Question

Error Analysis question

Answer 1

Given dimensions \[1.32 \pm 0.01 ~~~~~ by~~~~~~ 3.25 \pm 0.01 \]
and thickness \[8.0 \pm 0.1 \]

\[\rm using ~the~ propagation~ of~ error~ relationship ~: (\frac{ \delta D }{ D })^2= (\frac{m \delta X}{X})^2 +(\frac{n \delta Y}{Y})^2 +(\frac{p \delta Z}{Z})^2\]
write the expression for the relative error in the volume (deltaV/V) calculated from the measured data

Answer 2

\[\rm (\frac{ \delta D }{ D })^2 = (\frac{ m \delta X}{X})^2 +(\frac{n \delta Y}{Y})^2 +(\frac{p \delta Z}{Z})^2\] *

Answer 3

@mathmale

Answer 4

Do yew still need help?

Answer 5

yeah....??

Answer 6

I can't upload a powerpoint thing for some reason.
Add the fractional errors in each measurement separately. Multiply by 100 for the %ge error. Calculate the volume and quote the volume with the calculated %ge error is one way of doing this.

Answer 7

For Volume(??).....it's unclear......try:

\(V = xyz\)

\(\ln V = \ln xyz = \ln x + \ln y + \ln z\)

And then, by differentials....
\(\dfrac{dV}{V} = \dfrac{dx}{x} + \dfrac{dy}{y} + \dfrac{dz}{z}\)

For Surface Area (mmmmm....????.....squared terms), that should be easier.

Answer 8

hahah i think i know you

Answer 9

actually 100% lol

Answer 10

:/ ...

Answer 11

and......so what ?!?!

Answer 12

Nothing.
Thanks for the help.

Answer 13

😂

mp, ma'am

Answer 14

mp= my problem ?

Answer 15

mp = my pleasure/privilege

:)

Answer 16

ohh ok.

Answer 17

V= LWH \[\frac{ \delta V }{ V }=(\frac{\delta L}{L} )^2 +(\frac{\delta W}{W})^2+(\frac{\delta H }{H})^2\]

Answer 18

\[(\frac{ \delta V }{ v })^2\]**

Answer 19

\[\frac{ \delta V }{ V } =\sqrt{( \frac{\delta L}{L})^2+(\frac{\delta W}{W})^2+(\frac{\delta H}{H}^2)}\]

Answer 20

\[ \delta V=V\sqrt{(\frac{ 0.01 }{ 1.32 })^2+(\frac{ 0.01 }{ 3.25})^2+(\frac{ 0.1 }{ 8.0 })^2}\]