Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 2 / 16 / a

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

i

Solution

In Cartesian coordinates, the divergence is the sum of the three component derivatives. Therefore

\nabla \cdot r = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 3.

ii

Solution

For

r = x^{2} + y^{2} + z^{2} > 0

, the chain rule gives

\partial_{x} (r^{- 1}) = - x / r^{3}

, with analogous formulas for the other components. Hence

\nabla (1/ r) = - \frac{r}{r ^{3}} .

On the right the numerator denotes the position vector, while its denominator is the cube of its magnitude.

iii

Solution

m

is constant, so

\nabla (m \cdot r) = m

. Also

\nabla (r^{- 3}) = - 3 r / r^{5}

, by the same chain rule. The product rule therefore yields

\nabla (\frac{m \cdot r}{r ^{3}}) = \frac{m}{r ^{3}} - \frac{3 ( m \cdot r ) r}{r ^{5}}, r > 0.

The three components follow by differentiating

(αx + β y + γ z) (x^{2} + y^{2} + z^{2})^{- 3/2}

.

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