Tracking significant figures through a complicated problem

[ago01]

Homework Statement

calculating the magnitude of the sum of vector A who has magnitude 13.0 and angle 38.7 degrees off the horizontal and a vector who has magnitude 5.0 and angle 26.0 degrees off the horizontal. Use proper sig figs.

Relevant Equations

$|V| = sqrt(V_x^2 + V_y^2)$

In general I think I follow significant figures. On one and two step problems I don't really have such a problem but I seem to constantly miss points because I get lost in complicated ones. I'm hoping I can understand more as getting into science means getting friendly with these.

Let's suppose I am calculating the magnitude of the sum of vector A who has magnitude 13.0 and angle 38.7 degrees off the horizontal and a vector who has magnitude 5.0 and angle 26.0 degrees off the horizontal.

Then:

$R_x = (13.0)cos(38.7) + (5.0)cos(26.0)$

So the first product maintains 3 sig figs, and the second retains 2 sig figs.

$R_x = 7.014 + 4.49$

retaining one sig fig extra (my professor says to do this instead of holding the whole number).

$R_x = 11.508$

Retained to 1 extra sig fig (so now we have 2 decimals of precision available).

To the correct decimals of precision it's 11.51.

The same logic applies to #R_y#

$R_y = (13.0)sin(38.7) + (5.0)sin(26.0)$

$R_y = 10.94 + 3.81$

$R_y = 14.75$

In this case the extra retained SF in both the products ended up producing the right amount of degrees of precision here.

Then the magnitude of R

$|R| = \sqrt{11.508^2 + 14.75^2}$

Since the equations above say we really don't "have" the 8 in the first square we want 4 SF, and we don't really "have" the 5 in the second square so we want 3 SF there.

$|R| = \sqrt{132.43 + 217.7}$

Retaining one extra sig fig. So we have 1 decimal of precision available to us in the sum, so we will retain 2 to reduce rounding errors.

$|R| = \sqrt{350.13}$

and we have 4 sig figs here (since the 3 was carried over)

$|R| = 18.71$

But this can't be right. The original question only had 2 SF available (5.0)! So this should be 19 to our problem's precision no?I *think* I followed the logic correctly but I cannot figure out why I get more precision in the answer than I started with. That doesn't add up. Should I always round to the sig figs given to the problem or did I make another critical mistake here I am missing?

Thank you!

 

[jbriggs444]

Let's suppose I am calculating the magnitude of the sum of vector A who has magnitude 13.0 and angle 38.7 degrees off the horizontal and a vector who has magnitude 5.0 and angle 26.0 degrees off the horizontal.

Then:

$R_x = (13.0)cos(38.7) + (5.0)cos(26.0)$

So the first product maintains 3 sig figs, and the second retains 2 sig figs.

$R_x = 7.014 + 4.49$

So far this is pretty standard. Though another piece of advice is to avoid rounding or truncation of intermediate results at all. You can track significant figures without actually rounding.

retaining one sig fig extra (my professor says to do this instead of holding the whole number).

$R_x = 11.508$

So your professor would have you calculate with one more significant figure than is actually there.

Right here you have added two significant figures. One because the professor wants you to use intermediate results with one more significant figure than can be justified. That's fine as long as you trim it back off for the final result. The other because 7 plus 4 add to 11 (one sig fig plus one sig fig yields a result with two sig figs).

Since you never trim the one extra digit of precision off when you get a final result instead of an intermediate result, you end up with a result that is at least one more digit more precise than is proper and two more than you'd expected.

Note that two hash tags (##) are needed to trigger LaTeX processing here.

Significant figures are looked down on a bit by practicing scientists. They are quite crude as a means of reporting error bounds and tracking the way errors propagate through a calculation. However the more sophisticated approaches require background knowledge of calculus (things like partial derivatives), so significant figures are what get taught.

 

[ago01]

Right here you have added two significant figures. One because the professor wants you to use intermediate results with one more significant figure than can be justified. That's fine as long as you trim it back off for the final result. The other because 7 plus 4 add to 11 (one sig fig plus one sig fig yields a result with two sig figs).

Ok, so I'm following you to here.

We start with

$R_x = 7.014 + 4.49$

Which retained 4 SF in the first product because the previous had 3 SF (and we retained an extra) and the same for the second product, so we retain 3 (2 SF exist in reality).

The decimal rule says retain the least precision, and so we see that in *reality* we are looking at

$7.01 + 4.5$

and we see we only have one extra decimal of precision. So the result would be

$11.5$

because we only have two significant figures available to us (from the first product), plus one degree of precision from the sum.

Then we have to realize that we also only had 2 SF when we started calculating the products. So this answer would be

$12$

To signify this. This would explain the "sig fig check" rule we learned where the input sig figs and output sig figs should agree.

It sure is a nightmare tracking these. I'm happy to hear practicing scientists have methods that don't rely on arbitrary rules.

 

[mfb]

Sums work differently from products.

• What is the sum of 12345.6 + 0.1? It's 12345.7 with six significant figures, even though "0.1" just had one.
• What is the sum of 12.3 and 0.11472463534? It's 12.4, all other digits are not significant because the first number doesn't have them.

Instead of significant figures you need to keep the same significant decimal places. That means R_x should come with the last digit removed, and the sum under the square root needs its last digit removed as well. Once you do that, and cut the last digit in the last step, you get 18.7.

A quick way to check your results: How does the answer change if you replace 5.0 by 4.9 or 5.1? The two vectors are going in similar directions, so the magnitude of the sum will change by about 0.1 as well. That means 18.7 is a good answer.

 

[Steve4Physics]

From a practical point of view, remember scientific calculators provide nested brackets and memories. With a bit of thought/practice you effectively avoid rounding errors.

For example $R_x = (13.0)cos(38.7) + (5.0)cos(26.0)$ can easily be evaluated in a single calculation and the value stored in memory. (If required, this intermediate value could be written down rounded).

Similarly for $R_y$.

Then $|R| = \sqrt{R_x^2 + R_y^2}$ can be evaluated in another single calculation using the values stored in memory.

It’s even better if you have a spreadsheet.,

 

[DaveE]

You may find it helpful to remember that the significant digits imply an error range.

So, if x = 26 it has two significant digits. ⇒ 25.5 ≤ x < 26.5.
if x = 26.0 it has three significant digits. ⇒ 25.05* 25.95 ≤ x < 26.05.
[using the rounding up convention]This doesn't work for numbers expressed like 12300 which could be 1.23e4, 1.230e4, or 1.2300e4, but can't be the same as 1.2e4 or 1.23000e4.

* oops!

PS: I suppose this is all in the engineering/scientist context. Many mathematicians wouldn't care, most of the time. 26 is sometimes just the name of a concept, a member of a set, not a measurement.

 

[Mark44]

if x = 26.0 it has three significant digits. ⇒ 25.05 ≤ x < 26.05.

I think your lower end of the range has a typo, and should be 25.95.