Help a kid with Calculus!

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tckthomas
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Help a kid with Calculus!

Postby tckthomas » Thu Oct 21, 2010 12:17 pm UTC

EDIT: Subject originally "Calculus (more specifically Integral Calculus)" which is not true at all.

I understand Derivative Calculus, which is taking the slope of a graph, then Integral Calculus should be taking the graph from the slopes, but then weird stuff like finding the area under the curve and Fundamental Theorem of Calculus pops out and I don't even know what I'm talking about.

I'm skipping a lot because I am bored, I haven't learnt trigonometry in school yet but I taught myself and I understand the 3 basic trig functions on the triangle, but once I step into calculus and then trig in calculus suddenly becomes so scary. So I have a few questions.

1 ) Why would the derivative of sin(3x^2+1) become 9x*cos(3x2+1)?
2 ) Why is [imath]\int_0^1 dx f(x) = F(1) - F(0)[/imath] and what the heck is the capital F?
3 ) They say that you can get more accurate approximations of the area under the curve but can't you just figure out another function with the width of the box as one parameter and find the limit as that approaches 0? How hard is finding such a function?
Last edited by tckthomas on Sat Oct 23, 2010 8:52 am UTC, edited 1 time in total.

Syrin
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Re: Calculus (more specifically Integral Calculus)

Postby Syrin » Thu Oct 21, 2010 12:35 pm UTC

tckthomas wrote:I'm skipping a lot because I am bored

I seem to have identified your problem.

1 ) Why would the derivative of sin(3x^2+1) become 9x*cos(3x2+1)?

Chain rule.
2 ) Why is [imath]\int_0^1 dx f(x) = F(1) - F(0)[/imath] and what the heck is the capital F?

Convention is that you place dx after the f(x), not before. F is the antiderivative of f, the function such that F'(x) = f(x).
3 ) They say that you can get more accurate approximations of the area under the curve but can't you just figure out another function with the width of the box as one parameter and find the limit as that approaches 0? How hard is finding such a function?

I'm not really sure what you're saying here.

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Re: Calculus (more specifically Integral Calculus)

Postby BlackSails » Thu Oct 21, 2010 12:44 pm UTC

Syrin wrote:
2 ) Why is [imath]\int_0^1 dx f(x) = F(1) - F(0)[/imath] and what the heck is the capital F?

Convention is that you place dx after the f(x), not before. F is the antiderivative of f.


Its actually done both ways

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Re: Calculus (more specifically Integral Calculus)

Postby tckthomas » Thu Oct 21, 2010 12:47 pm UTC

Syrin wrote:
3 ) They say that you can get more accurate approximations of the area under the curve but can't you just figure out another function with the width of the box as one parameter and find the limit as that approaches 0? How hard is finding such a function?

I'm not really sure what you're saying here.


Wikipedia: Integral wrote:Notice that we are taking a sum of finitely many function values of f, multiplied with the differences of two subsequent approximation points. We can easily see that the approximation is still too large. Using more steps produces a closer approximation, but will never be exact: replacing the 5 subintervals by twelve as depicted, we will get an approximate value for the area of 0.6203, which is too small.


Then I said can't I make a function that sums up the rectangles with the width as a parameter and then let that parameter approach 0.

Also,
4 ) Am I doing fine?

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Re: Calculus (more specifically Integral Calculus)

Postby Syrin » Thu Oct 21, 2010 12:56 pm UTC

BlackSails wrote:
Syrin wrote:
2 ) Why is [imath]\int_0^1 dx f(x) = F(1) - F(0)[/imath] and what the heck is the capital F?

Convention is that you place dx after the f(x), not before. F is the antiderivative of f.


Its actually done both ways

Oh yeah?

Well, learn something new every day I guess!

tckthomas wrote:Then I said can't I make a function that sums up the rectangles with the width as a parameter and then let that parameter approach 0.

You certainly could. I'm not sure why you'd want to. If you're asking if it magically solves something - not really, you still end up with the same limit.

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Re: Calculus (more specifically Integral Calculus)

Postby tckthomas » Thu Oct 21, 2010 1:00 pm UTC

Thanks, the "F" really cleared things up, but why would F(1)-F(0) = the area?

so am I doing fine?

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Re: Calculus (more specifically Integral Calculus)

Postby Syrin » Thu Oct 21, 2010 1:03 pm UTC

tckthomas wrote:Thanks, the "F" really cleared things up, but why would F(1)-F(0) = the area?

Because of the fundamental theorem of calculus, of course!

so am I doing fine?

