Did You Know?

Did you know I can see your classroom from two windows?!

My first window is your blog. I am excited by what I see and hear! I never cease to be amazed by the quality and sophistication of your scribes; you constantly achieve new heights in illustrating and annotating your scribes. More than that I am so impressed when you celebrate each others’ learning, when you are creative, and when you critically reflect upon your learning in your BOBs.

Did you know Mr. K’s blog is my second window? I admire and respect what I see and hear here too! Did you know that Mr. K celebrates your learning on his blog? that he reflects upon what best helped you to learn and why? that he unselfishly shares all he knows with those who read his blog? that he learns from the conversations on his blog? that he writes with passion and is creative? and that he commits many random acts of kindness by honoring other teachers’ accomplishments in his posts? Did you know all he expects of you, he shares those same expectations for himself?

Did you know that because of all that and more, Mr. K.’s blog has been nominated for “Best Teacher Blog 2006” on the Edublog Awards website?

I just happen to think that no one deserves this honor more than Mr. K.

What about you?

Scribe Post

Hello everybody! Sorry I'm kinda late, I had work.

But anyways today we had a substitute. We didn't do much except a quiz. A quiz that normally would take 15 or 20 minutes, ended taking up the whole period class.

There was three questions on the quiz; finding the rate of a radius, local linerlization, and newtons law. I think everybody did okay on it, since we took a really long time.

Then we were supposed to get a sheet of questions, but I don't think anybody got them.

Remember everybody keep up with the homework and you'll keep up with the great marks.

The next scribe will be suzanne :D

Oh, and Happy Birthday Charlene!! Don't stay out too late :)

Scribe POst

Hey Ladies and Gents, this is Mark, your voluntary scribe for this fine evening.

Well, today we started with two questions and worked through both for the class. Mr. K showed us how to upload pics to the other blogs, seemingly as we have RUN OUT OF SPACE on this blog =).

1)Find the global max of f(x) = 2x³-9x²+12x+2 on the interval [0, 3].

f'(x) = 6x² -18x + 12 Differentiating
= 6(x²- 3x + 2)
= 6[(x - 2)(x - 1)]

x= 2, x =1

End Points:
f(0) = 12
f(3) = 11

Critical Numbers:
f(1) = 7
f(2) = 6

From here, we can use the First Derivative test to see if there is a max or a min:
1) Draw a number line
2) Put endpoint and critical numbers in increasing order, left to right.
3) Find values in between each point on the number line and look for 'changes in sign' (maxima and minima)

2) Assume P is a positive constant f(x) = x^3/7 - px^10/7
A) Find f'(x)
B) Find all critical number of f (answer may involve p)
C) Classify each critical number (max, min , neither)
D) Does f have a global max in it's domain? JUSTIFY.

Solutions
A) f'(x)= 3/7x^(-4/7) - 10/7px^(-3/7)
= 1/7x^(-4/7)(3-10px)
= (3 - 10px)/(7x^4/7)

B) f'(x)= 0 when 3 - 10px = 0
3 = 10px
x= 3/(10px)

f' (x) is undefined when

7x^(4/7)
x = 0

Looking at the number line, f' is positive for all x values less than 0. between 0 and 3/10p all values are also positive. All values to the right of 3/10p are negative.



C) 3/10p there is a maxima
At x = 0 there is a vertical line on the parent function.

D)


We will use: f(x) = x³ - 2x²
f'(x) = x(3x-4)

+ - +
f' -----|-----|-----
0 4/3

f''(x) = 6x - 4
=2(3x-2)

second derivative test
- +
-----|-----
2/3

Inflection point at 2/3

There is a min at f (0), because
f' (4/3) = 0
f"(4/3) > 0

Here is a Short tutorial for uploading pictures =p






Upload Tutorial

1) Log into Blogger



2) Go to Pedro the Pi Rate Panda Project



3) Make a new post and click the upload button



4) Upload the file you want



5) Go back to the editor, click on "edit html" and then copy the code. Take this code and post it into your composer. Note: if your not good with html, paste the code into the compose tab so you know where your placing your image.


HW is 5.2 all odds and #2

Next scribe is...Cryyystalll =)

BOB

Whoo.. I almost forgot about this thing. Well, this unit was really easy at the beginning. I wasn't having much trouble until we got to the evil related rates problems. My biggest problem was that I don't remember things from previous math classes. And another thing is that there are a lot of places you can go wrong while doing a related rates problem. It would have been easier if the back of the book had more than just the answer. I know that the full answers are available in the classroom but it's kind of difficult to find enough time to sit in there and look through it. I think it would be easier if the solutions were available online or something.. I wish the publishers would do that. Anyway, today's test day and I think I'll pass.. hopefully..

BOB

Another unit past and yet another test to be written =S. Surprisingly when the unit first started everything seemed so easy and I understood everything that we were learning. I figured "alright! this unit is gonna help my mark soooo much!". My excitement basically ended there. As the unit went on I found things more complicating and much harder to understand, I wasn't getting it. This is definately a tough unit and I pretty sure the test is going to be a tough one too. Learning the Differentiating Rules weren't too bad, it was actually keeping them all in your brain and knowing when and how to apply each of them to questions is what got me off whack a bit. Then we went into related rates problems and everything went down hill from there. So I'm praying a good amount of studying will help me produce a better result then I'm predicting coming into the test tomorrow. I'm more or less clinging at the end of the cliff desperately trying to pull myself back up. Hopefully I got enough strength.

