Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability (resit) / 4 / b

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

i

Solution

Taking the partial derivatives of the parabolic coordinates transformation, holding the other coordinate fixed each time, gives

x_{s} = t, x_{t} = s, y_{s} = - s, y_{t} = t .

Thus its Jacobian matrix is

J = (t - s s t)

.

ii

Solution

The determinant of the Jacobian matrix is

s^{2} + t^{2}

. Wherever

(s, t) \neq = (0, 0)

, the inverse function theorem permits a local inverse of the parabolic coordinates transformation, and

\frac{\partial ( s , t )}{\partial ( x , y )} = J^{- 1} = \frac{1}{s ^{2} + t ^{2}} (t s - s t) .

Reading off its entries gives

s_{x} = \frac{t}{s ^{2} + t ^{2}}, s_{y} = - \frac{s}{s ^{2} + t ^{2}}, t_{x} = \frac{s}{s ^{2} + t ^{2}}, t_{y} = \frac{t}{s ^{2} + t ^{2}} .

These are local formulas on a chosen branch:

(s, t)

and

(- s, - t)

have the same image, and at the origin the Jacobian matrix is singular, so this inverse-derivative formula does not apply there.

iii

Solution

Apply the multivariable chain rule to a differentiable function

f (s, t)

on the chosen local branch of the parabolic coordinates:

\frac{\partial f}{\partial x} = \frac{\partial f}{\partial s} \frac{\partial s}{\partial x} + \frac{\partial f}{\partial t} \frac{\partial t}{\partial x} .

Substituting part (ii) gives

\frac{\partial f}{\partial x} = \frac{t f _{s} + s f _{t}}{s ^{2} + t ^{2}}, (s, t) \neq = (0, 0) .

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