Well, that should be a judgement call that you should make, not us - what level of education are you in? How are you learning these things? Why are you learning these things? If you're in first year university and you don't know trig yet, and you're taking math courses because you want to be a mathematician - I would say no, you're not doing fine. Other than that, are you asking if you're learning the subject well? Well, again, that's something you should know (attend class, sir), not us. We don't know what the subject matter is or how you're being taught. Are you aware of the fundamental theorem of calculus? Have you proved it? Et cetera.

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Re: Calculus (more specifically Integral Calculus)

Postby tckthomas » Thu Oct 21, 2010 1:05 pm UTC

Oh, I just began Secondary 2 education, that's 13 years old. Guess I'll go searching the nets for the proof of that theorem... So am I doing fine 13 years, 20 days, 18 hours of age?

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Re: Calculus (more specifically Integral Calculus)

Postby mike-l » Thu Oct 21, 2010 1:09 pm UTC

tckthomas wrote:
Wikipedia: Integral wrote:Notice that we are taking a sum of finitely many function values of f, multiplied with the differences of two subsequent approximation points. We can easily see that the approximation is still too large. Using more steps produces a closer approximation, but will never be exact: replacing the 5 subintervals by twelve as depicted, we will get an approximate value for the area of 0.6203, which is too small.


Then I said can't I make a function that sums up the rectangles with the width as a parameter and then let that parameter approach 0.


Of course you can. But for all but the most basic functions it's going to be unwieldly. Which is why the Fundamental Theorem of Calculus is useful.

The fundamental theorem also tells us that 'graphs from slopes' is the same as 'area under the slope curve'.

tckthomas wrote:so am I doing fine?


Don't get too far ahead of yourself. It's good to look at things you don't know yet, but you've skipped some rather important stuff, and there's a reason those things are taught first. You'll find it hard to get a good idea of how things all fit together, why we do the things we do, and why they're interesting. So yeah, keep looking at new stuff, but don't ignore the stuff you should learn first.
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Re: Calculus (more specifically Integral Calculus)

Postby Syrin » Thu Oct 21, 2010 1:11 pm UTC

Sure! Education differs quite radically depending on location, so I still can't quite comment, but you seem well enough off if you're interested in math. Some advice, though - don't try to learn through Wikipedia. It's a great repository for mathematical knowledge when you already vaguely know the subject matter, not so much when you want to learn Calculus. Head over to the library and find some decent books! It seems that you're trying to learn a lot of definitions without some of the necessary foundations for those definitions. Don't skip ahead, despite how trivial some things might appear - limits, for example, are really quite important.

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Re: Calculus (more specifically Integral Calculus)

Postby tckthomas » Thu Oct 21, 2010 1:16 pm UTC

mike-l wrote:
tckthomas wrote:so am I doing fine?


Don't get too far ahead of yourself. It's good to look at things you don't know yet, but you've skipped some rather important stuff, and there's a reason those things are taught first. You'll find it hard to get a good idea of how things all fit together, why we do the things we do, and why they're interesting. So yeah, keep looking at new stuff, but don't ignore the stuff you should learn first.


I skipped stuff? I thought Trig was right before Calculus and I thought I learned the 3 functions, the sine rule, the cosine rule, what have I missed that I can devour? :D

Syrin wrote:Sure! Education differs quite radically depending on location, so I still can't quite comment, but you seem well enough off if you're interested in math. Some advice, though - don't try to learn through Wikipedia. It's a great repository for mathematical knowledge when you already vaguely know the subject matter, not so much when you want to learn Calculus. Head over to the library and find some decent books! It seems that you're trying to learn a lot of definitions without some of the necessary foundations for those definitions. Don't skip ahead, despite how trivial some things might appear - limits, for example, are really quite important.


Oh, learned limits when I wanted to learn calculus. it was in a web "tutorial" on calculus. I'm definitely not learning through Wikipedia though, it uses funny words that are not worth reading before I know what I'm eating. I can't go over to the library because all the books are in chinese and I am really bad at my chinese. Please tell me what I am missing so I can go over and learn them! :D

EDIT: I am going through the Singapore Syllabus

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Re: Calculus (more specifically Integral Calculus)

Postby mike-l » Thu Oct 21, 2010 1:43 pm UTC

tckthomas wrote:I skipped stuff?


I wasn't referring to Trig, I was referring to earlier stuff in calculus... eg the following 2 things:

Syrin wrote:
tckthomas wrote:I'm skipping a lot because I am bored

I seem to have identified your problem.

1 ) Why would the derivative of sin(3x^2+1) become 9x*cos(3x2+1)?

