A point is moving on y=4-2x^2. The x-coordina | vedclass

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(D) The given equation of the curve is $y = 4 - 2x^2$. On differentiating both sides with respect to time $t$,we get: $\frac{dy}{dt} = -4x \frac{dx}{d

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$20 	ext{ units/s}$

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(D) The given equation of the curve is $y = 4 - 2x^2$. On differentiating both sides with respect to time $t$,we get: $\frac{dy}{dt} = -4x \frac{dx}{dt}$. Given that the $x$-coordinate is decreasing at the rate of $5 	ext{ units/s}$,we have $\frac{dx}{dt} = -5 	ext{ units/s}$. At the point $(1, 2)$,we substitute $x = 1$ and $\frac{dx}{dt} = -5$ into the derivative equation: $\frac{dy}{dt} = -4(1)(-5) = 20 	ext{ units/s}$. Thus,the $y$-coordinate is increasing at the rate of $20 	ext{ units/s}$.

$A$ point is moving on $y=4-2x^2$. The $x$-coordinate of the point is decreasing at the rate of $5 \text{ units/s}$. Then,the rate at which the $y$-coordinate of the point is changing when the point is at $(1,2)$ is

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