Using implicit differentiation, it is found that the area of the square is increasing at a rate of:
The area of a square of side length l is given by:
[tex]A = l^2[/tex]
Applying implicit differentiation, the rate of change is given by:
[tex]\frac{dA}{dt} = 2l\frac{dl}{dt}[/tex]
- The side is 3 m in length, hence [tex]l = 3[/tex].
- Growing at a rate of 0.8 m/min, hence [tex]\frac{dl}{dt} = 0.8[/tex]
Then:
[tex]\frac{dA}{dt} = 2l\frac{dl}{dt} = 2(3)(0.8) = 4.8[/tex]
Hence, option A is correct.
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