Chain rule.
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Re: Calculus (more specifically Integral Calculus)

Postby tckthomas » Thu Oct 21, 2010 1:45 pm UTC

oh, the chain rule thing i mean the sin becoming cos, i get the chain rule. what other calculus goodies are there? I will try to decipher quotient and product rule soon

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Re: Calculus (more specifically Integral Calculus)

Postby mike-l » Thu Oct 21, 2010 1:56 pm UTC

tckthomas wrote:oh, the chain rule thing i mean the sin becoming cos, i get the chain rule. what other calculus goodies are there? I will try to decipher quotient and product rule soon


Do you know:
How to maximize/minimize functions?
Solve related rates problems?
Implicit differentiation?
The Intermediate Value Theorem?
The Mean Value Theorem?

These are all things that actually show why we do calculus, and you can learn about them as soon as you understand derivatives (actually you can do the Intermediate Value Theorem as soon as you learn continuity)

As for sin becoming cos, by looking at the graphs you should be able to convince yourself that the derivative of sin behaves like cos, and that the derivative of cos behaves like -sin. To show that they are actually identical involves the addition formula for sin and the limit [math]\lim_{h\to 0} \frac{\sin h}h = 1.[/math]
You can prove the latter in a number of ways (though don't fall into the trap of using l'Hopital's rule on it, because you're using the derivative to find the derivative... bad!) and there are actually a few threads about it on this board.

Speaking of which, do you know l'Hopital's rule?
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Re: Calculus (more specifically Integral Calculus)

Postby tckthomas » Thu Oct 21, 2010 1:58 pm UTC

mike-l wrote:Do you know:
How to maximize/minimize functions?
Solve related rates problems?
Implicit differentiation?
The Intermediate Value Theorem?
The Mean Value Theorem?

These are all things that actually show why we do calculus, and you can learn about them as soon as you understand derivatives (actually you can do the Intermediate Value Theorem as soon as you learn continuity)

Speaking of which, do you know l'Hopital's rule?


nonononononononono! :D time to search!

EDIT: will get back to answer those questions and ask more questions on answering the questions so as to answer more questions, creating more questions to answer.

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Re: Calculus (more specifically Integral Calculus)

Postby tckthomas » Thu Oct 21, 2010 2:46 pm UTC

1 ) How to maximize/minimize functions?
I think, find the derivative, let that equal 0, then solve the equation.

2 ) Solve related rates problems?
Find the relationship between the rates, differentiate everything, substitute.

3 ) Implicit differentiation?
umm, make it "explicit" like [imath]x^2 + y^2 = r^2[/imath] into [imath]y = +-\sqrt{r^2-x^2}[/imath] then differentiate?

4 ) The Intermediate Value Theorem?
If the line has no breaks between 2 points on the y axis, I extend out from the y axis to meet 2 points on the line, and get their 2 x values, then when I find a y between the 2 points, and I extend to meet a point on the line, it's x value is between the 2 x values. Not so sure about this because this seems quite useless...

5 ) The Mean Value Theorem?
About the same as above, except it says that there is a tangent between 2 points of 2 values of the x axis that is parallel to the line drawn from the 2 points on the line above the 2 x values

6 ) L'Hopital's rule?
Again I don't see the significance of this... if the limit as x->anything f(x) and g(x) are the same, then limit as x->anything f(x)/g(x) = limit as x->same anything f'(x)/g'(x)

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Re: Calculus (more specifically Integral Calculus)

Postby Dark567 » Thu Oct 21, 2010 3:09 pm UTC

Syrin wrote: limits, for example, are reallyquite important absolutely essential to even the most basic understanding of Calculus.


Fix'd. Limits are probably the single most essential piece to understanding calculus, the other major parts of calculus: integrals, derivatives, infinite series etc. are all based off the concepts of infinitesimal rates of change i.e. limits.
tckthomas wrote:4 ) The Intermediate Value Theorem?
If the line has no breaks between 2 points on the y axis, I extend out from the y axis to meet 2 points on the line, and get their 2 x values, then when I find a y between the 2 points, and I extend to meet a point on the line, it's x value is between the 2 x values. Not so sure about this because this seems quite useless...

The intermediate value theorem isn't useless, its just obvious.

tckthomas wrote:6 ) L'Hopital's rule?
Again I don't see the significance of this... if the limit as x->anything f(x) and g(x) are the same, then limit as x->anything f(x)/g(x) = limit as x->same anything f'(x)/g'(x)

f'(x)/g'(x) is often simpler to solve than f(x)/g(x). Also it doesn't work for "if the limit as x->anything f(x) and g(x) are the same", the limits have both to be 0, +infinity or -infinity.
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Re: Calculus (more specifically Integral Calculus)

Postby tckthomas » Thu Oct 21, 2010 3:27 pm UTC

okay, but is there any problem where the L'Hospital rule is useful?

