Why did you divide it by 1/3 after you calculated delta q in the first example calculation? Is it because you are adding three quantities to calculate the average?
@@aqsaamin96 See that is the part that I was confused about. If we divide by the quantities being added, then our uncertainty becomes less than the uncertainty that those quantities started with. So how could an average uncertainty be less than the uncertainty of the starting values.
We took the partial derivative of xe^-y with respect to x. Therefore, we treat e^-y as a constant multiple of x resulting in just e^-y after differentiating (like how 2x differentiates to 2). Then we substituted the values x=20 and y=1 resulting in dq/dx = e^-1 = 0.3679 at the point of interest.
Thank you. You are the only one who explain error propagation the way my lecturer wants in exams.
What an eye-opening lecture this was. Thank you for this explanation.
the beginning is like how uncertainty propagates
You are a life saver, thanks!
Why did you divide it by 1/3 after you calculated delta q in the first example calculation? Is it because you are adding three quantities to calculate the average?
Should we always divide the uncertainty after squaring and rooting it with the number of quantities involved?
@@aqsaamin96 See that is the part that I was confused about. If we divide by the quantities being added, then our uncertainty becomes less than the uncertainty that those quantities started with. So how could an average uncertainty be less than the uncertainty of the starting values.
My life saver
I'm not understanding how you got e^-1 when you took the partial with respect to x of e^-y... wouldn't the partial of e^-y with respect to x be 0?
We took the partial derivative of xe^-y with respect to x. Therefore, we treat e^-y as a constant multiple of x resulting in just e^-y after differentiating (like how 2x differentiates to 2). Then we substituted the values x=20 and y=1 resulting in dq/dx = e^-1 = 0.3679 at the point of interest.