(D) Given the 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}{

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

Class 12 Mathematics Applications of Derivatives Rate of Change of Quantities

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$A$ 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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A
$5 \text{ units/s}$
B
$10 \text{ units/s}$
C
$15 \text{ units/s}$
D
$20 \text{ units/s}$

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Applications of Derivatives

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Rate of Change of Quantities

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$A$ 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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If $V = \frac{4}{3}\pi r^3$,find the rate of change of $V$ with respect to $t$ when $r = 10$ and $\frac{dr}{dt} = 0.01$.

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