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Re: Calculus (more specifically Integral Calculus)

Postby gorcee » Thu Oct 21, 2010 3:33 pm UTC

tckthomas wrote:3 ) Implicit differentiation?
umm, make it "explicit" like [imath]x^2 + y^2 = r^2[/imath] into [imath]y = +-\sqrt{r^2-x^2}[/imath] then differentiate?


Not every function can be made explicit. Try [imath]x e^y = y^2 \sqrt{x}+y[/imath]

tckthomas wrote:6 ) L'Hopital's rule?
Again I don't see the significance of this... if the limit as x->anything f(x) and g(x) are the same, then limit as x->anything f(x)/g(x) = limit as x->same anything f'(x)/g'(x)


This makes computing limits substantially easier, and it is not quite as straightforward as you might think. Compute
[math]\lim_{x\rightarrow 0} \frac{\sin x}{x}[/math]
without using L'Hopital's rule (it's not hard, just somewhat annoying). Now compute it with L'Hopital's rule.

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Re: Calculus (more specifically Integral Calculus)

Postby 314man » Thu Oct 21, 2010 3:51 pm UTC

tckthomas wrote:I understand Derivative Calculus, which is taking the slope of a graph, then Integral Calculus should be taking the graph from the slopes, but then weird stuff like finding the area under the curve and Fundamental Theorem of Calculus pops out and I don't even know what I'm talking about.

I'm skipping a lot because I am bored, I haven't learnt trigonometry in school yet but I taught myself and I understand the 3 basic trig functions on the triangle, but once I step into calculus and then trig in calculus suddenly becomes so scary. So I have a few questions.

1 ) Why would the derivative of sin(3x^2+1) become 9x*cos(3x2+1)?
2 ) Why is [imath]\int_0^1 dx f(x) = F(1) - F(0)[/imath] and what the heck is the capital F?
3 ) They say that you can get more accurate approximations of the area under the curve but can't you just figure out another function with the width of the box as one parameter and find the limit as that approaches 0? How hard is finding such a function?



1) It's actually 6x in front, not 9x
2) Capital F just means the antiderivative. The reason it is F(1) - F(0) is because you are taking the integral in that area. What it does is take the area up to x = 1, and subtract it from the area up to x = 0 (it's not exactly that because by themselves it isn't an area (eg F(1) by itself isn't an area... but when put together like F(1) - F(0), it is). So in your case, you are taking the area under the curve between 0 and 1
3) Not exactly sure what you're saying. Sounds like you're just doing the antiderivative by using the limits. It's not that important to know as long as you have the general idea. I never use it to solve any problems.

Anyways I've read you skipped a lot, and I'd recommend learning these things before going on:

logarithms (and the natural logarithm)
trig identities (btw also I've found that I've never used the sine and cosine law past high school so far. Identities is it's own thing)
general function knowledge (like knowing how it'll look like just by looking at it, asymptotes, domain, how you can change a function and what it will do eg. stretching and transformations)
factorials (they'll start popping up often, it's not hard to learn. It's just learning how to deal with them and reducing them)

and the stuff you should learn about in calculus right now:
implicit differentiation
chain rule/product rule/quotient rule
squeeze theorem (it comes up now and then. It's not hard once you know the general knowledge about functions)
afterwards, I'd either do the spins on calculus that are neat but not too important (ie area by rotation), or going into series

but for a 13 year old.... wow good job!

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Re: Calculus (more specifically Integral Calculus)

Postby Nitrodon » Thu Oct 21, 2010 5:40 pm UTC

gorcee wrote:This makes computing limits substantially easier, and it is not quite as straightforward as you might think. Compute
[math]\lim_{x\rightarrow 0} \frac{\sin x}{x}[/math]
without using L'Hopital's rule (it's not hard, just somewhat annoying). Now compute it with L'Hopital's rule.


If you already know the derivative of sine, note that [imath]\frac {\sin x}{x} = \frac {\sin x - \sin 0}{x - 0}[/imath], so L'Hopital's rule isn't needed. If you don't know the derivative of sine, L'Hopital's rule won't be of much use when computing that limit.

In addition, I don't know of any way to find the derivative of sine that doesn't involve computing that limit first.

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Re: Calculus (more specifically Integral Calculus)

Postby mike-l » Thu Oct 21, 2010 6:11 pm UTC

Nitrodon wrote:In addition, I don't know of any way to find the derivative of sine that doesn't involve computing that limit first.


