Question

A particle moves along the curve $y=x^{3/2}$ in the first quadrant in such a way that its distance from the origin increases at the rate of 11 units per second. The value of $\frac{dx}{dt}$ when $x=3$ is

• A. $4$
• B. $\frac{9}{2}$
• C. $\frac{3\sqrt{3}}{2}$
• D. none

Answer 1

Answer: (A)

$y=x^{3/2};\frac{dr}{dt}=11$
$\frac{dx}{dt}$ when $x=3$
when $x=3;y=3\sqrt{3}$
$r^2=x^2+y^2$
$r\frac{dr}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}\dots \dots (1)$
$\text{ also }\frac{dy}{dt}=\frac{3}{2}\sqrt{x}\frac{dx}{dt}\dots \dots (2)$
$\therefore r\frac{dr}{dt}=x\frac{dx}{dt}+y\frac{3}{2}\sqrt{x}\frac{dx}{dt}$
$r\frac{dr}{dt}=\left(x+\frac{3y\sqrt{x}}{2}\right)\frac{dx}{dt}$
$6\cdot 11=\left(3+\frac{3\cdot 3\sqrt{3}\cdot \sqrt{3}}{2}\right)\frac{dx}{dt}\Rightarrow 66=\left(3+\frac{27}{2}\right)\frac{dx}{dt}\Rightarrow 66=\left(\frac{33}{2}\right)\frac{dx}{dt}$
$\Rightarrow \frac{dx}{dt}=4$