The side of a square sheet is increasing at the rate of 4cm per minute. At what rate is the area increasing when the side is 8cm long?
Answer
361.8k+ views
Hint – In this question the rate of increase of the side of a square sheet is given to us. We need to find the rate change of the area. Use the basic formula for area of square in terms of side and differentiate both sides with respect to time. This will help to get the solution.
Complete step-by-step answer:
Let A be the area of the square.
We need to find the rate of change of area w.r.t. time when the side of the square is 8 cm.
Let the side of square be x.
$ \Rightarrow x = 8{\text{ cm}}{\text{.}}$
Therefore we have to find $\dfrac{{dA}}{{dt}}$ at $x = 8{\text{ cm}}{\text{.}}$
As we know that the area (A) of the square is side square.
$ \Rightarrow A = {x^2}$ Square meter.
Differentiate it w.r.t. time
$ \Rightarrow \dfrac{{dA}}{{dt}} = 2x\dfrac{{dx}}{{dt}}$……………….. (1)
Now it is given that the side of the square is increasing at the rate of 4 cm/min.
$ \Rightarrow \dfrac{{dx}}{{dt}} = + 4$ Cm/min. (plus sign indicates it is increasing)
So, put this value in equation (1) we have,
$ \Rightarrow \dfrac{{dA}}{{dt}} = 2\left( 8 \right)\left( 4 \right) = + 64$ Cm/min.
So the rate at which the area is increasing is 64 cm/min.
Note – Whenever we face such types of problems the key concept is to make sure that whether the side is increasing or decreasing with respect to time, as it directly affects the signs of rate change that we need to take into consideration. This concept along with the gist of the basic formula for area will help you get on the right track to get the answer.
Complete step-by-step answer:
Let A be the area of the square.
We need to find the rate of change of area w.r.t. time when the side of the square is 8 cm.
Let the side of square be x.
$ \Rightarrow x = 8{\text{ cm}}{\text{.}}$
Therefore we have to find $\dfrac{{dA}}{{dt}}$ at $x = 8{\text{ cm}}{\text{.}}$
As we know that the area (A) of the square is side square.
$ \Rightarrow A = {x^2}$ Square meter.
Differentiate it w.r.t. time
$ \Rightarrow \dfrac{{dA}}{{dt}} = 2x\dfrac{{dx}}{{dt}}$……………….. (1)
Now it is given that the side of the square is increasing at the rate of 4 cm/min.
$ \Rightarrow \dfrac{{dx}}{{dt}} = + 4$ Cm/min. (plus sign indicates it is increasing)
So, put this value in equation (1) we have,
$ \Rightarrow \dfrac{{dA}}{{dt}} = 2\left( 8 \right)\left( 4 \right) = + 64$ Cm/min.
So the rate at which the area is increasing is 64 cm/min.
Note – Whenever we face such types of problems the key concept is to make sure that whether the side is increasing or decreasing with respect to time, as it directly affects the signs of rate change that we need to take into consideration. This concept along with the gist of the basic formula for area will help you get on the right track to get the answer.
Last updated date: 28th Sep 2023
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Total views: 361.8k
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