True, but the concept of a tangent line to visualize a slope can be a useful notion. It's hard for people to wrap their head around an instantaneous slope.

Besides, nobody builds a stair that approaches zero length.

 

Hi, wow thanks for all the replies.

Think I’m going to have to assign a day a week to studying Calculus. I'm not a natural to mathematics you might say, but have a great interest in it.
Semi retired Medical Gas fitter by the way GopherT. Thanks for the link nsaspook, have downloaded it cheers. Got myself 40 hours off utube, a course in Circuit Analysis from the Colorado School of mines. Great lecturer, working the mathematics out as I go (formulae translation, Matrix solutions etc). Again thanks to everyone and take this opportunity to say thanks to the people running the site, wonderful learning resource.

Cheers.
Newcastle, England.

 

Hi,

I might be able to offer a little more insight into the dy and dx and slope and the derivative and the like.

As you know by now, dy/dx means "The change in y with respect to the change in x", or as often quoted, "The change in y with respect to x". So we are trying to quantify change and what it causes, and that is what calculus is all about. There are several ways to look at this.

First of all, why do we care about dy/dx? Many times we want to know the slope of a function like y=x^2 at a certain point and that tells us how the function CHANGES at that point, not what the value of the function itself is. The change itself is a very important quantity in physics and engineering, and so calculus sort of mixes with those areas of interest and makes things easier to calculate later. Knowing how things change allows us to calculate more important things that we would have a very hard time calculating without that knowledge, and because we 'loose' the function value itself in the process, some equations generalize to a wider class of applications.

Since the derivative is the simplest of them, we can start by taking a brief look at that first. One catch though is that when we start to do things like look for change, other little concepts come into play also such as the concept of infinity. Infinity is a number which is considered to be so high that it swamps the other numbers in an equation, which makes them all look insignificant. We use certain rules to figure out how to handle these situations that involve infinity and that is how we arrive at an answer, if there is one that is.

Without understanding infinity however, it is still possible to understand the derivative by using a numerical approximation of a derivative. Remember in dy/dx we are looking to find out what happens to y when we change x, and we want to keep the change in x (dx) small so we can get a reasonable approximation near the x we need.
The simplest approximation for the derivative is:
dy/dx=(y2-y1)/(x2-x1)

So you see right away that the numerator is using two values of y and finding the difference, and the denominator is the difference in the two x's that correspond to the two y's. When we keep the differences small we get a reasonable approximation to the derivative at the point between x1 and x2.
If you have studied the point-slope form of a line, you might note that this is nothing more than 'm' in the equation of a line:
y=m*x+b

So the derivative can be looked at as a line that runs through the point (x,y) where we are finding the derivative, and the slope 'm' is the derivative at that point. This is often called the 'tangent' line.

Numerical examples help to gain insight as in the following example...
Start with the function:
y=x^2+1

Calculate y1 and y2 from x1 and x2 where x1 and x2 are close to the point in question where we want the derivative. Let's say we want it at the x=2.
Starting with x1=2 and taking x2 to be slightly higher x2=2.1, we get:
y1=2^2+1=5
y2=2.1^2=5.41
Now we have x1, y1, and x2 and y2, so we use the simple formula above:
dy/dx=(y2-y1)/(x2-x1)=(5.41-5)/(2.1-2)=0.41/0.1=4.1

so already we have an approximation to the derivative near x=2.
Checking with the exact derivative which is 2*x, we get 4, so we are close to the right value already with the 4.1
and we only need to take one more example to get a clear picture of how this works...

Notice when we chose the second x, x2, we made it 0.1 higher than x1, which was 2. This time, lets make it just 0.01 higher than x1:
x1=2 (again)
x2=2.01 (an increment 10 times smaller than before)
Now calculate again y2 and y1:
y1=2^2+1=5 (again as before)
y2=2.01^2+1=5.0401
Now find the ratio as before:
dy/dx=(y2-y1)/(x2-x1)=(5.0401-5)/(2.01=2)=0.0401/0.01=4.01
NOW we see an even closer approximation to the exact answer, which was 4. This new result is only 0.01 off when before it was 0.1 off, so we got CLOSER by making the difference (or increment) in x SMALLER.

This is the whole basis of differential calculus. When we make the difference in x so small that everything else pales in comparison, we get an exact derivative, which would be 4 in our example.

In the literature this is usually given as:
dy/dx=[F(x+h)-F(x)]/h

where F(x) is the function and h is the increment:
h=x2-x1
F(x1)=y1
F(x1+increment)=y2
so again we get:
dy/dx=(y2-y1)/(x2-x1)

as the approximation. There are other ways to do this, but that is the most used way.

Comparing the two examples should show you how this process works. You might note that calculus is almost like a process in itself rather than a single calculation because we are always concerned with what happens when something changes. We do get a single result though unless there is more than one solution to the entire problem.

 

y(x) = equation, so dy / dx is the 'rate' at which 'y' changes when 'x' changes, as apposed to the value that 'y' has when 'x' has a value.
This is useful information in any context.
Even though dy and dx can be manipulated as divisor and dividend in an algebraic expression, they cannot be evaluated, as they are infinitely small.
The thing is, y(x) can take on an infinite number of values. So it is possible, 'Mathematically', to find the slope at an infinitely precise point.
So, once the algebra has been done, and you have manipulated dy and dx to where you want them to be, keeping everything on the left side of the equals sign equal to the right side, use 'integration' to get a finite value. No good on a building site.
Now, the square root of -1, symbol 'i' or 'j' is an equally un-magical 'value', and 'disappears' once a solution is found.
Great and useful tools. My hat off to those that made them.

 

Here is an ebook that explains all.

http://www.ibiblio.org/kuphaldt/socratic/sinst/ebooks/Calculus_Made_Easy.pdf

Priceless, read a couple of chapters over the weekend, thank you so much for that absolute gem of a book. Should have its own link in the education section. Loved the prologue and his description of "d”

CHAPTER I
TO DELIVER YOU FROM THE PRELIMINARY TERRORS.

The preliminary terror, which chokes off most fifth-form boys from

even attempting to learn how to calculate, can be abolished once for

all by simply stating what is the meaning in common-sense terms of

The two principal symbols that are used in calculating.


These dreadful symbols are:


(1) d which merely means “a little bit of."

…….

Thanks again for a wonderful link, going to see if I can get an old copy of this.

Cheers

 

While saying d simply means "a little bit of" there is a subtle matter of importance that should be brought to your attention.

The question is - what is one's "a little bit of"?

In calculus, dy/dx, where dx is "a little bit of x" means "a zero amount of x".

Lets go back to where dy/dx comes from.

We want to determine a change of y, written as Δy, divided by a change in x, written as Δx.

Slope = Δy/Δx

You can do this for practically any type of measurement.
What we do is we make Δx as small as possible. We write this as Δx→0, meaning as Δx approaches zero.

Hence as Δx→0

Δy/Δx → dy/dx

that is dy/dx becomes a specific value at the instant x. There is no if, and, but about how little of x is chosen. We have chosen zero amount of x.

Note that we have been taught that dividing by zero gives us a very large result, i.e. infinity, which really means that there is no valid mathematical number to represent a division by zero. In our case, this does not happen, Δy/Δx as Δx→0 still gives a valid result. That's the beauty of calculus.

btw - this is not a "dreadful symbol" but is an important useful part in the vocabulary of mathematics. As you move on to learning about the useful mathematics of electronics, you will encounter another symbol "j" which we will leave for another lesson.

(The other thing that I intentionally left out is how differentiation "dy/dx" relates to electronics and where it becomes useful. I didn't want to muddy the discussion.)