Depends on how you define sine. I define it as the solution to y'' = -y with y(0) = 0, y'(0)=1, and I define cos similarly. Then it's trivial. You can then show that (cos t, sin t) is uniform circular motion and that the argument is how many radians you've gone, to show that these are equal to the triangle definitions.
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Re: Calculus (more specifically Integral Calculus)

Postby gorcee » Thu Oct 21, 2010 6:27 pm UTC

Nitrodon wrote:
gorcee wrote:This makes computing limits substantially easier, and it is not quite as straightforward as you might think. Compute
[math]\lim_{x\rightarrow 0} \frac{\sin x}{x}[/math]
without using L'Hopital's rule (it's not hard, just somewhat annoying). Now compute it with L'Hopital's rule.


If you already know the derivative of sine, note that [imath]\frac {\sin x}{x} = \frac {\sin x - \sin 0}{x - 0}[/imath], so L'Hopital's rule isn't needed. If you don't know the derivative of sine, L'Hopital's rule won't be of much use when computing that limit.

In addition, I don't know of any way to find the derivative of sine that doesn't involve computing that limit first.


You can know the derivative of sin without constructing it formally through limits... for example, by looking it up in a table. The point I was trying to make is that if you know the derivative of something, then L'Hopital's rule can make taking the limit easier. Sure, there are other implications and applications of the rule, but in the context of first-year, first-semester Calculus, an example like sin(x)/x is fairly illustrative. I suppose I could have picked some rational function, but the point is the same.

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Re: Calculus (more specifically Integral Calculus)

Postby iamacow » Thu Oct 21, 2010 6:43 pm UTC

tckthomas wrote:okay, but is there any problem where the L'Hospital rule is useful?


http://en.wikipedia.org/wiki/L'H%C3%B4p ... e#Examples

Wikipedia has a some good examples of l'hopital's in action. It can make some pretty non-obvious limits far more clear.

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Re: Calculus (more specifically Integral Calculus)

Postby Dopefish » Thu Oct 21, 2010 6:55 pm UTC

I'd be a little bit hesitant in becoming too attached to l'hopitals rule early on.

In eary year stuff, many professors tend to be pretty keen to have you focus on limit taking without using methods like that, and it's something that you could easily become too attached to (also, you can look at some things backwards, such as the sinx/x limit). In applications it's great, but I wouldn't worry to much about that just yet.

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Re: Calculus (more specifically Integral Calculus)

Postby mmmcannibalism » Thu Oct 21, 2010 9:07 pm UTC

So everyone else seems to have the part about what you need to learn covered pretty well. One thing to add, your doing very well for your age; don't get discouraged if you have to learn some simpler stuff because of gaps.
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Re: Calculus (more specifically Integral Calculus)

Postby whaatt » Thu Oct 21, 2010 11:17 pm UTC

tckthomas wrote:Why would the derivative of sin(3x^2+1) become 9x*cos(3x^2+1)?


Hmm... forgive me if I am seriously spacing out, but isn't the derivative of sin(3x2+1) equal to 6x*cos(3x2 + 1) NOT 9x*cos(3x2 + 1) ?
echo "Dang. Forgot Semicolon."

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Re: Calculus (more specifically Integral Calculus)

Postby Dason » Fri Oct 22, 2010 12:06 am UTC

whaatt wrote:
tckthomas wrote:Why would the derivative of sin(3x^2+1) become 9x*cos(3x^2+1)?


Hmm... forgive me if I am seriously spacing out, but isn't the derivative of sin(3x2+1) equal to 6x*cos(3x2 + 1) NOT 9x*cos(3x2 + 1) ?

You're right but 314man already brought that up. I don't know why the OP used that as the specific example though when they mentioned that the part they didn't understand was d/dx sin(x) = cos(x).
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Re: Calculus (more specifically Integral Calculus)

Postby whaatt » Fri Oct 22, 2010 3:23 am UTC

Dason wrote:You're right but 314man already brought that up.


I should've figured that somebody on these forums would have gotten to it already. :D
echo "Dang. Forgot Semicolon."

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Re: Calculus (more specifically Integral Calculus)

Postby BlackSails » Fri Oct 22, 2010 4:06 am UTC

As for the intermediate value theorem being obvious - thats true, but obvious things still need proof. For instance, the proof that a circle, and any shape you can smoothly deform a circle into divides the plane into an "inside" and "outside" is rather difficult, and evaded Gauss himself.

