Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 6

Table of contents
- a 6
  - Solution a
- b 6
  - Solution b

a

Solution

For

r > 0

, the partial derivatives of radius are

r_{x} = x / r

,

r_{y} = y / r

and

r_{z} = z / r

. Applying the chain rule to the outer sine and then the radial profile gives the Cartesian coordinates of the gradient:

\nabla sin [f (r)] = \frac{f ^{'} ( r ) cos [ f ( r )]}{r} (x, y, z) .

At the origin these coordinate formulas require a differentiable radial extension; differentiability of

f

as a one-variable profile alone does not guarantee it.

b

Solution

The vector field has only a vertical component, namely

f (r)

. Its divergence is therefore its vertical partial derivative:

\nabla \cdot [k f (r)] = \frac{\partial f ( r )}{\partial z} = \frac{z}{r} f^{'} (r), r > 0.

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