An object is moving in the clockwise direction around the unit circle $x^{2} + y^{2} = 1$ . As it passes through the point $(1 / 2, \sqrt{3} / 2)$ , its y-coordinate is decreasing at the rate of 3 units per second. The rate at which the x-coordinate changes at this point is (in units per second)

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$3 \sqrt{3}$

$\sqrt{3}$

$2 \sqrt{3}$

answer is B.

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Detailed Solution

We find $\frac{d x}{d t}$ when $x = \frac{1}{2}$ and $y = \frac{\sqrt{3}}{2}$ given

that $\frac{d y}{d t} = - 3$ units/s and $x^{2} + y^{2} = 1$ .

Differentiating $x^{2} + y^{2} = 1$ , we have

$2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0 .$

Putting $x = 1 / 2, y = \sqrt{3} / 2$ and $d y / d t = - 3$ ,we have

$\frac{1}{2} \frac{d x}{d t} + \frac{\sqrt{3}}{2} (- 3) = 0 \Rightarrow \frac{d x}{d t} = 3 \sqrt{3}$ .

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