Derivatives

Anakin · 2009-12-10 19:53:46

Alright, I just have a few general question about this concept. I'm only into Pre-Calculus at the moment so we haven't come across this.

I know that derivatives can be used to find the average rate of change, or was it the instantaneous rate of change.. (I somehow used some function on the calculator and was able to find one of them.)

But I was just wondering, can someone tell me how they work? I asked my teacher if I could borrow a Calculus book for the year but he said I'd have to wait til next week.

So until then, any basic knowledge would be helpful, and it will give me something to think about.

Thank you!

bobbym · 2009-12-10 21:03:21

Hi anakin;

The derivative is the instantaneous rate of change at a point. Best described in what they can do. They can find the slope of a point,they can find the tangent to a curve at a point. They find roots of equations, they can approximate functions with polynomials. They solve maxima and minima problems. They form differential equations that describe physical processes. They find limits, etc. It's safe to say they are the backbone of calculus.

http://www.mathsisfun.com/calculus/deri … ction.html

http://www.mathsisfun.com/calculus/deri … dy-dx.html

Here is some vids:

http://www.5min.com/Video/What-is-a-Der … 1-51063290

The definition of a derivative as a limit:

http://www.youtube.com/watch?v=cRfNOg9Q22U

http://www.youtube.com/watch?v=rAof9Ld5sOg

Anakin · 2009-12-12 04:32:27

Thanks for all those links Bobby and the definition. So from what I've understood, this is what a derivative is: The slope of a tangent at any given point on a curve (or maybe even line).

http://www.youtube.com/watch?v=ay8838UZ4nM I even ended up watching that link for the extra examples and I'm sort of getting the basic understanding of the things.

Mind asking me 2 or 3 questions (any basic sort) to see if I get the hang of things?

bobbym · 2009-12-12 06:22:42

Hi Anakin;

Great! Glad you got a start! So cool to be learning math now. The internet is the greatest thing this civilization has ever done.

I prefer understanding things by examples and problems and going easy on the jargon.
Using what you know about derivatives what can you say about the lines y = 3x -5 and y=5 - x / 15?

Anakin · 2009-12-12 06:56:52

The derivative of any point on the line y=3x-5 is 3. The derivative of any point on the line y=5-x/15 is equal to -1/15. So basically those are the slopes of the line.

Hmm I get the feeling that I'm missing something.

bobbym · 2009-12-12 07:21:29

That is correct. You are not missing anything. How about the lines y=3x-5 and y=7-x/3? What can you say about them?

Anakin · 2009-12-12 07:24:03

Hurray!

And y=3x-5 has a slope or derivative of 3. The line y=7-x/3 has a slope or derivative of -1/3.

bobbym · 2009-12-12 07:35:45

True, but you missed an application of derivatives that I didn't list in post #2.

Anakin · 2009-12-12 08:05:09

I tried graphing them but the only differences I notice between y=5-x/15 and y=7-x/3 are the different y and x-intercepts. And the different slopes. Nothing else.

And for y=3x-5 and y=7-x/3, the derivatives of the two are negative reciprocals.

What am I missing?

Last edited by Anakin (2009-12-12 08:09:56)

bobbym · 2009-12-12 08:28:04

Hi Anakin;

Look closely at the two lines, if you graphed them at the url I gave you what property do they possess?

Anakin · 2009-12-12 08:58:52

http://www.mathsisfun.com/graph/function-grapher.php?func1=7-x/3&func2=3x-5&xmin=-20&xmax=20&ymin=-12.8333333333333&ymax=13.8333333333333

That's what I get when I graph them. But for the life of me, I can't seem to notice anything significant.

Wait, is it because they are perpendicular to each other? Which would be due to the negative reciprocal of the slope, right?

bobbym · 2009-12-12 09:04:20

Wait, is it because they are perpendicular to each other? Which would be due to the negative reciprocal of the slope, right?

Correct! Another use of the derivative.

Anakin · 2009-12-12 09:19:36

Excellent! That can only apply for lines?

