Calculus

[ComradeRed]

Note that all my equations are in LaTeX syntax, so you can just plug it into a LaTeX compiler or IDE to see it better ;)

Basics and Refresher
OK, so the basics of calculus is quite simple. Remember polynomials? Those fun things? We are going to be using a hell of lot of those things!

An example problem is: (3x+2y)(5x-7y)=???

First off, we can put this in the form of several distributive equations: (3x+2y)(5x-7y)=3x(5x-7y)+2y(5x-7y).

Now carry the multiplication through, we get 15x^{2} - 21 xy +10xy -14y^{2}. Then we add up all the components to get 15x^{2} - 11 xy -14y^{2}.

Remember:
1. Put it in distributive form
2. Multiply by every term
3. Add it up!

Derivatives: What You Need to Know
Well, permit me to introduce something called infinitesimal (if that is spelled correctly :P). If we have a variable, x, then the slightest change is denoted as dx. Now, if we have any large number n, then n*dx will always be less than 1.

Think of it as the smallest number greater than zero.

This is somewhat important to understand, but extremely important in derivatives.

The other important component in derivatives are the limits. The only thing that you will really need to understand is that it takes the form:
\lim_{z\rightarrow n}
for some variable z and some constant n. Note that n can include infinity.

What it tells us is that for every z we plug in its place n.

So consider:
\lim_{z\rightarrow 3}z^{z} = 3^{3} = 27.

"What does this all have to do with derivatives???"

Well, time to put the two together. Suppose we had a function f(x)=x^{3}. It's derivative is:

\lim_{dx\rightarrow 0} \frac{f(x+dx)-f(x)}{dx}

and through the polynomial exercise, we get

\lim_{dx\rightarrow 0} \frac{x^{3}+3x^{2}dx+3xdx^{2}+dx^{3}-x^{3}}{dx}

but this reduces to

\lim_{dx\rightarrow 0} \frac{3x^{2}dx+3xdx^{2}+dx^{3}}{dx}

then we divide out by dx reducing it to the form

\lim_{dx\rightarrow 0} 3x^{2}+3xdx+dx^{2}.

Now we substitute 0 in for dx, giving us

3x^{2}+3x(0)+(0)^{2} = 3x^{2}.

This is the derivative of the original function f(x)=x^{3}.

But Isn't there A Faster/Lazier Way???

Short answer: yes!!!

Long answer: what we do is for a function f(x)=x^{n}, we simply take it and make it

f'(x)=nx^{n-1}.

There are some exceptions for trig functions, etc. But that's the basic outline.

"But what if I am taking multiple derivatives???"

Easy! If we want the mth derivative, we use the formula:

\frac{n!}{(n-m)!} x^{n-m}

So the 2nd derivative of f(x)=x^{3} is 6x. Simple!

Summary Thus far:
1. Arrange polynomials as the addition of a number of distributive problems.
2. A derivative is simply a polynomial problem with "add ons".
3. If all else fails, use the algebra (lazy) way to do it!

I don't know if this is how far you have gotten but this is another way to think of it.

[Tormented by Treachery]

ComradeRed, I seem to have a lot of problems with Related Rates... for instance (gets his calc book out...)

"A highway patrol airplane flies 3 mi above a level, straight road at a constant rate of 120 mph. The pilot sees an oncoming car and with radar determines that at the instant the line-of-sight distance from plane to car is 5 mi the line of sight distance is decreasing at the rate of 160 mph. Find the car's speed along the highway."


-----

Also, implicit differentiation:

"Find dy/dx: x + sin y = xy"

;) Any help is very welcome, comrade!

[ComradeRed]

What a stupid problem, I'm not teaching that crap :P

OK, so we have a pig in a plane 3 miles above the ground. I assume there will be no strafing? Thus his level will remain at 3 miles above the ground.

The pig flies at a velocity of 120 miles per hour.

The pig will see the car detected on radar in 5 miles.

The line of sight is decreasing at 160 miles per hour.

Now I haven't drunken coffee in over 12 hours, so I may be rusty.

If the velocity of the pig is denoted as dp/dt=120mi/h, and the velocity of the car is dc/dt, and IF the two are coming towards each other (since, otherwise, the car would somehow break the laws of physics, which although would be interesting it would cause more harm than good), thus dp/dt + dc/dt = 160mi/h.

