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increasing at the rate of 3 cm per minute. How fast issurface area of the sphere...

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jmg5639 | Student, Undergraduate | Honors

Posted June 18, 2012 at 5:21 AM via web

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increasing at the rate of 3 cm per minute. How fast issurface area of the sphere increasing, when the radius is 10 cm?(S = 4πR^2)

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The radius of a spherical balloon is increasing at the rate of 3 cm per minute. How fast is the surface
area of the sphere increasing, when the radius is 10 cm? Round to 1 decimal place. (S = 4πR^2)

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thilina-g | College Teacher | (Level 1) Educator

Posted June 18, 2012 at 9:58 AM (Answer #1)

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The surface area of a sphere `A = 4pir^2`

The rate of increase of radius is 3 per minute

`(dr)/(dt) = 3`

`(dA)/(dt) = 4pixx2rxx(dr)/(dt)`

`(dA)/(dt) = 8pir(dr)/(dt)`

At `r = 10,`

`(dA)/(dt) = 8pixx10xx3`

`(dA)/(dt) = 240pi `

 

Therefore at `r = 10` , the rate of area increase is `240pi m^2` per minute.

 

 

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vaaruni | High School Teacher | Salutatorian

Posted June 18, 2012 at 8:10 AM (Answer #2)

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The radius of the baloon = 10 cm.

The Surface area(S1) = 4 * (22/7) * R^2 [ pie=22/7]

                          = 4*(22/7)*(10)^2 = 100 * 4 * (22/7) sq.cm^2 

Aafter one minute the radius increased by 3 cm. The radius after 1 minute = 10 + 3 = 13

The surface area after one minute(S2) = 4 * (22/7) ( 13)^2

                                                  = 169 * 4 * (22/7) sq.cm^2

The ratio S2/S1 =(169 * 4 * (22/7)) / (100 * 4 * (22/7))

               S2/S1= 169 / 100 = 1.69 sq.cm^2

  Area will increase by 1.7 sq.cm^2  per minute.  Answer           

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