# Vectors, Component Addition, and Significant Figures

I have two vectors $\vec{A}$ and $\vec{B}$ and I need to find the x- and y-components of $\vec{C} = \vec{A} + \vec{B}$. Here's what I have so far:

$$|\vec{A}| = 50.0 \mathrm{m}, \theta = -20.0^\circ$$ $$|\vec{B}| = 70.0 \mathrm{m}, \theta = 50.0^\circ$$

$$C_x = |\vec{A}| \cos (\theta) + |\vec{B}| \sin (\theta)$$ $$C_y = |\vec{A}| \sin (\theta) + |\vec{B}| \sin (\theta)$$

Now, according to my professor, this is the solution for C_x:

$$C_x = 50.0 \cos(-20.0^\circ) + 70.0 \cos(50.0%^\circ)$$ $$C_x = 46.98^\circ + 45.0^\circ$$ $$C_x = 92.0^\circ$$

What I'm wondering is how the rounding works here. I got $46.98$ for $50.0 \cos(-20.0^\circ)$ and $44.99$ for $70.0 \cos(50.0^\circ)$. Why is $44.99$ rounded to $45.0$? If anything, shouldn't it be rounded to $45.00$? What am I missing here?

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The instructor is being slightly inconsistent, by giving one part to 4 places, and the other to 3. You can be similarly inconsistent too, so long as you don't keep six figures of accuracy without justification. – Ron Maimon Sep 10 '11 at 23:04
The truth is that no one explicitly uses the significant figures rules beyond high school. Typically one regards experimental measurements as defining a particular probability distribution for the true value of the quantity. Ultimately the result is expressed as the best estimate of the mean (or most likely) value of the quantity, with error bars describing the uncertainty (i.e. the standard deviation of the probability distribution). The main purpose of 'sig fig' exercises is to bring home the point that it is pointless to include digits beyond the measured precision. – nibot Sep 12 '11 at 1:25

I think the real question is actually posed most directly in your comment (so you might want to consider editing some of this into the original question):

I'm more concerned about understanding whats going on and making sure that I know how to do it. I've heard that you shouldn't worry about significant figure rules until you have your final answer. How many decimal places should you round a number like 50.0 cos(-20.0) to? Do you always round to 2 decimals as in my problem?

Yes, you are correct that you should never actually round a number off until you are done with the calculation. However, when you are writing out your intermediate steps, it's common practice to write rounded values, rather than copying every digit your calculator shows you, just to avoid burdening the reader with a lot of extra digits that don't really add anything interesting. Keep in mind that this convention only affects what you write. You still keep the number to full precision in your calculator.

As for choosing the number of digits to write out, you can use the significant figure rules, which go like so:

• Addition and subtraction: find the last significant digit of each number, and choose the one with the larger place value. That place should be the last significant digit of your result. Another way to think of this rule is that a digit in the sum (or difference) is not significant unless both digits that were added to produce it are significant. So, using gray shading to designate insignificant digits:

\begin{align*}3&.146309\\+2&.71\\ =5&.85\color{red}{6309}\end{align*}

So you would round to the last significant digit in this case, i.e. you would write out $5.86$. But if you use this result again:

\begin{align*}5&.85\color{red}{6309}\\+4&.93101\\ =10&.78\color{red}{7319}\end{align*}

This time you would again round to the last significant digit, and write out $10.79$.

• Multiplication and division: your result should have the fewest number of significant digits of either of the numbers you're multiplying or dividing.

\begin{align*}&253.1\\\div &45\\ = &5.6\color{red}{2444\ldots}\end{align*}

and if you multiply this by the earlier result,

\begin{align*}&5.6\color{red}{2444\ldots}\\\times &10.78\color{red}{7319}\\ = &60.\color{red}{67268\ldots}\end{align*}

These rules are a simplification of a slightly more complex (but more accurate) system, the error propagation rules, which physicists normally use in research. Unfortunately, the error propagation rules for functions like the sine and cosine can't be simplified quite so easily, so in practice people often just use the multiplication/division rule (write out the fewest number of significant digits) for everything else not mentioned here. Of course, you have to remember that, except for final results, it's really not that important how many digits you write, since you should never be rounding "behind the scenes" in your calculator anyway.

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This is what I was looking for! Thanks :) – Brewer Sep 11 '11 at 2:52
Nice use of \color! – nibot Sep 12 '11 at 1:18

It's 44.9951 (source), which is closer to 45.00 than to 44.99.

45.00 and 45.0 are the same thing.

If you are concerned about accuracy, you can calculate the answer exactly in terms of trig functions. Use the law of cosines and the law of sines.

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I'm more concerned about understanding whats going on and making sure that I know how to do it. I've heard that you shouldn't worry about significant figure rules until you have your final answer. How many decimal places should you round a number like 50.0 cos(-20.0) to? Do you always round to 2 decimals as in my problem? – Brewer Sep 10 '11 at 22:32
@brewer The accuracy of your estimates depends on the accuracy desired in your answer. If you need a very accurate answer, keep lots of decimal places. If you need an order-of-magnitude answer, keep none. Any rule you hear to the effect of "always keep two decimal places" is ridiculous nonsense invented to keep schoolchildren quiet. In general, you will need to learn about error analysis. A good book is John Taylor's amazon.com/… – Mark Eichenlaub Sep 10 '11 at 22:39
Starting with an input like 50.0 you ought not to recommend using 45.00 as the output (especially on a physics website) as, absent an indication to the contrary, it has the implicit connotation of a precision to the last quoted digit (i.e. ±0.01 in your example) when in fact you probably don't know the answer to better than ±0.1. That is the heart of the question and I think this answer is misleading. The accuracy of the answer can never be greater than the accuracy of the inputs, regardless of how many digits you need (your word...). 45.0 is NOT the same (except numerically) as 45.00 – Floris Mar 13 '15 at 20:48