the vector field and taking the dot product gives

v (r (t)) \cdot r^{'} (t) = (t + 1) e^{t} + 5 t^{4} .

Hence the direct line integral is

\int_{C} v \cdot d r = \int_{0}^{1} [(t + 1) e^{t} + 5 t^{4}] d t = [t e^{t} + t^{5}]_{0}^{1} = e + 1.

ii

Solution

For the curved path

r (t) = (t, t^{2}, t^{3})

,

r^{'} (t) = (1, 2 t, 3 t^{2})

. The vector field along this parametric curve is

v (r (t)) = (t^{2} e^{t}, e^{t} + 2 t^{11}, 3 t^{10}) .

Its dot product with the tangent is

(t^{2} + 2 t) e^{t} + 13 t^{12}

. Therefore the direct line integral is

\int_{C} v \cdot d r = \int_{0}^{1} [(t^{2} + 2 t) e^{t} + 13 t^{12}] d t = [t^{2} e^{t} + t^{13}]_{0}^{1} = e + 1.

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