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If the length of the diagonal of a square is increasing at the rate of $0.1 cm/sec.$ What is the rate of increase of its area when the side is $\large\frac{15}{\sqrt{2}} $cm?

\[\begin{array}{1 1}(1)1.5 cm^{2}/sec&(2)\frac{1}{2\pi}\\(3)4\pi&(4)\frac{\pi}{3}\end{array}\]

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Let a be the length of the side , d the length of the diagonal and A be the area of the square at time t.
$d^2=a^2+a^2= 2a^2$
$d= \sqrt 2 a$
$\large\frac{d}{dt} $$(d) =\sqrt 2 \large\frac{da}{dt}$
$0.1 = \sqrt {2} \large\frac{da}{dt}$
$\large\frac{dA}{dt} =\large\frac{da}{dt}$
$A= a^2 \large\frac{15}{\sqrt 2} \frac{0.1}{\sqrt 2}$
$\qquad= 2a$
$\qquad= 2 \times \large\frac{da}{at} \times \large\frac{0.1}{\sqrt 2}$
$=>1.5 cm^{2}/sec$
Hence 1 is the correct answer.
answered May 19, 2014 by meena.p

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