Question:
An object is moving in the clockwise direction around the unit circle $x^{2}+y^{2}=1$. As it passes through the point $(\frac{1}{2},\frac{\sqrt{3{2})$, its y-coordinate is decreasing at the rate of 3 units per sec. The rate at which the x-coordinate changes at this point is
An object is moving in the clockwise direction around the unit circle $x^{2}+y^{2}=1$. As it passes through the point $(\frac{1}{2},\frac{\sqrt{3{2})$, its y-coordinate is decreasing at the rate of 3 units per sec. The rate at which the x-coordinate changes at this point is
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Logic Tip: In related rates problems, always ensure you assign the correct sign to the given rates. "Decreasing" implies a negative rate of change ($-3$), while "increasing" implies a positive rate.
Updated On: Apr 28, 2026
- 2 units/sec
- $3\sqrt{3}$ units/sec
- $\sqrt{3}$ units/sec
- $2\sqrt{3}$ units/sec
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The Correct Option is B
Solution and Explanation
Concept:
This is an application of derivatives involving related rates. The path of the object provides an implicit equation relating $x$ and $y$. Differentiating this equation with respect to time $t$ gives a relationship between the rates of change $\frac{dx}{dt}$ and $\frac{dy}{dt}$.
Step 1: Differentiate the path equation with respect to time.
The given equation is the unit circle: $$x^2 + y^2 = 1$$ Differentiate both sides implicitly with respect to time $t$: $$\frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(1)$$ $$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$$ Divide the entire equation by 2: $$x\frac{dx}{dt} + y\frac{dy}{dt} = 0$$
Step 2: Substitute the given instantaneous values.
We are given that at a specific instant, the point is $(x, y) = \left(\frac{1}{2}, \frac{\sqrt{3{2}\right)$. We are also given that the $y$-coordinate is decreasing at a rate of 3 units per sec, which means $\frac{dy}{dt} = -3$. Substitute these values into the differentiated equation: $$\left(\frac{1}{2}\right)\frac{dx}{dt} + \left(\frac{\sqrt{3{2}\right)(-3) = 0$$
Step 3: Solve for the unknown rate of change.
Isolate $\frac{dx}{dt}$: $$\frac{1}{2} \cdot \frac{dx}{dt} - \frac{3\sqrt{3{2} = 0$$ $$\frac{1}{2} \cdot \frac{dx}{dt} = \frac{3\sqrt{3{2}$$ Multiply both sides by 2: $$\frac{dx}{dt} = 3\sqrt{3} \text{ units/sec}$$ Because the result is positive, the $x$-coordinate is increasing at this point, which matches the clockwise motion in the first quadrant.
This is an application of derivatives involving related rates. The path of the object provides an implicit equation relating $x$ and $y$. Differentiating this equation with respect to time $t$ gives a relationship between the rates of change $\frac{dx}{dt}$ and $\frac{dy}{dt}$.
Step 1: Differentiate the path equation with respect to time.
The given equation is the unit circle: $$x^2 + y^2 = 1$$ Differentiate both sides implicitly with respect to time $t$: $$\frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(1)$$ $$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$$ Divide the entire equation by 2: $$x\frac{dx}{dt} + y\frac{dy}{dt} = 0$$
Step 2: Substitute the given instantaneous values.
We are given that at a specific instant, the point is $(x, y) = \left(\frac{1}{2}, \frac{\sqrt{3{2}\right)$. We are also given that the $y$-coordinate is decreasing at a rate of 3 units per sec, which means $\frac{dy}{dt} = -3$. Substitute these values into the differentiated equation: $$\left(\frac{1}{2}\right)\frac{dx}{dt} + \left(\frac{\sqrt{3{2}\right)(-3) = 0$$
Step 3: Solve for the unknown rate of change.
Isolate $\frac{dx}{dt}$: $$\frac{1}{2} \cdot \frac{dx}{dt} - \frac{3\sqrt{3{2} = 0$$ $$\frac{1}{2} \cdot \frac{dx}{dt} = \frac{3\sqrt{3{2}$$ Multiply both sides by 2: $$\frac{dx}{dt} = 3\sqrt{3} \text{ units/sec}$$ Because the result is positive, the $x$-coordinate is increasing at this point, which matches the clockwise motion in the first quadrant.
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