Differentials and approximations

[mg0stisha]

Homework Statement

Use differentials to approximate the value of \sqrt{15}

Homework Equations

N/A

The Attempt at a Solution

I can't really figure out how to start this, so any help would be appreciated. I know how I'd do it by Newton-Raphson Method, but this is what the sample exam asked for.

 

[Hurkyl]

I can't really figure out how to start this

Definitions are always good. In general, what is a differential approximation?

Once you have recalled that, can you plug things in (both for the variables and for the functions) to make differential approximation say something relevant?

 

[mg0stisha]

All I could find in my notes/book that would be relevant is that dy=\frac{dy}{dx} dx. Using this, I get dy=2xdx, but I'm still not too sure that this helps. I am a using a wrong definition, or interpretting this the wrong way?

 

[Mark44]

If \Delta x is suitably "small,"
\frac{\Delta y}{\Delta x}~\approx~\frac{dy}{dx}~=~f'(x_0)
\Rightarrow~\Delta y~\approx~f'(x_0)~\Delta x
\Rightarrow~y~-~y_0~\approx~f'(x_0)~\Delta x
\Rightarrow~y~\approx~y_0~+~f'(x_0)~\Delta x

 

[Hurkyl]

Did you mean

\Delta y \approx \frac{dy}{dx} \Delta x​

? We should be careful about what we mean here! It's hard to do calculations if we're sloppy and imprecise about what we're manipulating!

You did something else you didn't say -- you decided to insist that x and y are related by the equation

y = x²​

(or something similar) Any particular reason?^*

Again, use your definitions. What do things like \Delta y mean? Can you make them relate to \sqrt{15} somehow?



*: I don't mean to imply that it's wrong -- I just want to make sure you're thinking. And if you can explain why you did that, maybe it will help you figure out what's next. It's not what I thought was most obvious, but I'm pretty sure this approach still works.

 

[mg0stisha]

Yes, I did mean that, thank you for the correction!

To be honest, I don't know exactly why i chose y=x^2 in this case. I mainly did it because it was what I used when i was asked to find the the root using Newton's Method. However, \Delta y means a small change in y, probably caused by a change in x, which could be defined as \Delta x.

 

[Hurkyl]

Yes, I did mean that, thank you for the correction!

To be honest, I don't know exactly why i chose y=x^2 in this case. I mainly did it because it was what I used when i was asked to find the the root using Newton's Method. However, \Delta y means a small change in y, probably caused by a change in x, which could be defined as \Delta x.

What is a "small change in y", or a "small change in x"? To do algebra, you need an algebraic expression, not an English expression. Introduce new variables (and declare what they are) if you need to...

(Not to put down the English expression -- it helps us organize our thoughts and know what we intend for our symbols mean. You just can't do algebra with English)

 

[Hurkyl]

Or, as an alternative, can you make "small change in x" or "small change in y" say anything about \sqrt{15}? If we can do that directly, then maybe we won't need to analyze those phrases further. (But then again, maybe we will)

(Or can you come up with some other idea)

 

[mg0stisha]

\Delta y=(x+\Delta x)^2-(x)^2
Is this putting me on the right track?

 

[Hurkyl]

Well, it's something. Let's see where we can run with it. It's not so much about being on the "right" track than being on any track at all.

(P.S. have you considered other forms? like y + \Delta y = (x + \Delta x)^2 or maybe introducing variables like y_0 and rewriting \Delta y = y - y_0? Well, never mind, let's keep going with what you have, unless one of these really strikes your fancy)

Can you make \sqrt{15} appear in that equation in any useful way? Remember, while you have already insisted that y = x^2, you still have some amount of freedom to make other specific choices.

 

[mg0stisha]

The only thing i can think of is defining y=15 and then algebraically manipulating our equation to be \sqrt{y+\Delta y}=x+\Delta x.

I think the reason that I'm finding this hard is because i don't exactly see the point in finding the answer this way, or at least in where it's going to get us without eventually plugging \sqrt{15} into a calculator.

 

[Hurkyl]

The only thing i can think of is defining y=15 and then algebraically manipulating our equation to be \sqrt{y+\Delta y}=x+\Delta x.

I think the reason that I'm finding this hard is because i don't exactly see the point in finding the answer this way, or at least in where it's going to get us without eventually plugging \sqrt{15} into a calculator.

Well, actually, you're almost there. If we do what you suggested, then we get

\sqrt{15 + \Delta y} = \sqrt{15} + \Delta x​

and

\Delta y \approx 2 \sqrt{15} \Delta x​

right? You still have some free variables there -- are there any especially nice choices for \Delta y or for \Delta x?

But, alas, it won't work out completely nicely. But don't despair! We don't always get things exactly right on the first try -- but often we still learn something that will be useful. With the equations you've worked out, is there something else you might try setting to 15?

 

[Hurkyl]

I was hoping to stay up until you worked all the steps out yourself, but I'm getting sleepy, so I'll just say what I have on my mind. I hope you work through each part rather than just reading it all at once.




The thing I hope you saw was to try \Delta y = 1, so that you'd get the relatively simple expression \sqrt{15 + 1} = \sqrt{15} + \Delta x, which is unfortunately tainted by the "complex" expression 1 \approx 2 \sqrt{15} \Delta x and doesn't simplify well.




Then, the thing I hoped you'd see next is that you might try y + \Delta y = 15. But now what else to set?




Either by chance, or by thoughtfully avoiding the problems with the previous attempt, I hope you'd try y = 16.




The main thing I was trying to demonstrate is that you often solve problems simply by blazing forward! Most of the steps were mostly your ideas -- you just had to be prompted to go forward. If you learn how to prompt yourself, then you'll be able to do a lot. When you reach an obstacle, you then study it to see if the obstacle can be avoided. Or just try slightly tweaking things until you get around the obstacle. I like doing that a lot -- it often works quickly and without having to put a lot of thought into it.

(sometimes, our line of attack really doesn't work and we have to go back to square 1. But that didn't happen here)



Anyways, now that you have the final answer, can you find a shorter and clearer route to get there?

 

[mg0stisha]

Well, if we use \sqrt{15 + \Delta y} = \sqrt{15} + \Delta x, and say that \Delta y = 1, then it could become 4 = \sqrt{15} + \Delta x, or, \sqrt{15} = 4 - \Delta x. However, this still leaves me blank.

 

[HallsofIvy]

Well, I notice that 15 is not too far from 16 and 16 is a perfect square. Now what, would be convenient values for x and \Delta x in \sqrt{(x+ \Delta x)}\approx \sqrt{x}+ (d\sqrt{x})\Delta x?

 

[mg0stisha]

Well Hurkyl, I did exactly what you thought I would by using \Delta y = 1!

But your hint helped a lot. I set y=16, and got
\sqrt{16 + \Delta y} = \sqrt{16} + \Delta x

\Delta y = 2\sqrt{16} \Delta x

\Delta x = \frac{\Delta y}{8}

then... I set \Delta y = -1

\sqrt{15} = \sqrt{16} + \frac{-1}{8} \approx 3.875

Which I'd say is a good approximation to \sqrt{15}!

Thanks a lot for staying with me and being persistent with me!
Also, thanks to Mark44 and HallsofIvy for helping along the way.

And it wasn't even on the exam . Good to know anyways though!

 

[annoymage]

sorry, but, how to get this equation? y + \Delta y = (x + \Delta x)^2