Question

The volume of a cube is claimed to be 27 cu.in. correct to within + or -.027 cu.in. Use differentials to estimate the possible error in the measurement of the side of the cube

Answer 1

Similar to the last problem, but this time you want to find dw/dV, where w is the width of the side of the cube.

Answer 2

dV will be +/- 0.027 cubic in

Answer 3

and the formula is s^3?

Answer 4

and s=3 b/c i cubed 27

Answer 5

@agent0smith this one? :D

Answer 6

This one should be easy compared to the other. Yes s=3inches but remember we're only interested in the error in the measurement of the side. So we want ds/dV where ds is the error in the side, dV is the error in volume (+/- 0.027 cubic inches).

\[V=s ^{3}\]
so s is the cube root of V or V^(1/3)
\[s=\sqrt[3]{V} = V ^{\frac{ 1 }{ 3 }}\]

Answer 7

So differentiate that to get ds/dV
\[\frac{ ds }{ dV }=\frac{ 1 }{ 3 }V ^{\frac{ -2 }{ 3}}\]
\[ds=\frac{ 1 }{ 3 }V ^{\frac{ -2 }{ 3}}dV\]
And now you can just enter V and dV.

Answer 8

ahh okay ty <333

Answer 9

#agent0smith can you help me with 2 more problems? x-x

Answer 10

I might but i'm going to bed pretty soon. You should get ds = 0.001 btw, so the error in the side is +/- 0.001 inches.

Answer 11

aw. okay :(( it's been already past my bed time here xD 5:32 AM here :P

Answer 12

haha, 5:32 am here too

Answer 13

LOL and you're going to bed at this time? xD

Answer 14

haha yup. Almost always do.

Answer 15

eh sometimes I don't get sleep cause of my AP classes xD