(A) Given the curve equation: $y = x^3 - 3x^2 + 2x - 1$. Differentiating both sides with respect to time $t$,we get: $\frac{dy}{dt} = (3x^2 - 6x + 2)

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(A) Given the curve equation: $y = x^3 - 3x^2 + 2x - 1$. Differentiating both sides with respect to time $t$,we get: $\frac{dy}{dt} = (3x^2 - 6x + 2) \frac{dx}{dt}$. We are given that $\frac{dy}{dt} = 6$ units/sec and the point is $(2, -1)$,so $x = 2$. Substituting these values into the derivative equation: $6 = (3(2)^2 - 6(2) + 2) \frac{dx}{dt}$. $6 = (12 - 12 + 2) \frac{dx}{dt}$. $6 = 2 \frac{dx}{dt}$. Therefore,$\frac{dx}{dt} = \frac{6}{2} = 3$ units/sec.

Class 12 Mathematics Application of Derivatives Rate of Change of Quantities

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$A$ point is moving on the curve $y=x^3-3x^2+2x-1$ and the $y$-coordinate of the point is increasing at the rate of $6 \text{ units/sec}$. When the point is at $(2, -1)$,the rate of change of the $x$-coordinate of the point is:

AP EAMCET 2024 All AP EAMCET

A
$3$
B
$\frac{1}{2}$
C
$-\frac{1}{2}$
D
$-3$

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$A$ point is moving on the curve $y=x^3-3x^2+2x-1$ and the $y$-coordinate of the point is increasing at the rate of $6 \text{ units/sec}$. When the point is at $(2, -1)$,the rate of change of the $x$-coordinate of the point is:

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