At what rate is the area of an equilateral triangle increasing when each side is of length 20cm and each is increasing at a rate of 0.2cm/sec?

Question

Aurora Ayala · Accepted Answer

Area of an equilateral triangle is #A=sqrt3/4xxs^2#

Now, use implicit differentiation ...

#(dA)/(dt)=sqrt3/2sxxs'# Now insert the values from the problem ... #(dA)/(dt)=sqrt3/2(20)xx(0.2)=2sqrt3# hope that helped

James Atkins · Answer

undefined To find the rate at which the area of an equilateral triangle is increasing, we can use the formula for the area of an equilateral triangle and differentiate it with respect to time.

Let $ A $ be the area of the equilateral triangle, and let $ s $ be the length of each side. Given that $ \frac{ds}{dt} = 0.2 \, \text{cm/sec} $, the rate at which the side length is increasing, and $ s = 20 \, \text{cm} $, the length of each side.

The formula for the area $ A $ of an equilateral triangle is:

$$ A = \frac{\sqrt{3}}{4} s^2 $$

Differentiating both sides of this equation with respect to time $ t $, we get:

$$ \frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2s \cdot \frac{ds}{dt} $$

Substitute the given values:

$$ \frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2 \cdot 20 \cdot 0.2 $$

$$ \frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 40 \cdot 0.2 $$

$$ \frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 8 $$

$$ \frac{dA}{dt} = 2\sqrt{3} \, \text{cm}^2/\text{sec} $$

Therefore, the rate at which the area of the equilateral triangle is increasing is $ 2\sqrt{3} $ square centimeters per second.