Q. The diagonal of a square is changing at the rate of 0.5 cm $s^{- 1}$.Then, the rate of change of area , when the area is 400$\$cm^{2} is equal to
A) $20\sqrt{2}cm^{2}/s$
B) $10\sqrt{2}cm^{2}/s$
C) $\frac{1}{10\sqrt{2}}cm^{2}/s$
D) $\frac{10}{\sqrt{2}}cm^{2}/s$
Explanation
(b)
Let D denotes the diagonal of the square.
Given,
$\frac{dD}{dt} = \frac{0.5cm}{s}\ \ \ \ldots..(i)\$
Since, A=400$= \frac{1}{2}D^{2} \Rightarrow D = 20\sqrt{2}$ $\left( \therefore = \frac{D^{2}}{2} \right)$
On differentiating w. r. t. t, we get
$\frac{dA}{dt} = \frac{d}{dt}\left( \frac{D^{2}}{2} \right)$
$= 20\sqrt{2} \times 0.5$
=$10\sqrt{2}cm^{2}/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.