Question

The sides of an equilateral triangle are increasing at the rate of 2cm/s. the rate at which the area increases when the side is 10cm, is
$$\begin{gathered} A.\sqrt{3}c{m}^2/s \\ B.10c{m}^2/s \\ C.\text{ 10}\sqrt{3}c{m}^2/s \\ D.{\displaystyle \frac{10}{\sqrt{3}}}c{m}^2/s \end{gathered}$$

Answer 1

Hint: In order to solve this question we will use the general formula of area of equilateral triangle in terms of sides. We will differentiate the equation of area and substitute the value of change inside it along with the side given to find the rate of change of area at a particular instant.

Complete step-by-step solution:
Let x be the side of an equilateral triangle and A be the area of the equilateral triangle.
As we know the formula of area of equilateral triangle in terms of sides is given as:
$A={\displaystyle \frac{\sqrt{3}}{4}}{x}^2$
As we know that the sides of an equilateral triangle are increasing at the rate of 2cm/s.
This can be written mathematically as follows.
${\displaystyle \frac{dx}{dt}}=2cm/s$--- (1)
In order to find the change in area when side x = 10cm. --- (2)
Let us differentiate the equation of area with respect to time.
$$\Rightarrow {\displaystyle \frac{dA}{dt}}={\displaystyle \frac{d}{dt}}\left({\displaystyle \frac{\sqrt{3}}{4}}{x}^2\right)$$
Let us separate the constant term from RHS and use the formula for differentiation.
$$\begin{gathered} \Rightarrow {\displaystyle \frac{dA}{dt}}={\displaystyle \frac{\sqrt{3}}{4}}{\displaystyle \frac{d}{dt}}\left(x^2\right) \\ \Rightarrow {\displaystyle \frac{dA}{dt}}={\displaystyle \frac{\sqrt{3}}{4}}2x{\displaystyle \frac{dx}{dt}}\left[\because {\displaystyle \frac{d}{dt}}\left(x^n\right)=n{x}^{n-1}{\displaystyle \frac{dx}{dt}}\right] \end{gathered}$$
Now let us substitute the value from equation (1) and equation (2)
$$\begin{gathered} \Rightarrow {\displaystyle \frac{dA}{dt}}={\displaystyle \frac{\sqrt{3}}{4}}2\left(10\right)\left(2\right) \\ \Rightarrow {\displaystyle \frac{dA}{dt}}=10\sqrt{3}c{m}^2/s \end{gathered}$$
Hence, the rate at which the area increases is $$10\sqrt{3}c{m}^2/s$$
So, option C is the correct option.

Note: In order to solve such types of problems students must remember the rules for differentiation from one variable to another. Also the units given in the problem must be given a focus as sometimes a change in unit may be a cause for mistake. Also students must remember the formula for the area of equilateral triangle in terms of side.