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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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.

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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