SOLUTION: A sphere a radius of 60 meters is to be constructed. If the radius is made 0.01 centimeters too long, what is the approximate error in the surface area of the sphere? If the sph

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Question 820990: A sphere a radius of 60 meters is to be constructed.
If the radius is made 0.01 centimeters too long, what is the approximate error in the surface area of the sphere?
If the sphere cost $225/m^2, how much does this miscalculation cost the builder?
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
Simplest idea is to try to compare the two circles: The ideal one with the errored one.

Ideal 60 m radius: A=pi*6000^2 in square centimeters.
Too Big radius+0.01 cm: a=pi*6000.01^2 square centimeters.

Actually, since you are interested in money cost for area in square METERS, you may want this way:

Units of SQUARE METERS:
Ideal: A=pi%2A60%5E2
Too Big: a=pi%2A%2860%2B0.0001%29%5E2
Excess Area is then pi%2860.0001%5E2-60%5E2%29
'
Take advantage of Difference Of Squares.
Excess area is highlight_green%28pi%2860.0001%2B60%29%2860.0001-60%29%29
highlight_green%28pi%2A120.0001%2A0.0001%29
and having only two significant figures, an adequate area answer for the excess is highlight%280.12%29 square meters.




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The cost of this error is 27 dollars.