Q. A particle moves along the curve$y = x^{2} + 2x$. Then, The point on the curve such that x and y coordinates of the particle change with the same rate is
A) (1,3)
B) $\left( \frac{1}{2},\frac{5}{2} \right)$
C) $\left( - \frac{1}{2}, - \frac{3}{4} \right)$
D) $( - 1, - 1)$
Explanation
(c)
Since, $\frac{dx}{dt} = \frac{dy}{dt}$ … (i)
Given equation of curve is
$\ \ \ \ \ \ \ \ y = x^{2} + 2x$
$\Rightarrow \ \ \frac{dy}{dt} = (2x + 2)\frac{dx}{dt}$
$\Rightarrow \ \ 1 = 2x + 2$ [from Eq.(i)]
$\Rightarrow \ \ x = - 1/2,\$ $y = - 3/4$
$\therefore$point on the curve is $\left( - \frac{1}{2}, - \frac{3}{4} \right).$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.