1. Multiplying large numbers

I couldn't find an exact forum for this here - but it's very basic so I'd imagine anyone could help with it, and it stems from an algebra question so....

Basically when doing problems without using the calculator I've been finding that I'm having to do large long multiplications that seem like a pretty poor way to go about solving a problem.

My question is : am I going about things the correct way, and merely need to improve at them, or is there a different method that I haven't thought of for this?

So here's a multiplication of the type that I'm talking about.

As you can see, even though the answers is correct (hoorah...) the odds of me messing something up along the lines are fairly high. And I'm doing this with the luxury of a whiteboard space to spread out on....

How would you approach / do this problem?

Here's the problem in full if you'd like to see the question as presented

Thanks!

2. Re: Multiplying large numbers

You can try factoring, but in the end, if you are not using a calculator, you are going to be multiplying large numbers by hand, which is more error prone than using a calculator.

Here is how you might factor to simplify the problem:
\begin{align*}(-11)(12.2) + \dfrac{1}{2}(-9.81)(12.2)^2 & = -12.2\left(11+\dfrac{1}{2}(9.81)(12.2)\right) \\ & = -12.2(11+9.81\cdot 6.1)\end{align*}

There is no magic to this. You just have to plug in and multiply out.

3. Re: Multiplying large numbers

The obvious question is: why not use a calculator?

The second point I have is about significant digits - the problem statement says to round to two decimal places, so the answer to your multiplication is 1460.12 meters. But strictlty speaking that answer is wrong, because you have been given data that is good to only 3 significant digits. Hence the "correct" answer is 1.46 x 10^3 meters. To get this level of accuracy means you can round 12.2^2 to 149 and multiply that by 9.8, to get 1460. This is a much easier multiplication than what you did, much less prone to error, and more accurate in the sense that you're not tacking on extra digits that can't be justified.

4. Re: Multiplying large numbers

@SlipEternal -

Thanks for that... I didn't think to factor but both sides of the addition have 12.2 so that's a good way of simplifying a bit.

@Ebaines -

Why not... Well, 1: The book that I'm working from generally say's when to use a calculator, and 2: I'm learning so I figured I should be able to? Though I do lose track of when I should / shouldn't use a calculator sometimes, silly as that sounds...

the problem statement says to round to two decimal places, so the answer to your multiplication is 1460.12 meters
Yep

But strictly speaking that answer is wrong, because you have been given data that is good to only 3 significant digits
I'm not following you here - why would I assume the information given to me is inaccurate? Unless you're discussing thing's on some higher level theory (In which case it's going to be wasted on me... )

Hence the "correct" answer is 1.46 x 10^3 meters
Again, to me that looks worse than the original, given that the question specified answering with accuracy to 2 decimal places...

To get this level of accuracy means you can round 12.2^2 to 149 and multiply that by 9.8, to get 1460
I would have had to multiply 12.2^2 out by hand in order to know what it was... I know that 12^2 is 144 but wouldn't (couldn't) multiply out something like 12.2^2 without writing it down / using a calculator.

cheers

5. Re: Multiplying large numbers

My point about only going to three significant digits is that the data you were given - such as acceleration due to gravity of 9.81 m/s^2 - is accurate to only three digits, or one part in a thousand. Hence whatever answer you get is goint to be accurate to only three significant digits. But the answer of 1940.12 implies that your data is accurate to 6 digits, or one part in a million. In fact if you want that level of accuracy you would have to use g = 9.80665 m/s^2 instead of 9.81, and 12.2000 seconds instead of 12.2. Using this data gives an answer correct to six significant digits of 1459.62, not 1460.12.

This is a bit of a peave of mine. Of course I realize that the text is asking for 2 decimal points, so you have to proceed as you did. But in the "real" world giving an answer with more significant digits than the data you start with is wrong.

6. Re: Multiplying large numbers

My point about only going to three significant digits is that the data you were given - such as acceleration due to gravity of 9.81 m/s^2 -
Honestly - I'm sure that you know what you're talking about but I have no idea. I thought that from the topic and content of the post beginner / basic would have been assumed but perhaps I should have been more specific.

It sound's like you're talking about things outside of the question that I've been presented, which for me is confusing.

