Question

A ladder $AB$ of $10$ mts long moves with its ends on the axes. When the end $A$ is $6$ mts from the origin, it moves away from it at $2mts/minute$. The rate of increase of the area of the $△OAB$ is ..... sq. mts/ min.

• A. $\frac{4}{3}$
• B. $\frac{8}{3}$
• C. $\frac{14}{3}$
• D. $\frac{7}{2}$

Answer 1

The correct option is A $\frac{4}{3}$

Area of $\Delta OAB=\frac{1}{2}\times OA.OB$

We have $(OA{)}^2+(OB{)}^2=(10{)}^2$

$\frac{d\left(OA\right)}{dt}=2m/min$ when $OA=6,OB=8$

Hence, $\frac{d\left(OB\right)}{dt}=-\frac{OA}{OB}\frac{d\left(OA\right)}{OB}$

So,

$\frac{d\left(A\Delta OAB\right)}{dt}=\frac{1}{2}OA\frac{d\left(OB\right)}{dt}+\frac{1}{2}OB\frac{d\left(OA\right)}{dt}$

$=(\frac{-1}{2}\frac{O{A}^2}{OB}+\frac{1}{2}OB)\frac{dOA}{dt}$

$\frac{d}{dt}\left(AOAB\right)=\frac{1}{2}(-\frac{36}{8}+8)\times 2=\frac{28}{8}=\frac{7}{2}$