Mobius Strip Day!
Ryan Smith
Thursday, March 10, 2011
Today in class, we learned about the mobius strip, a one-sided, three-dimensional figure. We created own own mobius strips and tested the idea that it only had a single side. We then proceeded to cut the mobius strips that we made in various ways to see what shapes would form.
We then put all the mobius strips onto the ceiling, and hopefully they're still there.
This is a link that shows how to make a mobius strip and some experiments to do with it. You can always make your own experiment to. http://www.youtube.com/watch?v=4bcm-kPIuHE
Today, we started off the class with an activity on GSP, (Geometer's Sketchpad - which is conveniently available to you on all the OSCSS computers) and some review on increasing, decreasing and constant functions. We then did a worksheet where we sketched functions from their derivatives.
We can use the first derivative to determine the interval(s) over which the function is increasing or decreasing.
Ex 1: Find the intervals of increasing and decreasing for:
f (x) = 2x3 + 3x2 - 36x + 5 f ’(x) = 6x2+6x-36 0 = 6(x+3)(x-2) x = -2, 3 <- Called critical points
-3 > x
-3 < x < 2
x > 2
f ’ (x)
+
-
+
∴ f (x) is increasing when x > -2 and when x>3 (-∞, 3) f (x) is decreasing when -2 < x < 3 (2,∞)
Ex 2: Sketch a continuous function if f ' (x) > 0 when x < 0, and f ' (x) < 0 when x > 0 and when f (0) = 4
Remember that the derivative of a cubic function is a quadratic function, whereas the derivative of a quadratic function is a linear function, and the derivative of a linear function is a constant function. Knowing this is helpful when sketching functions and their derivatives.
- At inflection points, note that the slope of the tangent is equal to zero.
HOMEWORK: See http://archives.math.utk.edu/visual.calculus/3/graphing.5/index.html and complete the questions. We also have a curve sketching quest on April 1st. HANDOUTS: 1. Increasing / Decreasing / Constant Functions (front), GSP: When are Functions Increasing or Decreasing (back)
2. Sketching Functions from Derivatives
Today our first group task was to solve the speed challenge. The entire class was divided into groups of three or four members each and the challenge presented to us was:
- Place the numbers from 1 to 6 in the circles so that no two consecutive numbers are in circles joined by a line for the following diagram
Next, the topic we covered was optimization. For this activity, we worked in our groups again and using the following materials we made an open box: a pink sheet, scissors, ruler and tape.
Then, as a class we solved the following optimization problem: a piece of sheet metal 60cm by 30cm is to be used to make a rectangular box with an open top. Determine the dimensions that will make a box with the largest volume.
· A fxn is at its max or min at its critical points (when f’(x)=0) or at its end points · Our interval 0<x<15 is an open interval so we can’t test our end points V(x) = 4x^3-180x^2+1800 V’(x) = 12x^3-360x+1800 = 0
X= 23.6, 6.3 [23.6 is invalid because it would make width negative which is impossible, our original width is 30 cm so 30-2(23.6) of width would not let us make a box at all]
V (6.3) = 4(6.3)^3 -180 (6.3)^2 +1800 = 5196 cm3
So, the dimensions would be
Height = x = 6.3 cm Length = (60-2x) = (60-2(6.3)) = 47.4 cm Width = (30-2x) = (30-2(6.3)) = 17.4 cm
Next, each group figured out the steps to solve such optimization prlems. Then the class as a whole came up with the following concise set of steps to solve these problems 1.Draw a diagram for the problem and define a variable. 2. Determine the function and the interval. 3. Find the first derivative (f’(x)) and set it equal to 0 to find critical points. Also find the end points. 4. Test the critical points from the interval and end points (if closed interval) by inputting them into the function. 5. Find the highest and lowest # from step 4, it is the optimal point. 6. Refer back to the initial question and answer it.
After that we re-did the box making activity, this time using a yellow sheet and found the dimensions needed to get a box with the largest volume. (answer: height=4cm, width=13.6 cm, length=20cm)
Ryan Smith
Thursday, March 10, 2011
Today in class, we learned about the mobius strip, a one-sided, three-dimensional figure. We created own own mobius strips and tested the idea that it only had a single side. We then proceeded to cut the mobius strips that we made in various ways to see what shapes would form.
We then put all the mobius strips onto the ceiling, and hopefully they're still there.
This is a link that shows how to make a mobius strip and some experiments to do with it. You can always make your own experiment to.
http://www.youtube.com/watch?v=4bcm-kPIuHE
There was no homework today.
Moe Qureshi
Here is the note we took on March 18th
Here is a link to a good video if you need help on learning about oblique asymptotes:
http://patrickjmt.com/find-asymptotes-of-a-rational-function-vertical-and-obliqueslant/
March 21, 2011
Virginia Lee
Topic: Increasing and Decreasing Functions
Today, we started off the class with an activity on GSP, (Geometer's Sketchpad - which is conveniently available to you on all the OSCSS computers) and some review on increasing, decreasing and constant functions. We then did a worksheet where we sketched functions from their derivatives.
We can use the first derivative to determine the interval(s) over which the function is increasing or decreasing.
Ex 1: Find the intervals of increasing and decreasing for:
f (x) = 2x3 + 3x2 - 36x + 5
f ’(x) = 6x2+6x-36
0 = 6(x+3)(x-2)
x = -2, 3 <- Called critical points
∴ f (x) is increasing when x > -2 and when x>3 (-∞, 3)
f (x) is decreasing when -2 < x < 3 (2,∞)
Ex 2: Sketch a continuous function if f ' (x) > 0 when x < 0, and f ' (x) < 0 when x > 0 and when f (0) = 4
The derivative of a cubic function is a quadratic function!