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Re: Calculus (more specifically Integral Calculus)

Postby voidPtr » Fri Oct 22, 2010 4:36 am UTC

tckthomas wrote:I understand Derivative Calculus, which is taking the slope of a graph, then Integral Calculus should be taking the graph from the slopes, but then weird stuff like finding the area under the curve and Fundamental Theorem of Calculus pops out and I don't even know what I'm talking about.



Good on you for trying to understand some higher stuff..of course you'll probably have huge gaps but those will get filled in as you go along.


I don't want to discourage you or anything, but by the sounds of it your precalculus knowledge isn't quite there yet if you seriously want to tackle calculus rather than just vague notion of what it is.

I'll talk a little about the derivative and integral though ( it doesn't hurt):

Derivative:
I think you may have meant this, but a derivative is a function for the slope of a tangent line to any point on a continous curve (assuming the tangent line to the curve exists). A couple of questions for you to think about: i) What is a tangent line to a curve and why is it important to be able to find it? ii)Can you think of any real world examples of what the slope of a tangent line to a curve represents? iii) What is the precise limit definition of the tangent line? (iv) What types of functions have derivatives?

Antiderivative:
The *antiderivative* is the opposite of a derivative, and given a function f(x), that is a slope of a tangent line to any point on a continuous curve, the antiderivative F(x) is the family of curves. Questions: i) Why family of curves rather than single curve? ii) Does every function that has a derivative also have an antiderivative?

Integral:

Integral calculus starts out completely separate from derivative calculus and has actually been studied a lot longer. The definite integral of a function from point a to bl is, by definition, is the area under a curve between and b. Area under a curve is hard to precisely define, but it is the same as our normal geometric intution of area, with the exception that if the curve is in the 2nd or 4th quadrant the value is negative. The oldest way and still very relevant way of estimating this area is to do what I think you were suggesting..divide the function into rectangles and add up the rectangles. Questions: i) Is the definite integral a real number or a function? ii) How many rectangles do we need for the definite integral to be precise? iii) Using your knowledge of limits, can you find a limit definition for this?

The Fundamental Theorem of Calculus ties together the relationship between the antiderivative and the integral (area under the curve). There are two parts to the theorem. You stated the second part of of the fundamental theorem, and that's the part that is mostly used in practice, but most of the work deriving the fundamental theorem is to derive the first part. Anyways, nevermind this for now..

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Re: Calculus (more specifically Integral Calculus)

Postby tckthomas » Fri Oct 22, 2010 5:49 am UTC

Oh, I get L'Hopital rule now... Is it like splitting that fraction into 2, seeing if they both make the same 3 values, and deriving them one by one and seeing if they both exist, then finding the answer to that and knowing that is the correct answer to the fraction...

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Re: Calculus (more specifically Integral Calculus)

Postby 314man » Fri Oct 22, 2010 6:38 am UTC

whaatt wrote:
Dason wrote:You're right but 314man already brought that up.


I should've figured that somebody on these forums would have gotten to it already. :D


It was my first time being the first to bring something up. It makes me feel so helpful lol :D


tckthomas wrote:Oh, I get L'Hopital rule now... Is it like splitting that fraction into 2, seeing if they both make the same 3 values, and deriving them one by one and seeing if they both exist, then finding the answer to that and knowing that is the correct answer to the fraction...


I don't exactly what you mean the same 3 values... but if the limit is 0/0, or infinite/infinite (the signs on infinite don't matter, so infinite/-infinite is fine), then L'Hopital's rule can be applied. Also you don't derive them one by one, but both of them at the same time (the top being its own function and the bottom being its own function). Then yeah you know the rest. Take the limit of the new function and that's your answer. If it's still in undetermined form, just do L'Hopital again, and repeat until you get a real number.

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Re: Calculus (more specifically Integral Calculus)

Postby tckthomas » Fri Oct 22, 2010 7:31 am UTC

314man wrote:Anyways I've read you skipped a lot, and I'd recommend learning these things before going on:

logarithms (and the natural logarithm)
trig identities (btw also I've found that I've never used the sine and cosine law past high school so far. Identities is it's own thing)
general function knowledge (like knowing how it'll look like just by looking at it, asymptotes, domain, how you can change a function and what it will do eg. stretching and transformations)
factorials (they'll start popping up often, it's not hard to learn. It's just learning how to deal with them and reducing them)

and the stuff you should learn about in calculus right now:
implicit differentiation
chain rule/product rule/quotient rule
squeeze theorem (it comes up now and then. It's not hard once you know the general knowledge about functions)
afterwards, I'd either do the spins on calculus that are neat but not too important (ie area by rotation), or going into series

but for a 13 year old.... wow good job!


