A hypothetical square grows so that the length of its diagonals are increasing at a rate of 4 m/min. How fast is the area of the square increasing when the diagonals are 14 m each?

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Answer 1 · 2016-10-29T20:32:20

$(dA)/dt = (7sqrt(6))/6$ $m^2$ /min

Let the length of one side be $l$ , the diagonal be $z$ and the area be $A$

Then By Pythagoras, $z^2=l^2 + l^2$
$:. z^2=2l^2$ .......... [1]

And, the Area is, $A=l^2$ , Substituting this into [1] gives us;
$z^2=2A^2$ .......... [2]

We are told that $dz/dt=4$ (constant), and we want to find $(dA)/dt$ when $z=14$ .

By the chain rule, $(dA)/dt = (dA)/(dz).(dz)/(dt)$ .......... [3]

Differentiating [2] wrt z (implicitly) gives:
$2A^2 = z^2$
$:. 2z = 2(2A)(dA)/(dz)$
$:. 2A(dA)/(dz) = z$
$:. (dA)/(dz) = z/(2A)$

Substituting this into [3] gives us:
$(dA)/dt = (z/(2A)).(4)$
$(dA)/dt = (2z)/A$ .......... [4]

So, When $z=14$ , using [2] we have
$2A^2 = 14^2$
$2A^2 = 196$
$A^2 = 98$
$A = sqrt(98 )$

Substituting $z=14$ and $A = sqrt(98)$ into [4] we have:
$(dA)/dt = (2(14))/sqrt(98)$
$(dA)/dt = (7sqrt(6))/6$