Taken from: http://www.contracosta.edu/legacycontent/math/mathlinks.htm
Remember that the derivative of a cubic function is a quadratic function, whereas the derivative of a quadratic function is a linear function, and the derivative of a linear function is a constant function. Knowing this is helpful when sketching functions and their derivatives.
- At inflection points, note that the slope of the tangent is equal to zero.
HOMEWORK: See http://archives.math.utk.edu/visual.calculus/3/graphing.5/index.html and complete the questions. We also have a curve sketching quest on April 1st.
HANDOUTS: 1. Increasing / Decreasing / Constant Functions (front), GSP: When are Functions Increasing or Decreasing (back)
2. Sketching Functions from Derivatives
Additional Links: See http://biobio.loc.edu/raja/web/FirstDerivativetest.htm for a concise explanation about increasing / decreasing / constant functions and the first derivative test (which we need to know in order to do the homework. http://www.ltcconline.net/greenl/courses/115/applications/increasedecrease.htm also has a great explaination, with lots of examples!
Maxima and Minima
March 22, 2011 by Rosetta D'souza
- Max and min points happen at either end points* or critical points.
*Function must be defined over an interval.f’(x) = 0
Ex) Find the extreme values (absolute max and min points) of f(x) = -2x3 + 9x2 + 4, x ϵ [-1, 5]
end points - x = -1, 5
f’(x) = 0 = -6x2 + 18x
∴ -6x (x - 3) = 0
∴ x = 0, 3 - critical points
f(-1) = 15
f(5) = -21 – absolute min
f(0) = 4
f(3) = 31 – absolute max
Sketching a Function Graph From a Derivative Graph Handout
Homework: N/A
Extra link: http://www.csp.science.ubc.ca/life/StudentSamples/Website1/theory.html explains the above in more detail.
Rabeea Fatima
March 23, 2011
Minima and maxima (contd.)
Helpful Links:
http://www.youtube.com/watch?v=QtXCIxB6kW8
http://www.youtube.com/watch?v=Z7QWpBU1ePU&feature=related
Homework sheet handed out.
Notes:
Gurpriya Kaberwal
March 29, 2011
Teamwork Tuesday
Today our first group task was to solve the speed challenge. The entire class was divided into groups of three or four members each and the challenge presented to us was:- Place the numbers from 1 to 6 in the circles so that no two consecutive numbers are in circles joined by a line for the following diagram
Next, the topic we covered was optimization. For this activity, we worked in our groups again and using the following materials we made an open box: a pink sheet, scissors, ruler and tape.
Then, as a class we solved the following optimization problem: a piece of sheet metal 60cm by 30cm is to be used to make a rectangular box with an open top. Determine the dimensions that will make a box with the largest volume.
Volume = X (60-2x) (30-2x)
= 4x^3-180x^2+1800
· A fxn is at its max or min at its critical points (when f’(x)=0) or at its end points
· Our interval 0<x<15 is an open interval so we can’t test our end points
V(x) = 4x^3-180x^2+1800
V’(x) = 12x^3-360x+1800 = 0
X= 23.6, 6.3 [23.6 is invalid because it would make width negative which is impossible, our original width is 30 cm so 30-2(23.6) of width would not let us make a box at all]
V (6.3) = 4(6.3)^3 -180 (6.3)^2 +1800
= 5196 cm3
So, the dimensions would be
Height = x = 6.3 cm
Length = (60-2x) = (60-2(6.3)) = 47.4 cm
Width = (30-2x) = (30-2(6.3)) = 17.4 cm
Next, each group figured out the steps to solve such optimization prlems. Then the class as a whole came up with the following concise set of steps to solve these problems
1. Draw a diagram for the problem and define a variable.
2. Determine the function and the interval.
3. Find the first derivative (f’(x)) and set it equal to 0 to find critical points. Also find the end points.
4. Test the critical points from the interval and end points (if closed interval) by inputting them into the function.
5. Find the highest and lowest # from step 4, it is the optimal point.
6. Refer back to the initial question and answer it.
After that we re-did the box making activity, this time using a yellow sheet and found the dimensions needed to get a box with the largest volume. (answer: height=4cm, width=13.6 cm, length=20cm)
Then we spent the rest of the class working on the “Practice with optimization” sheet to understand how to set up functions for such problems.(Answers:
1) A(x) = x(50-x)
2) P(x) = 2x + 128/x
3) SA(x) = 2x^2 + 4000/x
4) D(t) = √ (√12.5-4t^2) + (√12.5-6t^2)
5) Cost(r) = 2πr^2a +2vol/r
6) P(x) = (50+x)(30-0.25x)
Useful links:
http://www.vitutor.com/calculus/graphs/optimization_problems.html
http://www.johnkerl.org/teaching/fa08-124-026/oprr.pdf
http://www.youtube.com/watch?v=Ef22yTJDUZI
Homework:
- Curve sketching assignment is due tomorrow March 30, 2011
- Optimization homework sheet is due Thursday March 31, 2011
Work Period
Ryan Smith
March 30th, 2011
Today was a work period, so we had a chance to work on the Curve Sketching assignment, the Optimization homework, and study for the test on Friday.Here are some useful links to help you with curve sketching and optimization.
http://www.youtube.com/watch?v=vOTTuZflAIM
http://www.youtube.com/watch?v=3GYv-BZYYdg&feature=fvwrel
http://www.youtube.com/watch?v=i8Wtu-kdDC4&feature=fvwrel
http://www.youtube.com/watch?v=T8sG4Sb3g7Y&feature=fvwrel
Homework:
Remember that the Optimization homework is due by 9:00 am on Thursday, March 31st and the test is on Friday, April 1st!
STUDY HARD!!