If the rate of change of the slope of the tan | vedclass

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(D) Given the curve $y = x^3 - 2x^2 + 3x - 2$. The slope of the tangent $m$ is given by $\frac{dy}{dx} = 3x^2 - 4x + 3$. We need to find the rate of c

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$8$

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(D) Given the curve $y = x^3 - 2x^2 + 3x - 2$. The slope of the tangent $m$ is given by $\frac{dy}{dx} = 3x^2 - 4x + 3$. We need to find the rate of change of the slope $m$ with respect to time $t$, which is $\frac{dm}{dt}$. Differentiating $m$ with respect to $x$, we get $\frac{dm}{dx} = \frac{d}{dx}(3x^2 - 4x + 3) = 6x - 4$. By the chain rule, $\frac{dm}{dt} = \frac{dm}{dx} \cdot \frac{dx}{dt} = (6x - 4) \frac{dx}{dt}$. At the point $(2, 4)$, $x = 2$. Substituting $x = 2$ into the expression, we get $\frac{dm}{dt} = (6(2) - 4) \frac{dx}{dt} = (12 - 4) \frac{dx}{dt} = 8 \frac{dx}{dt}$. The problem states that $\frac{dm}{dt} = k \cdot \frac{dx}{dt}$. Comparing the two expressions, we find $k = 8$.

If the rate of change of the slope of the tangent drawn to the curve $y=x^3-2x^2+3x-2$ at the point $(2,4)$ is $k$ times the rate of change of its abscissa, then $k=$

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