Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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?

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

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App