Question

Side of an equilateral triangle expands at the rate of $2\, cm / \sec$. If the rate of increase of its area when each side is $10\, cm$ is $k \sqrt{3} \,cm ^{2} / \sec$ then find $k$.

Answer 1

$[2]$ If $x$ is a side of equilateral triangle then area
$(A)=\frac{\sqrt{3}}{4} x^{2}$
$\frac{ dA }{ dt } =\frac{\sqrt{3}}{4} \times 2 x \frac{ dx }{ dt }$
$=\frac{\sqrt{3}}{2} \times 10 \times 2$
$=10 \sqrt{3}\, cm ^{2} / \sec$