# How fast is the area increasing for an equilateral triangle under the given conditions?

Question:. The sides of an equilateral triangle are increasing at the rate of $\sqrt{3}$ cm/sec. How fast is the area increasing when its side is $6$ cm?

I have done this problem using calculus. Is there any alternative way to solve the above problem?

Given the sides the equilateral triangle = $a$ cm. Area of the equilateral triangle is $A=\dfrac{\sqrt{3}}{4}a^2$
differentiating w.r.t $t$ we get,
$\dfrac{dA}{dt}=\dfrac{\sqrt{3}}{4}\times 2a\times \dfrac{da}{dt}$
But it is given by
$\dfrac{da}{dt}=\sqrt{3}$ cm/sec
when $a=6$ cm.
$\therefore \left[\dfrac{dA}{dt}\right]_{a=6}=\dfrac{\sqrt{3}}{4}\times 2\times 6\times \sqrt{3}=9\; cm^2/sec$.

• Calculus is the art of rates of change. I'd say it's a tool developed especially to deal with this question (and others like it). I don't think you can avoid it and at the same time get a comparably nice proof. – Arthur Sep 12 '18 at 5:26

An object is accelerating at constant $a = \sqrt{3}\ \text{cm/sec}^2$. Using the formula for constant acceleration we find that for initial velocity $0$ that $x = \frac{1}{2}at^2$ where $x$ is the distance traveled. We know $x = 6\ \text{cm}$ and $a = \sqrt{3}\ \text{cm/sec}^2$ so we solve for $t$:
$$6 = \frac{1}{2} \cdot \sqrt 3 \cdot t^2$$ $$t = \sqrt{\frac{2 \cdot 6}{\sqrt 3}} = 2\sqrt[4]{3}$$
Now we can calculate the velocity at $t = 2\sqrt[4]{3}$ using $v = at$ to find $v = 2 \cdot 3^{3/4}$.
So we know that the side of the equilateral triangle when it reaches $6\ \text{cm}$ is increasing by $2 \cdot 3^{3/4}\ \text{cm/sec}$, and thus the area of the equilateral triangle is increasing by $9\ \text{cm}^2/\text{sec}$