 Solution

ID: 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/ii/solution

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.

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