We can infer 120mi/h + dc/dt = 160 mi/h, thus by subtraction we have dc/dt = 40 mi/h. :)

HOWEVER this is assuming that the pig is flying on the freeway, i.e. with an altitude of 0 feet! :o

Sadly, I am exhausted from typing this out, so I will finish the rest tomorrow -_-

[Tormented by Treachery]

Sorry, it is kind of hard to read, but here is the direct scan of my book... the right reads:

above at
the pilot
at the
5 mi
160 mph.


The answer listed is 80 mph :blink:

[ComradeRed]

OK, it's a changing triangle with respect to time, that is to say that the hypoteneuse (spelling?) is 5 miles, and we know that one of the legs is 3 miles, de facto by the Pythagorean theorem, the remaining leg is 4 miles long (the 3-4-5 triangle).

The 3 mile leg is the only leg that is constant (this is given).

Now the 4 mile leg changes at a rate of dc/dt. We do not know what dc/dt is yet, that is what we are looking for!

We know that the aeroplane apparatus is travelling at 120 miles per hour.

We also know that the hypoteneuse is changing at 160 miles per hour. What we can deduce from this is that the car is moving.

So we know now that dc/dt>0 (though I know this helps none! :P).

The x(t)=sqrt(s(t)^2 +9). Now dx/dt = (dx/ds)(ds/dt).

On the other hand we do know that ds/dt = 160 mi/h. Thus we can rearrange this to take the form:

ds/dt = (ds/dx)(dx/dt)

So what you have to do is multiply (ds/dt)(dx/ds) and you have your answer.

I did most of the work for you though ;) I just need more coffee...

Side note on Related Rates

Use the chain rule, so if we have f(x)=y, and we want to find out when they'd meet or something, then:

dy/dt = (dy/dx)(dx/dt)

C'est ca!

[Tormented by Treachery]

Originally posted by [email protected] 5 2006, 11:00 PM
The x(t)=sqrt(s(t)^2 +9). Now dx/dt = (dx/ds)(ds/dt).
You lost me somewhere around there -- and also, where are you getting your variables? What are you labelling s and c?

[ComradeRed]

OK, I'm sorry, I kept referring back to the diagram after using variables I had written done in scratchwork (oops :rolleyes: )

s refers to the pig, and c refers to the car.

Remember this thing is a right (enough) triangle! You can use the pythagorean theorem (rather a variant of it appropriate for this exercise).

Have you learned about the chain rule? It's a nifty rule that says if we have two derivatives (dx/dy) and (dy/dz) then multiply them together we get

(dx/dy)(dy/dz) = Dx/dz.

What we need to do is figure out in this exercise presented what x and s are. The rest is a cakewalk.

Now completely start from scratch.

The pig is represented by a point p(t). The car is represented by a point x(t). The distance between them is represented by a distance s(t).

The point p(t) is 3 miles above the ground flying at 120 mi/h.

The point x(t) is on the ground cruising at some speed.

The distance s(t) is changing with a velocity of 160 mi/h.

We need to know how to derive x(t) from the knowledge of p(t) and s(t). THat is the problem.

We know s(t) is the distance from p(t) to x(t), with the pythagorean theorem this is of the form

s(t) = sqrt{p(t)^2 - x(t)^2}.

This is from the pythagorean theorem.

We could alternatively write s(x(t),p(t)) if we really wanted to. This however gives us the liberty to write ds(t)/dp(t) and ds(t)/dx(t).

We are solving for dx(t)/dt, right?

So, what we need to do is introduce our friend: the chain rule!

What we do is set up the equation as

dx(t)/dt = (dx(t)/ds(t))(ds(t)/dt)

What I would do is set up another equation to solve for ds(t)/dt:

ds(t)/dt = (ds(t)/dp(t))(dp(t)/dt)

Which is really simple to solve. Do you see where I am going with this?

[Tormented by Treachery]

Sorry to suck so badly at mathematics, but perhaps if you solve it (inc all work) I could follow and attempt a similar problem on my own. My main problem is coming up with equations and assigning variables, and from that figuring out relations between them.

[ComradeRed]

OK, I picked up a copy of Calculus: Fourth Edition by James Stewart, so I'll see how he explains the chain rule and related rates ;)

Gimme a few minutes...