First, thanks!
I should give myself a test, then tell me what I have left to learn :D

d/dx ln(x) = 1/x
and you can convert any base log into natural log: base a log x = base e log x / base e log a

there seems to be too much trig identities here! i can only remember the reciprocal ones, sine(x) = 1/cosecant(x), cosine(x) = 1/secant(x), tangent(x) = 1/cotangent(x)

this i need more experience, this is the reason why Calculus in 20 minutes fails.

factorials? i think they are just too weird. is it impossible to derive a factorial if it isn't over another factorial?

implicit differentiation i think i get it. derive both sides with respect to x, then dy/dx magically appears, then put that on one side and done.

chain rule/product rule/quotient rule, i need to find out when to use the chain rule. I just don't know why to use the chain rule when sin(3x^2+1)... can you tell me more?

squeeze theorem, heh this is quite cool! squeezing functions into a point... its true when there is upper bound and lower bound functions and when they are equal to something, the middle is squeezed to also equal that.

spins on calculus, making it to a series, AAAHHHHH! freaky stuff!

thanks! this will help me the next 2 years when i do additional math. i would learn calculus secondary 3 or 4...

EDIT: yeah, either 0, -infinity, or infinity, 3 values...

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Re: Calculus (more specifically Integral Calculus)

Postby tckthomas » Fri Oct 22, 2010 8:03 am UTC

voidPtr wrote:
tckthomas wrote:I understand Derivative Calculus, which is taking the slope of a graph, then Integral Calculus should be taking the graph from the slopes, but then weird stuff like finding the area under the curve and Fundamental Theorem of Calculus pops out and I don't even know what I'm talking about.



Good on you for trying to understand some higher stuff..of course you'll probably have huge gaps but those will get filled in as you go along.
Thanks, that's the reason why I kept posting after my questions got answered, I wanted to fill those gaps.
voidPtr wrote:
I don't want to discourage you or anything, but by the sounds of it your precalculus knowledge isn't quite there yet if you seriously want to tackle calculus rather than just vague notion of what it is.
yeah, i never took precalculus...
voidPtr wrote:i) What is a tangent line to a curve and why is it important to be able to find it?
the "instantaneous rate of change"
voidPtr wrote:ii)Can you think of any real world examples of what the slope of a tangent line to a curve represents?
like throwing a ball onto a target and i want to know at what angle it would bounce off, i would need the tangent line at that point so i can bounce it off correctly.
voidPtr wrote:iii) What is the precise limit definition of the tangent line?
i... don't know... *search search* limit as x->0 of secants? i think that means slowly moving 2 points together, and finding out the line it makes, a secant line would be the average rate of change, and the tangent line is that with the distance approaching 0.
voidPtr wrote:iv) What types of functions have derivatives?
i don't know again... *search search* continuos functions at that point, and where the slopes "0" distance left and "0" distance right are the same, like y=|x| is not differentiable at x = 0 because immediately left is -1 slope, and immediately right is 1 slope. i think this means "inflection point"
voidPtr wrote:i) Why family of curves rather than single curve?
because you can always add a constant at the back and it would also be an antiderivative!
voidPtr wrote:ii) Does every function that has a derivative also have an antiderivative?
i don't know *search search* no, i think... i guess i still don't know
voidPtr wrote:i) Is the definite integral a real number or a function?
number, because an area is an area unless its not definite, then that area is a function...
voidPtr wrote:ii) How many rectangles do we need for the definite integral to be precise?
INFINITY! :D
voidPtr wrote:iii) Using your knowledge of limits, can you find a limit definition for this?
lim as x->infinity where x is the number of rectangles and then sum the areas up? i don't know... that's my intuition put into text as well as i can.

thanks everyone for helping!

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Re: Calculus (more specifically Integral Calculus)

Postby keeperofdakeys » Fri Oct 22, 2010 10:13 am UTC

tckthomas wrote:there seems to be too much trig identities here! i can only remember the reciprocal ones, sine(x) = 1/cosecant(x), cosine(x) = 1/secant(x), tangent(x) = 1/cotangent(x)

You'll find most people won't remember them, just try to recognise when there might be one that could apply to an equation, then look it up. Although if you use them all the time, then you'll easily remember them.

tckthomas wrote:factorials? i think they are just too weird. is it impossible to derive a factorial if it isn't over another factorial?

If by derive, you mean differentiate, then that is impossible. A factorial is not a continuous function, which is a prerequisite to be differentiated; but be warned, not every continuous function is differentiable. A good example of a non-differentiable continuous function is |x|, where |-3| = 3, |0| = 0, |1| = 1. If you graph this function, it looks like to lines that meet sharply at the point (0,0); it is defined everywhere and doesn't "jump"*, therefore it is continuous. You see from the graph that coming from the positive x, dy/dx = 1; then coming from the negative x, dy/dx = -1. There is no real clear definition for dy/dx at x=0, and going from first principles doesn't work.

tckthomas wrote:implicit differentiation i think i get it. derive both sides with respect to x, then dy/dx magically appears, then put that on one side and done.


