Mathematics: Asquare has a side length x cm and area a cm2. the area is increasing at a constant rate of 0.03 cm2s-1.find the rate of increase of x when a=25. ( solve and explain? )

Asquare has a side length x cm and area a cm2.

Xavier8247

11.03.2022

The rate of increase of x when A = 25 is 0.0006 cm/s.

Step-by-step explanation:

Area of a square A = $x^{2}$ sq. cm

So, x = $\sqrt{A}$ cm

$\frac{dx}{dt} =\frac{1}{2\sqrt{A} } (\frac{dA}{dt} )$ --- (1)

It is given that the area is increasing at the rate of 0.03 $cm^{2}$ /s.

Therefore,

$\frac{dA}{dt} =0.03 cm^{2} /s$

We need to find the rate of increase of x when A = 25.

Now, substituting the values of A and $\frac{dA}{dt}$ , (1) becomes

$\frac{dx}{dt} =\frac{1}{2\sqrt{25} } (0.03)$

= 0.0006

Hence, the rate of increase of x when A = 25 is 0.0006 cm/s.

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msanger7587

13.12.2020

answer:

Thank you so much

Step-by-step explanation:

I am joyous to assist you anytime.

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