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The radius of the circle is increasing at a rate of 2 meters per minute and the sides of the square are increasing at a rate of 1 meter per minute. When the radius is 6 meters, and the sides are 24 meters, then how fast is the AREA outside the circle but inside the square changing?

Respuesta :

braza909

Answer:

Change in area=24[tex]\pi[/tex]-48

Step-by-step explanation:

Let s will be the side of square and r will be the radius of circle.

Then two given conditions are

1)dr/dt=2 m/s

2)ds/dt=1 m/s

Area enclosed=(Area of square)-(Area of circle)

Area of square=[tex]s^{2}[/tex]

Area of circle=[tex]\pi r^{2}[/tex]

Area enclosed=[tex](\pi r^{2})-s^{2}[/tex]

dA/dt=2[tex]\pi[/tex]r(dr/dt)-2s(ds/dt)

At s=24,and r=6

dA/dt=2([tex]\pi[/tex])(6)(2)-2(24)(1)

Change in area=24[tex]\pi[/tex]-48

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