
Each side of a square is increasing at the rate of 6cm/s. At what rate will the area of the square increase if the area of the square is $16cm^2$?
Answer
491.1k+ views
Hint: This question is related to differentiation. Or we can say it is the application of differentiation. In which rate of the particular quantity is considered as the differentiation. Here also we will first use the formula of area of to find the length of the side of square and then we will find the rate of area of square and putting the respective values we will get the answer.
Complete step by step answer:
First we will note the given data.
The side of a square is increasing at the rate of 6cm/s. let s be the side of the square. Thus the rate will be,
\[\dfrac{{ds}}{{dt}} = 6cm/\sec \]
Now area of the square is 16cm2
But the area of a square is given by, \[Area = {s^2}\]
Thus the side of the square , \[16 = {s^2}\]
Taking roots on both sides, \[s = \sqrt {16} = 4cm\]
Now the rate of area is obtained by, \[\dfrac{{dA}}{{dt}} = 2s.\dfrac{{ds}}{{dt}}\]
Now putting the values in the equation above, \[\dfrac{{dA}}{{dt}} = 2 \times 4 \times 6 = 48c{m^2}/\sec \]
Thus the rate at which the area is increasing is \[48c{m^2}/\sec \].
Note:
Note that the rate of a particular quantity is the change in the quantity per unit time. If the rate is positive then it indicates the change is increasing and if it is negative then the change is decreasing. Also note that the unit of rate is per unit time. We found the rate of change of area of square by differentiating the formula of area of square. We do this in case of volume also. A single differentiation is the rate of change of that quantity.
Complete step by step answer:
First we will note the given data.
The side of a square is increasing at the rate of 6cm/s. let s be the side of the square. Thus the rate will be,
\[\dfrac{{ds}}{{dt}} = 6cm/\sec \]
Now area of the square is 16cm2
But the area of a square is given by, \[Area = {s^2}\]
Thus the side of the square , \[16 = {s^2}\]
Taking roots on both sides, \[s = \sqrt {16} = 4cm\]
Now the rate of area is obtained by, \[\dfrac{{dA}}{{dt}} = 2s.\dfrac{{ds}}{{dt}}\]
Now putting the values in the equation above, \[\dfrac{{dA}}{{dt}} = 2 \times 4 \times 6 = 48c{m^2}/\sec \]
Thus the rate at which the area is increasing is \[48c{m^2}/\sec \].
Note:
Note that the rate of a particular quantity is the change in the quantity per unit time. If the rate is positive then it indicates the change is increasing and if it is negative then the change is decreasing. Also note that the unit of rate is per unit time. We found the rate of change of area of square by differentiating the formula of area of square. We do this in case of volume also. A single differentiation is the rate of change of that quantity.
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