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(B) The equation of the circle is $x^2 + y^2 = 1$. Differentiating both sides with respect to time $t$,we get: $2x \frac{dx}{dt} + 2y \frac{dy}{dt} =

Vedclass · Accepted Answer

$3\sqrt{3} 	ext{ units/sec}$

Vedclass · Answer

(B) The equation of the circle is $x^2 + y^2 = 1$. Differentiating both sides with respect to time $t$,we get: $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$ $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$ Given that the $y$-coordinate is decreasing at the rate of $3 	ext{ units/sec}$,we have $\frac{dy}{dt} = -3 	ext{ units/sec}$. At the point $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}ight)$,we substitute $x = \frac{1}{2}$,$y = \frac{\sqrt{3}}{2}$,and $\frac{dy}{dt} = -3$: $\frac{1}{2} \frac{dx}{dt} + \left(\frac{\sqrt{3}}{2}ight)(-3) = 0$ $\frac{1}{2} \frac{dx}{dt} = \frac{3\sqrt{3}}{2}$ $\frac{dx}{dt} = 3\sqrt{3} 	ext{ units/sec}$. Thus,the $x$-coordinate is increasing at the rate of $3\sqrt{3} 	ext{ units/sec}$.

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

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