Hmm, and do you have some more thought-requiring questions?

bobbym · 2009-12-12 10:24:00

Nope, 2 curves can be perpendicular at a point.

Try this on, a problem in mechanics. A ball is thrown 100 per second straight up. The equation of the balls flight is 100 t - 16 t^2 (avoiding the metric sytem to keep it simple). What is the maximum height achieved by the ball, The maximum is found by setting f ' (t) = 0. Eyeball the graph to check your answer.

Anakin · 2009-12-12 10:38:08

By eyeballing, it looks like the max height is about 156.25.

As for f prime of t, how do I incorporate that into this?

bobbym · 2009-12-12 11:30:48

The maximum or minimum of a function occurs when the slope of the function = 0. In other words when f(t) = 100 t - 16 t^2 has a 0 slope at some point. Since the derivative represents the slope of a curve you just have to find where the derivative = 0 of that curve.

Anakin · 2009-12-12 11:42:40

Very well. But what I meant was how could I find the maximum/minimum by using that algebraically?

bobbym · 2009-12-12 11:50:35

You have to use the derivative. Please no shortcuts just yet. The problem is designed to get you acquainted with calculus in a simple way.

Anakin · 2009-12-12 11:56:16

Ok. Sorry.

So yes, the maximum height of the ball is 156.25 achieved at t=3.125 as I visually observed.

bobbym · 2009-12-12 11:59:33

Hi;

Where's your work?

Anakin · 2009-12-12 12:28:26

Um. You said to observe where the maximum was. I did so and got 156.25.

Then I took f(t) = 100 t - 16 t^2 and subbed in 156.25 for f(t) to solve for t:

156.25 = 100 t - 16 t^2
16t^2-100t+156.25=0
16(t-3.125)^2 = 0
t=3.125

Or I could have used completing the square but that does not involve calculus so I didn't post it.

Or did you want me to use a specific method in mind? I thought about using the lim (h→0)⁡〖(f(x+h)-f(x))/h 〗formula but did not know where to go.

bobbym · 2009-12-12 12:39:32

OK, first differentiate the function:

Set the derivative to 0:

Solve for t

t = 25 / 8 = 3.125 that is where t is a maximum or a minimum. Plug into f(t) and you get 156.25 as the maximum in this case it is the maximum height of the ball.

Anakin · 2009-12-12 13:03:47

Hmm, so that's how you differentiate it. I was just fooling around on paper and got that. I didn't think it was right so I didn't post it but here's how I did it:

f(t)=-16t^2+100t = -16(t-3.125)^2

df(t)/d(t) = -16(t-3.125+h)^2 - (-16(t-3.125)^2)
= [ -16h^2-32th+100h-16t^2+100t-156.25+16t^2-100t+156.25 ] / h
= [ -16h^2-32th+100h ] / h
= -16h-32t+100

But we're trying to find the limit as h approaches zero so -16(0) = 0.
=-32t+100

Did I get lucky or was that really how it should be done?
Also, I changed the f(t) = -16t^2+100t into -16(t-3.125)^2. Was that necessary in order to incorporate 'h' into the parenthesis? Or could I have kept it in expanded form?

Last edited by Anakin (2009-12-12 13:09:47)

bobbym · 2009-12-12 17:35:20

Hi Anakin;

I do the subs at the end.

You can start using the differentiation rules over the formula in A, which is really only for show.

Since you are differentiating a poly the power rule could do the whole thing much quicker.

where D stands for the differentiation operator.

Anakin · 2009-12-12 18:33:38

Hmm, the 2nd step is what I was concerned about. Thanks for clearing it up, that would make a bit easier to imagine. And the 3.125 was not even used, so no numbers.

And as for the last two formulas, I don't understand them. The D(x^n) = n * x^(n-1) : So for the first term we have -16(h+t)^2. Why isn't it -16(h+t)^1 then, since n-1 = 2-1 = 1.

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#1 2009-12-10 19:53:46

Derivatives

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