It is actually that when differentiating, in terms of x, a term of a function that involves y , you need to account for the fact that y changes differently to x. The rate at which y changes in terms of x sounds familiar... it is just dy/dx. So if you have the earlier function, the full calculation is:
x2+y2=r2
d/dx(x2+y2) = d/dx(r2)
d/dx(x2)+d/dx(y2) = d/dx(r2)
Now r is a constant, but y is a variable; so we have to differentiate y!
2*x + 2*y*(dy/dx) = 0
2*y*(dy/dx) = -2*x
dy/dx = -x/y

tckthomas wrote:chain rule/product rule/quotient rule, i need to find out when to use the chain rule. I just don't know why to use the chain rule when sin(3x^2+1)... can you tell me more?

The reason the chain rule exists is because it is impossible to differentiate the composition of two functions ie. one function inside another, ex. f(x) = sin(x), g(x) = 2*x2+1, f(g(x)) = sin(g(x)) = sin(2*x2+1).
In order to differentiate f(g(x)), we just differentiate f(x), leaving g(x) the same inside; then we work out g'(x) and multiply it onto the end. We can even have multiple chain rules:
f(x) = cos2(x3) = ( cos(x3) )2
This is really three functions: g(x) = x2, h(x) = cos(x), i(x) = x3, f(x) = g(h(i(x)))
f'(x) = g'(h(i(x)))*h'(i(x))*i'(x)
f'(x) = 2*cos(x3)*(-sin(x3))*(3*x2)

* The mathematical definition for continuous functions is a bit complex, if you want to try look at wikipedia. Basically, as the distance between two x values decreases, then the distance between the corresponding y-values is also decreasing; also this difference will approach zero.
You really need to understand the formal definition of a limit if you want to understand this. http://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit#Precise_statement

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Re: Calculus (more specifically Integral Calculus)

Postby tckthomas » Fri Oct 22, 2010 10:21 am UTC

but how would i know that the inside is another function? if i can't expand it out?

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Re: Calculus (more specifically Integral Calculus)

Postby keeperofdakeys » Fri Oct 22, 2010 10:50 am UTC

tckthomas wrote:but how would i know that the inside is another function? if i can't expand it out?

One of the most common uses is for powers, ex. (f(x))99. If you have any function that has sin, cos, ln, cf(x) or anything with a function inside, then you probably want the chain rule. It is one of those things that you have to practice, so you can recognise them. Try to answer the following:
1. d/dx(sin2(x)) =
2. d/dx(ln(x+3)) =
3. d/dx(sin(ln(3*x2))) =

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Re: Calculus (more specifically Integral Calculus)

Postby tckthomas » Fri Oct 22, 2010 11:15 am UTC

keeperofdakeys wrote:Try to answer the following:
1. d/dx(sin2(x)) =

d/dx((sin(x))2) = (cos(x))2
not even sure about the sin-->cos and cos-->sin thing, very unsure, just need more practice problems ack!
EDIT: got this wrong, should be 2sin(x)cos(x) from the outside in.
keeperofdakeys wrote:2. d/dx(ln(x+3)) =

1/(x+3)*1 = 1/(x+3)
Note: I checked this with grapher, and I got it right according to it! :D
keeperofdakeys wrote:3. d/dx(sin(ln(3*x2))) =

cos(ln(3x2))(d/dx(ln(3x2))) = cos(ln(3x2))(1/3x2)(d/dx(3x2) = cos(ln(3x2))(1/3x2)(6x) = 6x*cos(ln(3x2))/3x2 = 2cos(ln(3x2))/x
I failed at doing this, this is from W|A. I tried to do it from the inside out, but apparently its from outside in...
I learnt something!

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Re: Calculus (more specifically Integral Calculus)

Postby keeperofdakeys » Fri Oct 22, 2010 11:54 am UTC

Now try this, we will be doing implicit differentiation, so x and y are both variables.

1. d/dx(x*y2 + tan(32/x)) =
(hint: d/dx(tan(x)) = sec2(x) = 1/cos2(x) )

2. d/dx( y*ln(2*y3*x) ) =

Edit: forgot to specify implicit differentiation :shock:
Last edited by keeperofdakeys on Fri Oct 22, 2010 12:00 pm UTC, edited 1 time in total.


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