Question

The diagonal of a square is changing at the rate of $0.5\, cm/ sec$. Then the rate of change of area, when the area is $400\, cm^2$, is equal to

• A. $20\sqrt{2}\,cm^{2}/ sec$
• B. $10\sqrt{2}\,cm^{2}/ sec$
• C. $\frac{1}{10\sqrt{2}}cm^{2}/ sec$
• D. $\frac{10}{\sqrt{2}}cm^{2}/ sec$

Answer 1

Answer: (B)

$10\sqrt{2}\,cm^{2}/ sec$

Answer 2

Diagonal $D=\sqrt{2}a$ Differentiating w.r.t. t $\frac{dD}{dt}=\sqrt{2} \frac{da}{at}$ or $\frac{da}{dt}=\frac{1}{\sqrt{2}} \frac{da}{dt} =\frac{1}{\sqrt{2}}\times0.5\,cm/ s$ Let Area is denoted by $A$ $\frac{dA}{dt}=2a \frac{da}{dt}\,...\left(i\right)$ when area $A$ is $400\, cm^{2}$ then $a = 20$ $\therefore \frac{dA}{dt}=2\times20\times \frac{0.5}{\sqrt{2}}=10\sqrt{2}\,cm^{2}/sec$