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A tube is being stretched while maintaining its cylindrical shape. The height is increasing at the rate of 2 millimeters per second. At the instant that the radius of the tube is 6 millimeters, the volume is increasing at the rate of 96Ï€ cubic millimeters per second. Which of the following statements about the surface area of the tube is true at this instant? (The volume V of a cylinder with radius r and height h is V=Ï€r2h. The surface area S of a cylinder, not including the top and bottom of the cylinder, is S=2Ï€rh.)

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akiran007

Answer:

V = πr²h

S=2Ï€rh

h= 2/3 millimeters

new height= 8/3 millimeters

S= 8π  millimeters square

New surface area = 32π  millimeters square

Step-by-step explanation:

The volume of the cylinder is given by V = πr²h   where r is the radius and h is the height.

The surface area of the cylinder is given by  S= 2πrh + 2πr²

Where πr² gives the area of the base and 2πr² gives the area of the top and bottom surfaces. The surface area S of a cylinder, not including the top and bottom of the cylinder, is therefore S=2πrh.

V = πr²h

96π= π (6*6) (h+2)

96 = 36 (h+2)

96/36= h+2

h= 96/36-2

h= 96-72/36

h= 24/36

h= 4/6

h= 2/3 millimeters

New height

h + 2= 2/3 + 2

= 2+6/3= 8/3 millimeters

Now S =2Ï€rh

S = 2Ï€(6) (2/3)

S= 8π  millimeters square

New Surface area

S = 2Ï€(6) (8/3)

S= 32π  millimeters square

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