 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1/3/b/solution

Differentiate each coordinate with respect to the parameter

t

and use the chain rule:

\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{s e c ^{2} t}{- c s c ^{2} t} = - tan^{2} t .

Both coordinates are defined where

sin t \neq = 0

and

cos t \neq = 0

. As a check,

x y = 1

there, so the same derivative is

- 1/ x^{2}

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