Jedi (implicit)

Yoda

SCRIBE EPISODE 4:Return of the Jedi


So, Some people (those who are not in our class, or weren't there yesterday) reading this Scribe post may be a

little confused by the title.

The title is as such because Mr. K. actually started off the class with a couple "Jedi mind tricks"!!! Isn't that

awesome?!?!

I won't go into detail, but let's just say:

THERE ARE NO GREY ELEPHANTS IN DENMARK!!!

And so I begin the Calculus Related part of this Scribe post...


Well, Mr. K. then shoed us a couple pictures (man I wish the slides were up) with questions pertaining the existence of somethings that were not shown in the pictures, but were

eluded to by certain hints in these pictures.

i.e. You see a shadow of person on the ground, but have not turned around to see this person yet. Is there someone behind you???

When we gave an answer he asked us why? We said because it is implied. There aren't any visuals of the objects in question,

but certain evidence allows us to imply that they exist.

This leads to the topic of the day:IMPLICIT FUNCTIONS!!!


So we moved on to a quick review of composite functions and the chain rule, because we would have to use it later

on in the lesson.

In one example the answer was:


Then we were asked to identify THIS function:


It is, of course, a SEMI CIRCLE


Now what's the equation of the full circle???




And that can be drawn as...


2-2

2

-2


However, a circle is not a function. But, is there a function (besides √4-x2 and its opposite: -√4-x2) that is a

part of the circle???


Sure there is: There's this one...

2-2

2

-2


2-2

2

-2

And this one...


And this one...

2-2

2

-2


Do you understand that there are an infinite amount of functions with x as the

variable?

All of these functions hidden in the main graph of a circle, are

IMPLICIT FUNCTIONS!!!


Now think of a relation such as the circle given previously, and think how you would find the derivative

of any given point in the domain that relation?

There are infinite implicit functions, so who knows which one you should find the derivative of because there are

many of these functions that go through the same points!


The only way to find the derivative of a circle at any given point is by finding the

derivative of the relation using the following derivation:


First of all, let's use the following equation of a circle:


Now let's find the derivative:


First rewrite it as such:


Now we can recognize that the derivative of 25 is always 0.


Next, use the chain rule on the left side of the equation...

remember it is:


Identify ƒ(x), g(x), and their derivatives:

The reason why the inner function "g(x)" is simply "y" is because the inner function of y^2 is an implicit function. We do not know exactly what it is, but we know it is there.


Did everyone see the red text on the previous slide? It is very important!

The reason why the inner function "g(x)" is simply "y" is because the inner function of y^2 is an implicit function. We do not know exactly what it is, but we know it is there.


Anyway, now using the outer and inner functions, complete the chain rule:


Now combine it with the derivative of x2

(2x) and you have the derivative of the left side of the equation...


therefore...


Now solve for " y' "...


So now we know that the derivative (slope of the tangent line) at any given point on

the circle x2 + y2 = 25 is given by the following formula


So let's test it out.

Let's say we need to find the derivative of the previous function at x=3.


First, find the output of when x=3:


The next step is simple. To find the derivative at that point, plug the coordinates into the

equation of the derivative of the circle:


The derivative (slope of the tangent) at x = 3 is (-3)/4... Let's see if that is true.


So yes, the derivative (slope of the tangent) at any point of any circle with its center at the

origin (because the derivative of the constant will always be zero) is given by:


From there, Mr. K. gave us some examples to try and find the

derivative functions of relations (which could have been seen on

the slides if they were up).


Then one final point was made:

When the numerator of the derivative function of the relation

is 0, the tangent is a horizontal line.

When the denominator is 0, then the tangent is a vertical line.

(as seen on the next slide)



Yoda

About it, that is...

For reading, thank you...

Grey-M, the next scribe is...

Entertainment & Humor

Jedi (implicit)