A point P is moving on the curve x^3 y^4 = 2^ | vedclass

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(A) Given the curve equation: $x^3 y^4 = 2^7$. Differentiating both sides with respect to time $t$: $3x^2 y^4 \frac{dx}{dt} + 4x^3 y^3 \frac{dy}{dt} =

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increases at the rate of $6 	ext{ units per second}$

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(A) Given the curve equation: $x^3 y^4 = 2^7$. Differentiating both sides with respect to time $t$: $3x^2 y^4 \frac{dx}{dt} + 4x^3 y^3 \frac{dy}{dt} = 0$. We are given $\frac{dx}{dt} = -8 	ext{ units/sec}$ (since it is decreasing). At the point $(2, 2)$, substitute $x = 2, y = 2$ and $\frac{dx}{dt} = -8$: $3(2)^2 (2)^4 (-8) + 4(2)^3 (2)^3 \frac{dy}{dt} = 0$. $3(4)(16)(-8) + 4(8)(8) \frac{dy}{dt} = 0$. $-1536 + 256 \frac{dy}{dt} = 0$. $256 \frac{dy}{dt} = 1536$. $\frac{dy}{dt} = \frac{1536}{256} = 6$. Since $\frac{dy}{dt} > 0$, the $y$-coordinate increases at the rate of $6 	ext{ units per second}$.

$A$ point $P$ is moving on the curve $x^3 y^4 = 2^7$. The $x$-coordinate of $P$ is decreasing at the rate of $8 \text{ units per second}$. When the point $P$ is at $(2, 2)$, the $y$-coordinate of $P$:

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