WOW! Very, very sorry about the tardiness of this scribe... I worked both last and this evening and today was very busy.

So Wednesday's class started with a very unusual period in which we were given our tests back and allowed 10 minutes to continue them since we didn't get a chance last Thursday.

Then we briefly discussed the Developing Expert Voices projects as Mr. K. handed back the proposals.

Following that was a little conversation about "π day" (which is today actually). I'm bringing either a Strawberry-Mango or Blueberry-Peach pie =D.

Finally we started the lesson for the day. The first thing we did was a quick review of Differential Problems.

It dealt with a set amount of oil in a reserve, states that a function A(t) as the rate of consumption, and provides the derivative of the rate [ A'(t) ]. How long until the reserve is completely emptied.

Basically, to solve it, find the antiderivative, A(t)+C, of the the given function and then plug in t=0 to find the constant (C).

From there it is as simple as finding the positive root of that function and that will be the answer.

Next we moved onto the new content.

SLOPE FIELDS

these are kind of tough to explain... I'll let some pictures speak some of my words.

This is the FIELD in which we plot the SLOPES. Easy enough right?

So we first tried it with the expression ( ∂y/∂x = y ). This means that the derivative of any point in the function is the value of the function at that point (y-coordinate).

•Let's say we want to plot the point (1,1). The derivative (SLOPE) of the function at that point is 1, so we draw a small line with slope 1 at that point.

•Next, take the point (3,2). The derivative at that point is 2, so there's another small line, but with a slope 2 at the point (3,2).

Now, you understand how they are constructed, this is how the SLOPE FIELD of the function(s) that have ( ∂y/∂x = y ).

From here, you can solve for the point (1,1).

To do this, draw a curve that follows the slope of the line in either direction until the next whole number is reached. At that point, change the curve's slope to that of the new poitn and continue until the next whole number and so on... WHAT FUNCTION DOES THE CURVE LOOK LIKE???

Correct! It is easy to see that the function that has a derivative equal to the value of the function and passes through the point (1,1) is y=e^x.

However, you can also see that there are many other functions with a similar shape that could have been drawn, depending on the starting point.

This is because the SLOPE FIELD shows the entire family of functions [ ∫ƒ(x)∂x + C ].

So, by solving for a point, you can find the exact one function from that family.

We also did SLOPE FIELDS for ∂y/∂x = 2x and ∂y/∂x = -x/y... can you guess what functions they are???

This was the end of the lesson as we realized that any function can be determined by plotting it's derivative in a SLOPE FIELD and we can find which function out of a family given a point on the specific function.

This concludes my scribe post, once again, sorry it was so late.

The scribe for PI (π) Day will be...

VAN!!!!!!!!!!

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