Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 6

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

a

Solution

The maximum of sine is one. The allowed parameter values giving

sin (3 t) = 1

are

t = \frac{π}{6}, \frac{5 π}{6}, \frac{3 π}{2} .

Substituting each into

x = cos t

gives

y_{m a x} = 1, x \in {\frac{3}{2}, - \frac{3}{2}, 0} .

All three parameter values lie in the original interval, so none of these maxima is discarded.

b

Solution

For this parametric curve, the chain rule gives

\frac{d x}{d t} = - sin t, \frac{d y}{d t} = 3 cos (3 t), \frac{d y}{d x} = - \frac{3 cos ( 3 t )}{sin t} .

The quotient is a finite slope when

sin t \neq = 0

. At

t = 0

and

t = π

, the horizontal component of the tangent vanishes while the vertical component does not; these have vertical tangent lines, rather than missing finite values of the displayed quotient.

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