The surface of the cube is to be goldplated. If the length of 1 edge of the cube is 10 inches, use differentials to approximate the volume of gold to be used if the plating is 0.001 inches thick.
2007-01-30 16:50:14 · 2 answers · asked by Sammy Baby 1 in Science & Mathematics ➔ Mathematics
This doesn't require calculus.
g = volume gold
v = volume cube before gold plating
w = volume cube after gold plating
g = w - v = (10 + 2*0.001)³ - 10³ = (10 + 0.002)³ - 10³
= (10³ + 3*10²*.002 + 3*10*.002² + .002³) - 10³
= 3*10²*.002 + 3*10*.002² + .002³
= 300*.002 + 30*.000004 + .000000008
= 0.6 + 0.00012 + 0.000000008 = 0.600120008
Perhaps this is what you had in mind by differentials.
2007-01-30 17:52:33 · answer #1 · answered by Northstar 7 · 0⤊ 0⤋
I may be nit picking but ther is a slight error in the above solution.
If you take any edge of the cube the thckness of the gold plating is notr 0.001 but varies from 0.001 to sqrt(2)times 0.001
So at the edges you have to plate it in a form resembling a quarter of a cylinder.
well is it a cube then ? may be approximately?
well if you can tolerate an approximate shape
Then may be you can tolerate a slight deviation in the thickness too i.e. above answer
2007-01-30 18:08:12 · answer #2 · answered by pradeep p 2 · 0⤊ 0⤋