[edit:] After tinkering around, it turns out that the time derivative of the airplane seeing the car (the 160 mi/h rate) is merely the average of twice the velocity of the plane and twice the difference of the plane-car-visual rate (160 mi/h) and the velocity of the airplane. This may help you more, since I am looking at this from the relativistic perspective (which hurts more than it helps I am afraid :().

Ah wait I see it! Remember Pythagoras' theorem? a^2 + b^2 = c^2 ? Take the time derivative of all the components, and use it in this scenario.

We get:

2a(da/dt) + 2b(db/dt) = 2c(dc/dt).

This is actually very simple when you begin looking at it mathematically. We will call the phrase c(dc/dt) the 160 mi/h rate of the plane seeing the car, and b(db/dt) is the velocity of the airplane (120 mi/h).

Thus we need to find the velocity of the car (a(da/dt))! We just rearrange this to become:

2a(da/dt) = 2c(dc/dt) - 2b(db/dt)

and when we use substitution we get

2a(da/dt) = 2(160) - 2(120) = 320 - 240 = 80.

"But wait, ComradeRed, isn't that half of what we want!"

Ah, quite true, I failed to carry out the division property of equality which gives us (according to this equation) a(da/dt) = 40.

But wo! I also forgot to factor in the distances properly! We would actually have:

2a(da/dt)= 2c(dc/dt) - 2b(db/dt) = 2(5)(160) - 2(3)(120) = 1600 - 720 = 880

"But where do the coeffecients come from?"

Following the example, the distance between the car and the aeroplane apparatus is 5 miles. That's the hypoteneuse. The short leg is 3 miles long. Therefore the other leg is 4 miles long. We use these as the coeffecients for distance.

Following some more division we get:
a(da/dt)= 440

If we subtract out our original would-be-of-an-answer (40) then divide by the hypoteneuse, we get:

440-40 = 400/5 = 80.

Quite an ingenius alternative way to get the answer, although I am not sure if this is the exact method which your teacher would enjoy.

[bezdomni]

Figured I'd bring this up. I'm unfortunately in Algebra II, but I've been reading up on calculus. I decided that I would derive natural logs using calculus, since we were doing them in algebra anyway (just not with calc).

I think this is right...enjoy!

lim n->0 (n+1/n)^n = 2.718282...

e = 2.718282

therefore;

lim n->0 (n+1/n)^n = e

then;

A = P{lim n->0 (n+1/n)^n}^ rt/100

since; lim n->0 (n+1/n)^n = e

then;

A = Pe^rt/100

and letting rt/100 = x

then

A = Pe^x

so;

A/P = e^x

now we have an exponential function, which can easily turned into a logarithmic function...

loge(A/P) = x

Which is the natural log.

Let me know if I made a mistake. I don't think I did, but you never know.

[ComradeRed]

lim n->0 (n+1/n)^n = 2.718282...
It's actually lim n->infty not n->0, otherwise it would be 1 if it were n->0 :P

x^0 = 1 regardless of what x is.



loge(A/P) = x It can be written as ln(A/P) = x, where ln is the natural log rather than the (canonical) base 10 log.

It seems fairly straightforward, fairly simple substitution. It is fantastic mathematics, it has the "So what..?" factor and everything!

Now if you can only make it obscure with mathematical jargon like "let x be a well defined functional of the set of all real numbers" and whatnot, and you could very well be a mathematician!

Just don't get caught in the Platonism of math, it's a dangerous trap (look no further than Roger Penrose for mathematical Platonism).

[bezdomni]

My goal is to hopefully get extra credit from my algebra teacher by showing that I can use calculus. She thinks I'm a moron, which is annoying.


It's actually lim n->infty not n->0, otherwise it would be 1 if it were n->0
haha, thanks. I can't believe I didn't catch that.


It can be written as ln(A/P) = x, where ln is the natural log rather than the (canonical) base 10 log.

That was my goal. ^^

[ComradeRed]

My goal is to hopefully get extra credit from my algebra teacher by showing that I can use calculus. She thinks I'm a moron, which is annoying. Use tensors, and at the end of some elaborate equation, say "Owned, *****!" :lol: Be sure to use "teamspeak" while doing it :lol:

[bezdomni]

:lol:

I think maybe that would get credit taken away from me. :P