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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$18$

Vedclass · Answer

(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:

Similar Questions

The distance covered by a particle in $t$ seconds is given by $x = 3 + 8t - 4t^2$. After $1$ second, the velocity will be:

$A$ ladder of length $13 \ m$ has one end resting against a vertical wall and the other on the ground. If the lower end moves away from the wall at a speed of $2 \ m/min$, then the speed (in $m/min$) at which the upper end falls when the bottom is $5 \ m$ away from the wall is

If a metallic circular plate of radius $50 \, cm$ is heated so that its radius increases at the rate of $1 \, mm$ per hour,then the rate at which the area of the plate increases (in $cm^2/hour$) is (in $\pi$)

The distance $s$ in meters travelled by a particle in $t$ seconds is given by $s = \frac{2 t^3}{3} - 18 t + \frac{5}{3}$. The acceleration when the particle comes to rest is (in $m/s^2$)

$A$ 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 $P$:

Vedclass Test Series

Exam Paper Generator

Online Exam Module