A point moves along the arc of the parabola y | vedclass

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

Question

(B) Given,$\frac{dx}{dt} = 2 	ext{ units/sec}$.Given equation of the parabola is $y = 2x^2$.Differentiating with respect to $t$,we get $\frac{dy}{dt}

Vedclass · Accepted Answer

$\frac{18}{\sqrt{5}} 	ext{ units/sec}$

Vedclass · Answer

(B) Given,$\frac{dx}{dt} = 2 	ext{ units/sec}$.Given equation of the parabola is $y = 2x^2$.Differentiating with respect to $t$,we get $\frac{dy}{dt} = 4x \cdot \frac{dx}{dt}$.Substituting $\frac{dx}{dt} = 2$,we get $\frac{dy}{dt} = 4x(2) = 8x$ ... $(i)$.Let $r$ be the distance of the point $(x, y)$ from the origin $(0, 0)$,so $r = \sqrt{x^2 + y^2}$.The rate of change of distance is $\frac{dr}{dt} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot \frac{d}{dt}(x^2 + y^2) = \frac{2x \frac{dx}{dt} + 2y \frac{dy}{dt}}{2\sqrt{x^2 + y^2}} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{\sqrt{x^2 + y^2}}$.Substituting $\frac{dx}{dt} = 2$ and $\frac{dy}{dt} = 8x$,we get $\frac{dr}{dt} = \frac{x(2) + y(8x)}{\sqrt{x^2 + y^2}} = \frac{2x + 8xy}{\sqrt{x^2 + y^2}}$.At the point $(1, 2)$,$x = 1$ and $y = 2$.$\frac{dr}{dt} = \frac{2(1) + 8(1)(2)}{\sqrt{1^2 + 2^2}} = \frac{2 + 16}{\sqrt{1 + 4}} = \frac{18}{\sqrt{5}} 	ext{ units/sec}$.

$A$ point moves along the arc of the parabola $y = 2x^2$. Its abscissa increases uniformly at the rate of $2 \text{ units/sec}$. At the instant the point is passing through $(1, 2)$,its distance from the origin is increasing at the rate of

Similar Questions

$A$ ladder $20 \ ft$ long leans against a vertical wall. The top end slides downwards at the rate of $2 \ ft/sec$. The rate at which the lower end moves on a horizontal floor when it is $12 \ ft$ from the wall is

$A$ particle moves along the curve $y = x^2 + 2x$. At what point on the curve do the $x$ and $y$ coordinates of the particle change at the same rate?

The radius of a cylinder is increasing at the rate of $2 \text{ cm/sec}$ and its height is decreasing at the rate of $3 \text{ cm/sec}$. Find the rate of change of volume when the radius is $3 \text{ cm}$ and the height is $5 \text{ cm}$.

The volume of a spherical ball is increasing at a rate of $4 \pi \text{ cm}^3 \text{ s}^{-1}$. The rate at which its radius increases,when its volume is $288 \pi \text{ cm}^3$,is ....... $\text{cm s}^{-1}$.

If the radius of a circle increases at the rate of $7 \text{ cm/sec}$,then the rate of increase of its area after $10 \text{ minutes}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module