A particle moves along the curve y=x^2+2x. Th | vedclass

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

Question

(C) Given the equation of the curve is $y = x^2 + 2x$.Since the $x$ and $y$ coordinates of the particle change at the same rate, we have $\frac{dy}{dt

Vedclass · Accepted Answer

$\left(-\frac{1}{2}, -\frac{3}{4}\right)$

Vedclass · Answer

(C) Given the equation of the curve is $y = x^2 + 2x$.Since the $x$ and $y$ coordinates of the particle change at the same rate, we have $\frac{dy}{dt} = \frac{dx}{dt}$.Differentiating the equation of the curve with respect to $t$, we get:$\frac{dy}{dt} = (2x + 2) \frac{dx}{dt}$.Substituting $\frac{dy}{dt} = \frac{dx}{dt}$ into the equation, we get:$\frac{dx}{dt} = (2x + 2) \frac{dx}{dt}$.Assuming $\frac{dx}{dt} 
eq 0$, we can divide both sides by $\frac{dx}{dt}$:$1 = 2x + 2$$2x = -1$$x = -\frac{1}{2}$.Now, substitute $x = -\frac{1}{2}$ into the curve equation to find $y$:$y = \left(-\frac{1}{2}ight)^2 + 2\left(-\frac{1}{2}ight)$$y = \frac{1}{4} - 1 = -\frac{3}{4}$.Therefore, the required point is $\left(-\frac{1}{2}, -\frac{3}{4}ight)$.

$A$ particle moves along the curve $y=x^2+2x$. Then, the point on the curve such that the $x$ and $y$ coordinates of the particle change with the same rate is

Similar Questions

The volume of a sphere is increasing at the rate of $1200 \text{ cm}^3/\text{s}$. The rate of increase in its surface area when the radius is $10 \text{ cm}$ is: (in $\text{ cm}^2/\text{s}$)

If $pv = 81$,then $\frac{dp}{dv}$ at $v = 9$ is equal to

The volume of a cube is increasing at a rate of $9 \text{ cm}^3/\text{s}$. How fast is the surface area increasing when the length of an edge is $10 \text{ cm}$ (in $\text{ cm}^2/\text{s}$)?

The radius of a circular plate is increasing at the rate of $0.01 \text{ cm/sec}$. When the radius is $12 \text{ cm}$,the rate at which the area increases is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module