Question #281955

The sides of an equilateral triangle are increasing at the rate of 3 cm/min. Find: a) the rate of change of the perimeter.

b) the rate of change of the area when the side is 3 cm. long.


1
Expert's answer
2021-12-23T08:45:03-0500

Solution;

(a)

Perimeter of the triangle;

P=3aP=3a

Where a is the length of one side.

Hence;

dPdt=d(3a)dt\frac{dP}{dt}=\frac{d(3a)}{dt}

dPdt=3dadt\frac{dP}{dt}=3\frac{da}{dt}

dPdt=3×3cm/min\frac{dP}{dt}=3×3cm/min

Rate of change of perimeter is;

9cm/min9cm/min

(b)

Area of an equilateral triangle is;

A=34a2A=\frac{\sqrt3}{4}a^2

Differentiate with respect to t;

dAdt=43(2a)dadt\frac{dA}{dt}=\frac{\sqrt4}{3}(2a)\frac{da}{dt}

Substitute a=3cm and dadt=3cm/min\frac{da}{dt}=3cm/min

Hence;

dAdt=34×2×3×3cm/min\frac{dA}{dt}=\frac{\sqrt3}{4}×2×3×3cm/min

The rate of change of are is;

4.53cm2/min4.5\sqrt3cm^2/min





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