Question

Quick question: Yes or no type of answer. It's regarding propagated error.

Answer 1

It seems that question a is wanting me to do a plug and chug in p(9,6) on the original equation to compute for g. Correct? Simply, i can't get g after taking the partial derivative.

Answer 2

Nvm, this was a silly question

Answer 3

What i was thinking is correct.

Answer 4

Yes, for part \(a\), you simply plugin the measured values of L and T, and get the value of \(g\).

Answer 5

for part b, you may use total differential

Answer 6

may i see your work for part b ?

Answer 7

Thanks! Why would they want me to know what's g?

Answer 8

Calculating the propagated error doesn't require a value for g.

Answer 9

part a gives you an approximate value of \(g\)

part b gives you the error in that approximation

Answer 10

you're correct, you don't need part a to work out part b

Answer 11

\[\Delta g\approx g_L \Delta L + g_T\Delta T\]

Answer 12

we just need the partials and changes in L and T

Answer 13

No wait, you do need part \(a\) to calculate part \(b\). They want "relative error". So, you need to divide the error by the actual value of \(g\)

Answer 14

for part b, you need to compute \(\dfrac{\Delta g}{g}\)

Answer 15

Ah i see. I worked off b, c, but a just thre me off because it was way off topic lol xD

Answer 16

threw*

Answer 17

C is definitely the same :)
Actually, \(f_{xy}=f_{yx}\) always happen

Answer 18

Thank you @ganeshie8 . Stay awesome!

Answer 19

np :) just making sure... i hope you saw that \(\Delta L\ne 0.01\)

\(0.01\) is "relative error" in measuring \(L\). so \(\dfrac{\Delta L}{L}=0.01\)

Answer 20

http://mathworld.wolfram.com/RelativeError.html

Answer 21

I used this \(\dfrac{\Delta L}{L}=0.01\)
I think we are supposed to know that when calculating the ordinary derivative for g, it has to be \(\dfrac{d g}{g}\) and divide all the partial derivatives with g. Then the equation for g will get plugged in.... I think i should just type it

Answer 22

Do you guys know how to do a partial derivative symbol here? lol

Answer 23

looks nice

`\partial ` produces \(\partial\)

Answer 24

OHHHHH You mean the computed value for g should be in the denominator instead of plugging the whole equation?

Answer 25

Yes...

Answer 26

Here's what i did without using the computed value of g.
\(\Large \frac{\partial g}{\partial L}=\frac{4\pi^2}{T^2}\)
\(\Large \frac{\partial g}{\partial T}=\frac{-8\pi^2L}{T^3}\)

\(\LARGE \frac{dg}{g}=\frac{\frac{4\pi^2}{T^2}dL+\frac{-8\pi^2L}{T^3}dT}{\frac{4\pi^2L}{T^2}}\)
\(\LARGE \frac{dg}{g}=\frac{\frac{4\pi^2}{T^2}dL}{\frac{4\pi^2L}{T^2}}+\frac{\frac{-8\pi^2L}{T^3}dT}{\frac{4\pi^2L}{T^2}}\)

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

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

\(\LARGE \frac{dg}{g}=(\pm 0.01)-2(\pm 0.003)\)

Answer 27

\(\LARGE \frac{dg}{g}=\pm 0.016\)