Hi guys,
I'm stuck on a word problem and could use a little help. I did the work, just not real confident in it.
---> Ok, so what we're given is r = 1 cm and dr = +/- .01 cm (correct?)Company manufactures spherical ball bearing with radius 1 cm and percentage error in radius must be no more than 1%.
a) Find approximate percentage error for surface area of ball bearing?
b) Find range of radii possible.
c) What range of surface area possible.
a) Percentage error surface area
---> S=4(pi)r^2
--->dS=8(pi)r(dr) = 8(pi)(1cm)(+/- .01 cm) = +/- .25133 cm^2
---> Percent error = dS/S
-----------> S=4(pi)(1cm)^2= 12.56637 cm^2
----------->dS/S = (+/- .25133 cm^2)/(12.56637 cm^2) = .020000 = 2%
(Correct?)
b) Range of radii
-->would be: 1 cm + .01 cm = 1.01 and 1 cm - .01 = .99 cm
-->RANGE = .99 - 1.01 or +/ .01 cm (correct?...just seems a little simple)
c) Range Surface area
--> S = 4(pi)r^2
--> dS = 8(pi)r(dr) = 8(pi)(1cm)(+/- .01 cm) = +/- .25133 cm^2
(correct?)
EDIT: Perhaps on C, the range would be (12.56637 + .25133 = 12.8177 cm^2 and 12.56637 - .25133 = 12.31504 cm^2
Thanks for the help. Will return the favor if you'd post a link to your problem.