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[i2amroy]

I am definitely going to agree with Reelya in that a large part of math is simply the memorization of a variety of different formulas and templates, and then being able to use pattern recognition to recognize which particular formula or template (or set of formula or templates) is applicable in your current situation.

That's one way to do it, but not a particularly good one. More important than pure memorization is understanding where the formulas come from, how they are derived - that way you can adapt them if you hit a more complicated situation than you're used to.

While knowing derivations is definitely nice, I'm going to politely disagree that they are necessarily more important than the formula themselves. For example when I think of finding the volume of a sphere, I can't think of a single time (other than when the derivation was actually being taught) that I thought "why don't I construct an infinite number of increasing and decreasingly sized circles and add all of their areas together with an integral to find the volume"? Instead I simply think "Volume = 4/3 * pi * r^3". Similarly I don't care much about how the derivative/integral rules for trig functions are derived, I simply remember (and use) that "d(sin) = cos" and so forth. This also holds true for the vast majority of formula.

Knowing where formula comes from is often a nice thing to know, and can be a big help in understanding and remembering how a formula works. But ultimately unless you are actually working as a math person attempting to derive new formula (in which case it's very important for you to know how they are derived), in the vast majority of cases the derivations simply aren't important compared to the actual formula themselves.

And that's not even getting into all of the formula that exist out there that are used in real world applications like engineering or physics, where they often have no real derivation beyond "well we stuck stuff in a wind tunnel and that's the way that the world works!". Many of those formula just have to be memorized, because the way they were derived was just by gathering tremendous amounts of experimental data and plugging things in until someone found something that works.

[i2amroy]

When dealing with the world of (simple) arithmetic and calculus, sure, memorizing formulas is useful. It doesn't really extend too well outside of those two fields of math, though. Also, you will hate yourself if you try to memorize all of the general-solution cases of first-order linear differential equations. Those are the easiest kind!

Copious memorizing of formulas won't help you understand Lie groups any better, or the idea behind Clifford algebra. God forbid you start getting into p-adic numbers or fractal analysis.

Note that in my original post I was referring to templates as well as just straight formula. Heaven knows in many cases it's much better to memorize the basic template process used to generate a vast number of formulas rather than attempting to memorize the formula themselves. For example for first-order linear differential equations you shouldn't be memorizing the general-solution cases, you should be memorizing the template that is used to generate them (such as the basic steps of integrating factors).

My main idea is that the whole point of deriving formula is that so you can use them without needing to derive them all over again every single time you want to solve that type of problem. The derivation is important, and knowing how and why it works is important, but the thing that you are actually going to be using is going to be the formula itself, not the steps required to derive it. If you are looking at something and seeing that you still need to derive your "formula" from a more general case every single time you are using it, then that means that really what you should be thinking of and memorizing as the "formula" should be that general case, not whatever you are deriving each time.

[Arx]

Do they not give you a formula sheet in America? Hereabouts, knowing how to recognise patterns and manipulate formulae (including derivations) is far more important than memorising them, because in the exam you'll be given a formula sheet.

[3man75]

So I have a word problem that me and my tutor can't figure out.

An experienced bank auditor can check a bank's deposits twice as fast as a new auditor. Working together it takes the auditors 6 hours to do the job. How long would it take the experienced auditor working alone?

I've come out with the following to set up an equation to solve this and it just failed miserably.

1/X (This is the normal auditor in terms of time it takes).

1/2x (Experienced auditor and how he can do the job faster than the normal guy)

and I equate it with 6 which is the time it takes to get the job done when they work together. I've tried multiplying 2 to get it out of the way which results in: 1/x + 1/X = 12.

I add the fractions to get 2/x=12 but I got stumped. An I think i'm wrong tbh.

I also tried some other things but the math just didn't make sense.

The answers given to me are 6, 18, 12, and 9.

[frostshotgg]

Your set up seems a little weird for whatever reason, but I can't place why so I'll gloss over that for now.

The big issue I see is when you do 2 * (1/x + 1/2x). It should equal 2/x + 1/x, when you multiply one side of an equation and that side is the sum of terms, each individual term in it gets multiplied. You do 2 * 1/x + 2 * 1/2x. This gives you 3/x = 12 which gives you x = 1/4, so you know the rate at which the experienced auditor does work.

You then plug this rate back in to solve for how much work there was to do. (1/4)t + (1/4)t/2 = W, where t is the time worked and W is the amount of work to do. t is known to be 6, so you plug that in and get 6/4 + 6/8 = W = 9/4. So there's 9/4s mystery units of work to do.

Now that you know how much work to do and how long it takes to do, you just plug those in. W/x = t. (9/4)/(1/4) = t = 9. So it would take the experienced auditor 9 hours of work.
That's a bizarre way to approach the problem, though. What I would do is start by equating the two's rates. (2N = E) => (N = E/2), where N is the new auditor's work and E is the experienced auditor. Then, I would use this to make a system of equations with W = 6E +6N. Plug the first equation into the second one and get W = 6E + 3E. Simplify to W=9E, and then we have that it takes the experienced auditor alone 9 hours.

