Question

The side of a square sheet of metal is increasing at $3$ centimetre per minute. At what rate is the area increasing when the side is $10{\text{ cm}}$ long?

Answer 1

Hint : In order to find the rate of area increasing, first collect the information given in the question, and write them numerically. Find the Area of the square and then differentiate it with respect to time, then substitute the values in the equation and get the results. Also check for the positive and the negative results that proves the rate is increasing or decreasing.

Complete step by step solution:
Considering the sides of the square sheet to be $x{\text{ cm}}$ , time to be $t$ and area to be $A$ .
Therefore, the rate of change of side with respect to time can be written as $\dfrac{{dx}}{{dt}}$ .
We are given that the side of the square sheet is increasing at the rate of $3$ centimetre per minute. That can be numerically written as:
$\dfrac{{dx}}{{dt}} = 3$ ………(1)
As we know that Area of the square is the square of the side, that is:
$A = {\left( {side} \right)^2} \\\ \Rightarrow A = {x^2}{\text{ Square meter}} \;$
Differentiating the above equation with respect to time that is $'t'$ and we get:
$\Rightarrow \dfrac{{dA}}{{dt}} = \dfrac{{d{x^2}}}{{dt}}$ …..(2)
Since, we know that $\dfrac{{d{x^n}}}{{dt}} = n{x^{n - 1}}\dfrac{{dx}}{{dt}}$ .
Substituting this value in equation (2), we get:
$\Rightarrow \dfrac{{dA}}{{dt}} = 2x\dfrac{{dx}}{{dt}}$ ……(3)
Since, it’s given that the side is $10{\text{ cm}}$ long that is $x = 10{\text{ cm}}$ .
Substituting $x = 10{\text{ cm}}$ and the equation (1) in the equation (3), we get:
$\Rightarrow \dfrac{{dA}}{{dt}} = 2 \times 10 \times 3{\text{ }}\dfrac{{cm}}{{\min }}$
$\Rightarrow \dfrac{{dA}}{{dt}} = 60{\text{ }}\dfrac{{cm}}{{\min }}$
The $\dfrac{{dA}}{{dt}}$ represents the change in rate of Area.
Therefore, the rate at which the area is increasing at $60{\text{ }}\dfrac{{cm}}{{\min }}$ .
So, the correct answer is “ $60{\text{ }}\dfrac{{cm}}{{\min }}$ ”.

Note : Since, the result obtained is positive that shows the rate is increasing. If the result obtained is negative that shows the value/rate is decreasing.
Rate is nothing but the ratio between two quantities that belongs to two different units.
Examples of rate are $\dfrac{{cm}}{{\min }}$ , $\dfrac{{km}}{{hr}}$ etc.