1.

The area of an equilateral triangle of side'a! Hfeet is increasing at the rate of4 sq.ft./sec. The rate at which the perimeteris increasing is8v1)4)amoat and attains the speed318V32133)a2a​

Answer»

area \: = \frac{ \sqrt{3} }{4} {a}^{2}

\frac{d(area)}{dt} = 2 \times \frac{ \sqrt{3} }{4} \times a \times \: \frac{da}{dt} = 4

\frac{da}{dt} = \frac{8}{ \sqrt{3} a}

now,

perimeter(p) = 3a

\frac{dp}{dt} = 3 \times \frac{da}{dt} = 3 \times \frac{8}{ \sqrt{3} a} = \frac{8 \sqrt{3} }{a} ft \: per \: sec.


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