Error Propagation Sin Cos
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Error Propagation Sine
Physics Questions Tags Users Badges Unanswered Ask Question _ Physics Stack Exchange is a question and answer site for active researchers, error propagation exponential academics and students of physics. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to error propagation trig functions the top Error propagation estimations for sine and cosine up vote 0 down vote favorite My lab manual gives this: $B$ is a function of $A$, Greek are uncertainties... $$B + \beta = \sin(A + \alpha) = \sin(A)\cdot\cos(\alpha) + \sin(\alpha)\cdot\cos(A)$$ --> because $\alpha$ is taken to be (at least relatively) small, $\cos(\alpha) \to 1$, and $\sin(\alpha) \to \alpha$, measured in radians. I see that $\cos(\alpha) \to 1$, but I would
Uncertainty Of Sine
have expected that $\sin(\alpha)$ for a small would, by the same logic, go to 0. Has it got something to do with the rate of change of either function near zero? Why is $\cos(\alpha)$ of small $\alpha$ not also proportional or written by relation to $\alpha$? experimental-technique error-analysis calculus share|cite|improve this question edited Jan 17 '14 at 17:15 Kyle Kanos 18.8k103874 asked Jan 17 '14 at 16:22 user37464 684 1 Have a look at Taylor series. Normally you have to keep at least the linear term to have consistent results. –DarioP Jan 17 '14 at 16:33 add a comment| 3 Answers 3 active oldest votes up vote 1 down vote accepted The easiest way is to plot $x$, $\sin(x)$, and $\cos(x)$. Fortunately, Wikipedia has done that for us: From the first graph, when $x\lesssim0.2$ rad, $\sin(x)\simeq x$. From the second graph, the approximation that $\cos(x)\simeq1$ really only holds when $x\lesssim0.1$ rad; normally one writes it as $\cos(x)\approx1-x^2/2$. The reason for these approximations come from their series expansion: $$\sin(x)=x-\frac16x^3+\frac{1}{120}x^5+\cdots\\ \cos(x)=1-\frac12x^2+\frac{1}{24}x^4+\cdots $$ when $x$ is small, $x^3\approx0$, and all other higher terms are also zero, thus we eliminate them from the expansion. share|cite|improve this answer answered Jan 17 '14 at 16:35 Kyle Kanos 18.8k103874 As a
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Error Propagation Calculator
Quebec Singapore Taiwan Hong Kong Spain Thailand UK & Ireland Vietnam Espanol About About Answers Community Guidelines Leaderboard Knowledge Partners Points & Levels Blog Safety Tips Science & Mathematics Physics Next Error propagation of http://physics.stackexchange.com/questions/94110/error-propagation-estimations-for-sine-and-cosine sine? For my physics lab class. Find sin(theta), theta=.31 + or - .01 radians. What will the error be? 1 following 3 answers 3 Report Abuse Are you sure you want to delete this answer? Yes No Sorry, something has gone wrong. Trending Now Ali Wong Demi Lovato Isla Fisher Padma Lakshmi Bruce Springsteen Mortgage Calculator Conor McGregor Darth Vader Online Schools Dating Sites Answers Relevance Rating https://answers.yahoo.com/question/index?qid=20110926115447AAxjvqN Newest Oldest Best Answer: You can do it explicitly. Leaving out units for neatness and not worrying about significant figures: sin(0.31+0.01) = sin(0.32) = 0.3146 sin(0.31) =0.3051 sin(0.31-0.01) = sin(0.30) = 0.2955 So to a reasonable approximation, the error is +/- (0.3146-0.2955)/2 = +/- 0.00955 This is a percentage error of 100 x 0.00955/0.3051 = 3.1% The formal method is: y = sin(x) dy/dx = cos(x) Δy = (dy/dx)Δx = (cos(x))Δx So if x =0.31 and Δx =0.01, Δy =cos(0.31) * 0.01 = 0.00952 You might find the link useful. Source(s): http://www.rit.edu/cos/uphysics/uncertai... Steve4Physics · 5 years ago 2 Thumbs up 2 Thumbs down Comment Add a comment Submit · just now Asker's rating Report Abuse Error Propagation Formula Source(s): https://shrink.im/a0c3h casstevens · 1 week ago 0 Thumbs up 1 Thumbs down Comment Add a comment Submit · just now Report Abuse Since the variable with an attached uncertainty is within a sine function, it can be useful to apply the generalized propagation of error formula to it. Since the function contains a single term and will involve a single derivative this will be relatively simple and we do not have to distinguish whether it is a standard deviation or
