A particle is travelling on the circle with equation $x^{2} + y^{2} = 16$ - HSC - SSCE Mathematics Extension 2 - Question 9 - 2017 - Paper 1

Differentiate both sides with respect to time t:

$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$

This simplifies to:

$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$

Substitute ( \frac{dx}{dt} ):

$x(y) + y \frac{dy}{dt} = 0$

This simplifies to:

$y \frac{dy}{dt} = -x y$

Dividing by y (assuming (y \neq 0)):

$\frac{dy}{dt} = -\frac{x}{y}$

Using the equation for a circle, as the particle moves around the circle, we observe that:

• If ( \frac{dy}{dt} = -x ), the motion will correspond to a clockwise movement.

Thus, the correct statement is:

C. ( \frac{dy}{dt} = -x ) and the particle travels clockwise.