Question

If the rate of change of area of a square plate is equal to that of the rate of change of its perimeter, then the length of the side is?
A) 1 unit.
B) 2 units.
C) 3 units.
D) 4 units.

Answer 1

Hint: We will use the formula of area of a square which is given by A =$x^2$ and also the formula of perimeter of the square which is given by P = 4x, where x is the length of one of the sides.
Complete step-by-step answer:
Given that the rate of change of area of a square plate is equal to that of the rate of change of its perimeter, we have to find the length of the square.
Let at any time t, the length of a side of the square be x.
Then because the formula of area of the square is A= $side^2$ and the Perimeter P = 4side.
Then the area of the given square would be A = $x^2$ and the perimeter would be P = 4x.
We have,
A = $x^2$.
Differentiating both sides of the above expression with respect to t, we get,

& A={{x}^{2}} \\\ & \Rightarrow \dfrac{dA}{dt}=2x\dfrac{dx}{dt} \\\ \end{aligned}$$ Also, we have P = 4x. Differentiating both sides of the above expression with respect to t, we get, $$\begin{aligned} & P=4x \\\ & \Rightarrow \dfrac{dP}{dt}=4\dfrac{dx}{dt} \\\ \end{aligned}$$ Now we are given that the rate of change of area of a square plate is equal to that of the rate of change of its perimeter, which gives us, $$\dfrac{dA}{dt}=\dfrac{dP}{dt}$$ Substituting the values of both the expression obtained above we get, $$\begin{aligned} & \dfrac{dA}{dt}=\dfrac{dP}{dt} \\\ & \Rightarrow 2x\dfrac{dx}{dt}=4\dfrac{dx}{dt} \\\ \end{aligned}$$ Cancelling the term $$\dfrac{dx}{dt}$$ from both the sides of the above equation we get, $$\begin{aligned} & 2x=4 \\\ & \Rightarrow x=2 \\\ \end{aligned}$$ Therefore, we get the required length of the side as 2 units. Hence the correct option is option(B). Note: The possibility of error in this type of question can be calculating the value of $$\dfrac{dx}{dt}$$ which is not necessary and will definitely get you in trouble solving complex equations. So just go for cancelling $$\dfrac{dx}{dt}$$ present at both the sides of the equation to get the desired result.