Question #606a2

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Question #606a2

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Answer 1 · 2015-07-25T13:48:00

I found: $\frac{d A}{d t} = - 27 \frac{c m^{2}}{s}$ $(dA)/(dt)=-27(cm^2)/s$

Explanation:

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When $A = 27 \sqrt{3} c m^{2}$ $A=27sqrt(3)cm^2$ then using (1) $A = \frac{\sqrt{3}}{4} l^{2}$ $A=sqrt(3)/4l^2$ you get:
$27 \sqrt{3} = \frac{\sqrt{3}}{4} l^{2}$ $27sqrt(3)=sqrt(3)/4l^2$
$l = 10.4 c m$ $color(red)(l=10.4cm)$
Now you use (2) and get the change in the area as:
$\frac{d A}{d t} = 2 \frac{\sqrt{3}}{4} \times (10.4) \times (- 3) = - 27 \frac{c m^{2}}{s}$ $(dA)/(dt)=2sqrt(3)/4xx(10.4)xx(-3)=-27(cm^2)/s$

Answer 2 · 2015-07-25T13:53:22

$27$ $27$ cm^2/s

Explanation:

$A = (\frac{1}{2}) a \cdot a \cdot sin 60 = (\frac{\sqrt{3}}{4}) a^{2}$ $A = (1/2) a*a*sin60 =( (sqrt3)/4)a^2$

$\frac{d A}{d t} = (\frac{\sqrt{3}}{2}) a (\frac{d a}{d t})$ $(dA)/dt = ( (sqrt3)/2)a((da)/dt)$

If $A = 27 (\sqrt{3}$ $A = 27(sqrt3$ ) => $a = 6 (\sqrt{3})$ $a = 6(sqrt3)$

Then, $\frac{d A}{d t} = (\frac{\sqrt{3}}{2}) 6 (\sqrt{3}) (3) cm^2/s = 27 cm^2/s$ $(dA)/dt =( (sqrt3)/2)6(sqrt3)(3) " cm^2/s" = 27 " cm^2/s"$

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Question #606a2

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