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The edge of a cube was found to be 30 cm with a possible error in measurement of 0.1 cm. Use differentials to estimate the maximum possible error, relative error, and percentage error in computing (a) the volume of the cube and (b) the surface area if the cube.
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for volume v= x^3 dv/dx= 3x^2 dv= 3x^2 dx dv=3(30)^2 (.1)
relative error/percent dv/v= 3(30)^2 (.1) ------------ 30^3 for percent x100
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the surface area of the cube was the derivative of the volume of the cube right?
surface area of cube = 6x^2
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so then to find the surface area i would do 12xdx/6x^2?
yes
oh great i totally get it!
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