The x-coordinate changes on the curve y=3x^5+ | vedclass

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(C) Given the curve $y = 3x^5 + 15x - 8$. The rate of change of $x$ is $\frac{dx}{dt} = \frac{1}{5} 	ext{ units/sec}$. The rate of change of $y$ is $

Vedclass · Accepted Answer

$18$

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(C) Given the curve $y = 3x^5 + 15x - 8$. The rate of change of $x$ is $\frac{dx}{dt} = \frac{1}{5} 	ext{ units/sec}$. The rate of change of $y$ is $\frac{dy}{dt} = 6 	ext{ units/sec}$. Differentiating $y$ with respect to $t$: $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = (15x^4 + 15) \cdot \frac{dx}{dt}$. Substituting the given values: $6 = (15x^4 + 15) \cdot \frac{1}{5} = 3(x^4 + 1)$. $x^4 + 1 = 2 \Rightarrow x^4 = 1 \Rightarrow x = \pm 1$. For $x = 1$, $y = 3(1)^5 + 15(1) - 8 = 10$. So, $A = (1, 10)$. For $x = -1$, $y = 3(-1)^5 + 15(-1) - 8 = -3 - 15 - 8 = -26$. So, $B = (-1, -26)$. The slope of $AB$ is $\frac{y_2 - y_1}{x_2 - x_1} = \frac{-26 - 10}{-1 - 1} = \frac{-36}{-2} = 18$.

The $x$-coordinate changes on the curve $y=3x^5+15x-8$ at the rate of $\frac{1}{5} \text{ units/sec}$. If $A(x_1, y_1)$ and $B(x_2, y_2)$ are the points on the curve at which the $y$-coordinate changes at the rate of $6 \text{ units/sec}$, then the slope of $AB$ is:

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