Answer:
The rate of increase of x when A = 25 is 0.0006 cm/s.
Step-by-step explanation:
Area of a square A = [tex]x^{2}[/tex] sq. cm
So, x = [tex]\sqrt{A}[/tex] cm
[tex]\frac{dx}{dt} =\frac{1}{2\sqrt{A} } (\frac{dA}{dt} )[/tex] --- (1)
It is given that the area is increasing at the rate of 0.03 [tex]cm^{2}[/tex]/s.
Therefore,
[tex]\frac{dA}{dt} =0.03 cm^{2} /s[/tex]
We need to find the rate of increase of x when A = 25.
Now, substituting the values of A and [tex]\frac{dA}{dt}[/tex], (1) becomes
[tex]\frac{dx}{dt} =\frac{1}{2\sqrt{25} } (0.03)[/tex]
= 0.0006
Hence, the rate of increase of x when A = 25 is 0.0006 cm/s.