[Reelya]

For another approach using more english, you can work out how much of the total work the bank auditor did, then work out how long the remaining work would take him. Call the basic unit of work x, then guy 1 did 2x work, and guy 2 did 1x work. 2x work took the auditor 6 hours, but he's gotta do 3x work to finish, so it would be 3/2 times the hours, so 9 hours.

You can also cull the multiple choice by logic:

6 - impossible, since it's the same length of time. someone stopped working, so it must take longer
9 - possible
12 - impossible, since the slower person stopped, it will not double the time
18 - ridiculous, for the same reason, since the slower person quit

leaving only "9" as the answer

[Spehss _]

Never posted here before, don't see why not now.

Calculus homework. Chapters on derivatives. I occasionally get answers of 0/0 or something divided by 0. Pretty sure I'm doing something wrong. Current problem wants me to find the slope of the tangent line of the curve of the equation y=3+4x^(2)-2x^(3) where x=a. Then use the slope to find the equations for the tangent lines at points (1,5) and (2,3).

I get zero for the slope. That doesn't seem right at all. The tangent lines are just horizontal lines cutting through the curve of the first equation.

[MagmaMcFry]

The slope of the tangent at a is just the derivative at a.

[Arx]

Well, f'(x) = -6x²+8x, so logically you should just be able to solve m = f'(x) and sub in the values.

Do they expect you to derive from first principles?

Ninja'd by Fry with the same basic idea.

[Bauglir]

What do you find for the derivative of the function? Also, what is this "a" of which the problem speaks?

I'm seeing the derivative as being y' = 8x-6x².

Calculus is pretty much all about 0/0 stuff so don't be surprised if it's a result that crops up a lot. Anything divided by 0 is undefined, though, and represents a discontinuity in a function; these will often appear as well, so again don't be surprised if they do. A popular example is y=x²/x, which is y=x except at 0, where it's undefined (so when you make simplifications like that, you have to note it and make sure you never later allow x=0).

EDIT: I did, indeed, mess up some arithmetic.

[Spehss _]

Checked my math book and straightened out some inconsistencies in my notes.

I used the equation for calculating tangent slope in that problem. m=(f(x)-f(a)/(x-a))
That's what gave me 0.

I haven't tried the equation for derivatives. Based on everyone's response it sounds like I used the wrong equation for this. My teacher said that we could use either equation and should get the same result. Coincidentally in the first problem I tried, I got a different answer for each equation so I probably messed that up too. Guess I'll try the derivative equation and come back later.

[MagmaMcFry]

I used the equation for calculating tangent slope in that problem. m=(f(x)-f(a)/(x-a))
That's what gave me 0.

Yeah, that would be a secant slope for the secant through x and a. But when you need a tangent, you can't just take the secant through a and a, that wouldn't make sense. What you need to do is take the limit of the secant slope (f(x)-f(a)/(x-a)) as x approaches a.

[Helgoland]

I haven't tried the equation for derivatives. Based on everyone's response it sounds like I used the wrong equation for this. My teacher said that we could use either equation and should get the same result. Coincidentally in the first problem I tried, I got a different answer for each equation so I probably messed that up too. Guess I'll try the derivative equation and come back later.

That's why you should first learn the concept and visualization and only then memorize the formulae...

[Spehss _]

That's why you should first learn the concept and visualization and only then memorize the formulae...

Your snark is appreciated. Yes, I figured that out now, thanks. _{^{I didn't memorize the formula so much as I checked my notes for the formula written down...}}
I figured out wha tmy errors were in the problems I was asking about earlier, so I know what I'm doing now.

New problem, want to make sure I did this right. Equation of the tangent line of curve y=f(x) at the point where a=2 is y=4x-5. Find f(2) and f'(2). According to the example problem I have in my notes and textbook for this type of problem I can plug 2 into the tangent line equation so f(2)=4(2)-5=3. Then use the slope equation (y2-y1)/(x2-x1) to calculate f'(2). One point is (2,3) and the y intercept point is (0,-5). Plug the two points in the slope equation to get 8/2 which is 4 so that f'(2)=4. Sounds right?

[MagmaMcFry]

That's why you should first learn the concept and visualization and only then memorize the formulae...

Your snark is appreciated. Yes, I figured that out now, thanks. _{^{I didn't memorize the formula so much as I checked my notes for the formula written down...}}
I figured out wha tmy errors were in the problems I was asking about earlier, so I know what I'm doing now.

New problem, want to make sure I did this right. Equation of the tangent line of curve y=f(x) at the point where a=2 is y=4x-5. Find f(2) and f'(2). According to the example problem I have in my notes and textbook for this type of problem I can plug 2 into the tangent line equation so f(2)=4(2)-5=3. Then use the slope equation (y2-y1)/(x2-x1) to calculate f'(2). One point is (2,3) and the y intercept point is (0,-5). Plug the two points in the slope equation to get 8/2 which is 4 so that f'(2)=4. Sounds right?

Sounds right, yes. Although the slope of mx+c is always m, so you could have figured out f'(2) way easier.