Question

The side of an equilateral triangle is increasing at the rate of \[2\] cm/s. At what rate its area is increasing when the side of the triangle is \[20\] cm?

Answer 1

Hint: According to the question the side of an equilateral triangle is increasing at the rate of \[2\] cm/s and asking about what will be the rate of area of the equilateral triangle at particular length of side of the triangle. So, we can solve it with the concept of differentiation. We shall differentiate the area with respect to time and try to solve it.
Given: The side of an equilateral triangle is increasing at the rate of \[2\] cm/s and the length of the side at particular time is given \[20\] cm.

Step-by-step solution:
As we know the area of an equilateral triangle is $\dfrac{{\sqrt 3 }}{4} \times {(side)^2}$ . Using this formula we will proceed.
Let the area of an equilateral triangle be $A$ and side $x$ .
$\because A = \dfrac{{\sqrt 3 }}{4} \times {(side)^2}$
Now differentiating both sides with respect to time, we have
$\therefore \dfrac{{dA}}{{dt}} = \dfrac{{\sqrt 3 }}{4} \times \dfrac{{d{x^2}}}{{dx}} \times \dfrac{{dx}}{{dt}}$ (Using chain rule)
$ \Rightarrow \dfrac{{dA}}{{dt}} = \dfrac{{\sqrt 3 }}{4} \times 2x \times \dfrac{{dx}}{{dt}}$
$\dfrac{{dA}}{{dt}} = \dfrac{{\sqrt 3 }}{4} \times 2 \times 20 \times 2$ (Just putting the given values of $\dfrac{{dx}}{{dt}} = 2$ cm/s and $x = 20$ cm)
$\therefore \dfrac{{dA}}{{dt}} = 20\sqrt 3 $ cm2/s
Hence, Rate at which area increasing when the side of the triangle is $20$ cm is $20\sqrt 3 $ cm2/s.

Note: The question is based on the concept of chain rule where a student should be careful when differentiation is done. Hence, we differentiate the area with respect to time ( $t$ ). So, be careful when ${x^2}$ is differentiated with respect to $t$ . Hence apply chain rule in the proper way.