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

To determine the correct answer, we start by differentiating the circle's equation with respect to time. The equation of the circle is:

$x^2 + y^2 = 16$

Differentiating both sides gives:

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

From this, we can simplify to:

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

Given that ( \frac{dx}{dt} = y ), we can substitute this into the equation:

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

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

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

Thus, we find:

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

To understand the motion direction, substitute the values of ( x ) and ( y ) accordingly. If ( x ) is positive, ( \frac{dy}{dt} ) is negative, indicating the particle travels clockwise around the circle.

Hence, the correct answer is:

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