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Q.
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
Application of Derivatives
Solution:
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 $a = 20$
$\therefore \frac{dA}{dt}=2\times20\times \frac{0.5}{\sqrt{2}}=10\sqrt{2}\,cm^{2}/sec$