Question

Use differentials to approximate the maximum percentage error?

Answer 1

\(T=2\pi\sqrt{\Large\frac{L}{g}}\)
Given measurements: \(\frac{dL}{L}\)=0.5% and \(\frac{dg}{g}\)=0.1%
Solve for \(\frac{dT}{T}=?\)

I Started with taking the partial derivatives.
\(\Large \frac{∂T}{∂L}=\frac{2\pi}{\sqrt{g}}(\frac{1}{2})(L^{\frac{-1}{2}})=\frac{\pi}{\sqrt{gL}}\)

\(\Large\frac{∂T}{∂g}=\frac{2\pi}{\sqrt{L}}(\frac{-1}{2})(g^{\frac{-3}{2}})=-\frac{\pi\sqrt{L}}{g\sqrt{g}}\)

\(\Large\frac{dT}{T}=\frac{\frac{∂T}{∂L}}{T}(\frac{dL}{L})+\frac{\frac{∂T}{∂g}}{T}(\frac{dg}{g})\)

\(dT=\Large \frac{\pi}{\sqrt{gL}}(\frac{dL}{L})-\Large\frac{\pi\sqrt{L}}{g\sqrt{g}}(\frac{dg}{g})\)

\(\Large \frac{dT}{T}=\frac{\LARGE\frac{\pi}{\sqrt{gL}}}{\Large T}(\frac{dL}{L})-\frac{\LARGE\frac{\pi\sqrt{L}}{g\sqrt{g}}}{\Large T} (\frac{dg}{g})\)

We know that \(T=2\pi\sqrt{\Large\frac{L}{g}}\)
\(\Large \frac{dT}{T}=\frac{\LARGE\frac{\pi}{\sqrt{gL}}}{\Large 2\pi\sqrt{\Large\frac{L}{g}}}(\frac{dL}{L})-\frac{\LARGE\frac{\pi\sqrt{L}}{g\sqrt{g}}}{\Large 2\pi\sqrt{\Large\frac{L}{g}}} (\frac{dg}{g})\)

\(\Large\frac{dT}{T}=(\frac{\pi}{\sqrt{g}\sqrt{L}}* \frac{\sqrt{g}}{2\pi\sqrt{L}})

(\frac{dL}{L})-(\frac{p\sqrt{g}}{2\pi\sqrt{L}* \frac{\sqrt{g}}{2\pi\sqrt{L}}})(\frac{dg}{g})\)

\(\Large \frac{dT}{T}=\frac{1}{2}(\frac{dL}{L})-\frac{1}{2}(\frac{dg}{g})\)

\(\Large\frac{dT}{T}=\frac{1}{2}(\frac{dL}{L}-\frac{dg}{g})\)

\(\Large\frac{dT}{T}=\frac{1}{2}(\frac{dL}{L}-\frac{dg}{g})\)

We know that \(\frac{dL}{L}=\pm 0.005\) and \(\frac{dg}{g}=\pm 0.001\)

\(\Large{\frac{dT}{T}}=\frac{1}{2}\large ( 0.005 -0.001)=0.002\)


Final answer is 0.2%

Answer 2

But the book says the answer is 0.3%

Answer 3

There could be an error on my work but i have no idea of what it is.

Answer 4

@dan815 @ParthKohli

Answer 5

can yall folks help me

Answer 6

(0.005-(-0.001))/2 = 0.003

Answer 7

The only thing that will make the percentage error 0.003 is to change the sign to positive. From where did you get the positive sign? @dan815

Answer 8

i dunno lol it does that its +/- there so xD

Answer 9

do u have another example of these error calculations

Answer 10

\(\Large\frac{dT}{T}=\frac{\frac{∂T}{∂L}}{T}(\frac{dL}{L})+\frac{\frac{∂T}{∂g}}{T}(\frac{dg}{g}) \)
In this equation, we are adding. What caused the sign to change to negative is the partial derivative with respect to g

\(\Large\frac{∂T}{∂g}=\frac{2\pi}{\sqrt{L}}(\frac{-1}{2})(g^{\frac{-3}{2}})=-\frac{\pi\sqrt{L}}{g\sqrt{g}}\)

Answer 11

A simple sign change like this is very minor and the book would make an error if they added the measurement errors instead of subtracting them.

Answer 12

Correction on the 7th line:
.\(\Large\frac{dT}{T}=(\frac{\pi}{\sqrt{g}\sqrt{L}}* \frac{\sqrt{g}}{2\pi\sqrt{L}})-(\frac{ \pi \sqrt{L} }{ g \sqrt{g} }*\frac{ \sqrt{g} }{ 2 \pi \sqrt {L} })\)

Answer 13

i don't get this line at all:
\(\Large\frac{dT}{T}=\frac{\frac{∂T}{∂L}}{T}(\frac{dL}{L})+\frac{\frac{∂T}{∂g}}{T}(\frac{dg}{g})\)

probably my bad .....but i did it independently and got the same 0.2% result, FWIW.

ie by doing \(\Large\frac{dT}{T}=\frac{\frac{∂T}{∂L}dL}{T}+\frac{\frac{∂T}{∂g}dg}{T}\)

Answer 14

I used that process because it is the only way to plug in t and get the percentage error. Since i did ΔT/T, i had to do the same for the relative errors: ΔL/L and Δg/g.
It seems that the percentage error is 0.002. No doubt, the book is wrong
Thanks @IrishBoy123

Answer 15

Formula is similar to: http://www.rit.edu/cos/uphysics/uncertainties/differential.gif

Answer 16

thank you zale. i just stuffed it all in a typical differential [as per attachment, which does not help unravel the mystery].....:p

Answer 17

Awesome work @IrishBoy123 !

Answer 18

The reason our answer turns out to be 0.003 is that in error arithmetic, all errors are reported in \(\pm\) terms so carrying things to other sides sides does not change their signs to negative.

Answer 19

so the last line becomes \[\dfrac{dT}{T} = (\pm\dfrac{0.5}{2}) - (\pm \dfrac{0.1}{2})\] and you are looking for the biggest poss difference .... which should be \(\large (\dfrac{0.5}{2}) - (- \dfrac{0.1}{2}) = 0.3\%\)

??

Answer 20

OH That's right. Thanks guys!!!!!