Scribe Post: Day 54

Terribly sorry for having this up late and on the wrong day of all things =S But I'm going to cut to the chase here, on Friday we had a pre-test for "Differentiating Rules" here are the questions a solutions:


(1) To make things basic and easy all you need to do is put this equation into your calculator (since it's already a derivative) and substitute a very very small value like 0.000001 for your h and solve the equation. You should come up with an answer relatively close to 5.55 which is (C). I'm not completely sure if this applies for all questions like this, but I'm hoping it does. =S

(2) Differentiate f(x) = x - k / x + k by using the "Quotient Rule". This should give you the derivative:

f'(x) = (x + k)(1) - (x - k)(1) / (x + k)²

Then solve for f'(0):

f'(0) = (0 + k)(1) - (0 - k)(1) / (0 + k)²
f'(0) = k - (-k) / (0 + k)²
f'(0) = 2k / (0 + k)²
f'(0) = 2 / k


(3) Differentiate g(x) = f(f(x)) by using the "Chain Rule". This should give you the derivative:

g'(x) = f'(f(x)) f'(x)

You're trying to estimate g'(1) so your equation will look like this:

g'(1) = f'(f(1)) f'(1)

You already have your value for f(1) = 2. However you must find the derivative value for f'(2) and f'(1) in order to solve the equation. You do this by applying the "Symmetric Difference Quotient to the two points x = 1 and x = 2 on the table provided. Doing this you get the values f'(1) = 0.6 and f'(2) = 2. Now you can solve the equation:

g'(1) = f'(f(1)) f'(1)
g'(1) = f'(2) f'(1)
g'(1) = (2)(0.6)
g'(1) = 1.2


(4) You will be using the "Pythagorean Theorem" to solve this question. First you must find the value for side z:

4² + 3² = z²
16 + 9 = z²
5 = z

Then differentiate the Pythagorean Theorem which comes out to be this:

2x dx/dt + 2y dy/dt = 2z dz/dt (NOTE: you can cancel out the 2s)
x dx/dt + y dy/dt = z dz/dt

Using this you find the value for dy/dt:

(4) 3 dy/dt + 3 dy/dt = 5(1)
15 dy/dt = 5
dy/dt = 5/15 = 1/3

Now that we have dy/dt we can find dx/dt:

dx/dt = 3 (1/3)
dx/dt = 1


(5) Differentiate the function h(x) = f[g(x)] by using the Chain Rule this gives you:

h'(x) = f'(g(x)) x g'(x)

We are asked to find the horizontal tangent lines to the graph h, this will be where the derivative is equal to 0. So first we find out where g'(x) = 0, by looking at the graph that would be at x = -3, 0, 2. Then find out where f'(x) = 0, by looking at the graph that would be at x = -2, 1. Then since you have f'(g(x)) in the equation of h' you must find out where that would equal 0. This equals 0 where g(x) = -2, 1. Those values would be x = -4, -2. Add up all the values of x you found (making sure not to count a value twice) and you come up with 6 horizontal tangent lines.



Open Response: You will be using the equation for volume to solve this question. First differentiate the equation this gives you:

dV/dt = 4pir² dr/dt

Find the volume at 4 seconds by looking at the graph, V(4) = 5pi. Then using that value we can find r which is ³√15/4. Then we find the value of dV/dt from the graph. You draw a tangent line to the graph and just estimate by using the values on the graph. (NOTE: this value varies depending on how well and where you draw your tangent line) Mr. K got a value of 4pi/3 for dV/dt. Then we just plug all these values into the equation to find dr/dt which will be the approximate rate of the radius of the balloon changing after 4 seconds:

4pi/3 = 4pi(15/4)^1/3 dr/dt
1/3(15/4)^-1/3 = dr/dt
0.21455 ≈ dr/dt


And that is all folks, I'm hoping this isn't up tooo late so that it can be if you wish as studying material for the test which is TOMORROW. You all know Christian is the scribe for today, and he's probably waiting for me to put this up right now, sorry =S. Good luck to everyone on their test tomorrow, live long and prosper!

Scribe Post: Day 55

OKAAAY! Well, once again, I can't upload my graphs/pics. So I'll go early tomorrow morning at school, as usual, and complete this post, or go on again later and see if it works. Sorry for the inconvenience.

Today’s class was about Global Extrema and the Extreme Value Theorem.

TERMS

Maxima – plural for maximum
Minima – plural for minimum
Extreme – either a minimum or a maximum
Extrema – minima and maxima
Local – just a vicinity on a graph
Global – considers the entire interval
Critical Numbers – where f’ = 0 or undefined

In previous lessons, we learned that local extrema are the lowest or highest possible points in some vicinity on a graph. For example, given this graph (figure A), we’ll see that the points highlighted with green appear to be the maximum and minimum. If we consider the entire graph, however, we know that these points aren’t the extrema.



Global extrema are the maxima and minima over the entire interval. Notice that I said “over the entire interval”. The interval must be set to determine the global extrema.



To find the global extrema, do the following steps:
1) Find f’
2) Find the critical numbers
3) Evaluate f at the critical numbers and end points**
4) Determine the global extrema (the lowest and highest values on the interval)

**Always check where the function is undefined. This is the most overlooked place where the function has a min/max

This leads to a theorem in calculus called the Extreme Value Theorem (EVT), which states that if a function is continuous on a closed interval, [a, b], then it has a global max and global min. This is an existence theorem, meaning that if you can draw the graph of a function without lifting your pen (continuous), then global extrema must exist. It doesn’t say where these exist; just says they should exist.

Let’s do an example, following the steps above:



What we can take away from this is that from the interval [-1, 3], the global extrema are also the local extrema.

Due for Wednesday are Exercises 5.1, odds.
Also, due on Monday is the derivative photo assignment.