It's a pet hate for you, that's cool. But expressing your pet hates in a thread like this is helpful to no-one. It's only made me more confused.

7. Re: Multiplying large numbers

I don't mean to confuse you. This topic of significant digits is something appropriate for a beginning physics student (9th grade or above), which I assume you are given the nature of the equation you are working with. Yes, it's beyond the scope of the question you were asking - my apologies for trying to engage you to think about an important aspect of performing calculations.

8. Re: Multiplying large numbers

Yes, it's beyond the scope of the question you were asking
Which was my point, hence my confusion.

my apologies for trying to engage you to think about an important aspect of performing calculations
No need for the back hand

9. Re: Multiplying large numbers

Originally Posted by silverpen
Honestly - I'm sure that you know what you're talking about but I have no idea. I thought that from the topic and content of the post beginner / basic would have been assumed but perhaps I should have been more specific.

It sound's like you're talking about things outside of the question that I've been presented, which for me is confusing.

It's a pet hate for you, that's cool. But expressing your pet hates in a thread like this is helpful to no-one. It's only made me more confused.
In your first post, you ask about $(-11)(12.2) + \dfrac{1}{2}(-9.81)(12.2)^2 \approx -864.26$. Then, in your second picture, you show the formula $(11)(12.2) + \dfrac{1}{2}(-9.81)(12.2)^2 = -595.8602 \approx -595.86$

Anyway, if you count how many digits the number -595.86 has, it has 5 digits. Count the number of digits 9.81 has. It has 3 digits. Three is less than five. Any number that smaller than -9.805 and at least -9.815 will round to -9.81. So, if you were using a more accurate number for acceleration, a more accurate answer to your problem could be any number between $s_{min} = (11)(12.2)+\dfrac{1}{2}(-9.815)(12.2)^2 = -596.2323$ and $s_{max} = (11)(12.2)+\dfrac{1}{2}(-9.805)(12.2)^2 = -595.4881$. Note that rounding the first number to three significant digits gives $-596$ while the second gives $-595$. Thus, the answer is only accurate to two significant digits. This is an example that if you are given data that is only accurate to three digits, the outcome will only be accurate to at MOST three digits. And if 12.2 is actually a number between 12.15 and 12.25, and 11 is actually a number between 10.5 and 11.5, the potential for inaccuracy increases even more. So, the answer you obtain is only as good as the data you have. If you are given data with three significant figures, you know that your answer is correct to at most three significant figures. So, a more "accurate" answer that is correct to two decimal places AND three significant figures would be $-5.96 \times 10^2$ (this is called scientific notation).

So, ebaines answer is MORE than a personal pet peeve. It demonstrates potential INACCURACY in your textbook.

10. Re: Multiplying large numbers

The factoring was useful thanks I didn't think of that.

Using standard Index was something that I considered early on in this problem because I thought that It might be easier to adjust the number to the same 10^*** and work with them... Though after trying something along the lines of (1.42 * 10^1) * (7.82 * 10^1 ) instead of have 14.2 * 78.2, taking the decimals out, long multiplication and then putting them back in.... But I have to do the multiplication all the same, so I didn't really see the benefit to it.

Accuracy - I've been exposed to a bit of the 23 * 10^2 standard index notation, but I'm obviously not familiar with it... The only comparison that I could make is that a fraction is more accurate than a decimal or something along those lines? If that's the case then great, it makes sense but I'm still not connecting the dots to this problem.

So, if you were using a more accurate number for acceleration
I don't know what that sentence means
Note that rounding either number to three significant digits gives -1330, which is three significant digits
Are we talking about boundaries?? As in, 12.33 could be between 12.334 or 12.325 ?

Again, thats great but I'm not seeing the importance to this. To me it feels like you two have just gone off on some tangent that's no doubt true but not really useful to the situation.

So, the answer you obtain is only as good as the data you have
of course.... But that's what I got! Am I supposed to question every figure in every problem that's not given in scientific notation?!!

a more "accurate" answer that is correct to two decimal places AND three significant figures would be -1.33\times 10^3
-1.33 * 10^3 = 1330....

It demonstrates potential INACCURACY in your textbook
In what way?