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn http://math.stackexchange.com/questions/963803/relative-error-of-cos-and-sin-functions more about Stack Overflow the company Business Learn more about hiring developers or https://www.physicsforums.com/threads/calculating-uncertainty-with-a-sine-function.556268/ posting ads with us Mathematics Questions Tags Users Badges Unanswered Ask Question _ Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can error propagation answer The best answers are voted up and rise to the top Relative error % of cos and sin functions? up vote 0 down vote favorite I've been searching all over the net and I can't seem to find a definitive answer - perhaps I'm asking the wrong question. How does one calculate the relative error (%) of the cos/sin/tan of an angle in degrees? So, let's error propagation sin say that I have an angle of 30 degrees with an absolute error of ±2. The absolute error of the sin of 30 degrees would be: $sin (30+2)-sin(30) = 0.0299 $ Now what do I do to obtain the relative error? error-propagation share|cite|improve this question asked Oct 8 '14 at 14:20 Ursa Major 173212 Relative error is absolute error divided by function value. –Arthur Oct 8 '14 at 14:22 add a comment| 1 Answer 1 active oldest votes up vote 1 down vote accepted By definition relative error is given by $\delta f / f$ so f here is sin30 and the numerator is the difference you have written in your question. To calculate percentage error just multiply relative error by 100. share|cite|improve this answer answered Oct 8 '14 at 14:27 Jasser 1,523418 add a comment| Your Answer draft saved draft discarded Sign up or log in Sign up using Google Sign up using Facebook Sign up using Email and Password Post as a guest Name Email Post as a guest Name Email discard By posting your answer, you agree to the privacy policy and terms of service. Not the answer you're looking for? Browse other quest
Community Forums > Science Education > Homework and Coursework Questions > Introductory Physics Homework > Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors Dismiss Notice Dismiss Notice Join Physics Forums Today! The friendliest, high quality science and math community on the planet! Everyone who loves science is here! Calculating Uncertainty With A Sine Function Dec 2, 2011 #1 sunjay03 1. The problem statement, all variables and given/known data I am doing this calculation in my lab: h = sin(24.0°)[(180.0cm)(1m/100cm)] The uncertainty on the angle is ±0.5° and on the length it is also ±0.5cm. How can I go about calculating the uncertainty? If you know the answer, do you mind putting it in terms of a non-calculus student. I googled and found this, but could not understand a word of what they meant: http://www.sosmath.com/CBB/viewtopic.php?t=45581 I'm in more need of a simple formula I can plug numbers into to get the correct uncertainty. The understanding of the math behind it is not as relevant at this time. 2. Relevant equations N/A 3. The attempt at a solution I can get the answer, but not the uncertainty. Here is the answer: h = 0.732m Thanks for your help. sunjay03, Dec 2, 2011 Phys.org - latest science and technology news stories on Phys.org •Game over? Computer beats human champ in ancient Chinese game •Simplifying solar cells with a new mix of materials •Imaged 'jets' reveal cerium's post-shock inner strength Dec 2, 2011 #2 gneill Staff: Mentor Presumably the factor (1m/100cm) is simply a conversion factor that has infinite precision. You have only two variables with uncertainties attached, namely θ and x, where: h(θ,x) = sin(θ)*x*(1m/100cm) is the function that returns your result, and for which you want to propagate the uncertainties of the variables. Unfortunately, the way to compute the uncertainty in this situation does involve a bit of calculus. Have you had any calculus instruction at all? gneill, Dec 2, 2011 Dec 2, 2011 #3 sunjay03 gneill said: ↑ Presumably the factor (1m/100cm) is simply a conversion factor that has infinite precision. You have only two