Summer 2012
MATH 225: Multivariable Calculus and Geometry. II Branko Ćurgus
Thursday, August 9, 2012
The assigned problems for Section 20.4 are 1, 3, 5, 7, 13, 15, 17, 23, 25, 27, 28, 31.
The assigned problems for Section 20.5 are 1, 2, 7, 9, 11, 13, 15.
Wednesday, August 8, 2012
The assigned problems for Section 20.3 are 3, 7, 11, 12, 13, 14, 19, 21, 26, 27, 28, 29, 31, 36.
Tuesday, August 7, 2012
The assigned problems for Section 20.2 are 1, 3, 5, 7, 11, 17, 20, 21, 23, 27, 28, 29, 31, 33, 37, 43.
Wednesday, August 1, 2012
The assigned problems for Section 20.1 are 1, 2, 5, 7, 9, 12, 17, 19, 20, 21, 23, 29.
Wednesday, July 25, 2012
The assigned problems for Section 19.1 are 2, 3, 5, 7, 9, 11, 15, 21, 23, 25, 29, 34, 39, 45.
The assigned problems for Section 19.2 are 1, 3, 5, 9, 13, 15, 18, 19, 23, 25, 27.
The assigned problems for Section 19.3 are 1, 3, 5, 7, 8, 9, 10, 11, 12, 15.
Monday, July 23, 2012
The assigned problems for Review of Chapter 18 are 14, 15, 17-23, 30, 31, 34, 38, 42, 45, 50.
Here is the Mathematica file that I used today. It is called 20120705.nb. As before, right-click on the underlined word "Here"; in the pop-up menu that appears, your browser will offer you to save the file in your directory. Make sure that you save it with the exactly same name. After saving the file you can open it with Mathematica 5.2.
Thursday, July 19, 2012
The assigned problems for Section 18.4 are 1, 3, 5, 9, 13, 15, 17, 21, 25, 26, 27, 28, 30, 33, 35, 36, 37.
Tuesday, July 17, 2012
Today we did most of Section 18.3.
The assigned problems for 18.3 are 1,3,4,5,6,7,8,9,11,12,13,17,19,21,22,25,33,34,35,36,38,39,40,43,45.
Thursday, July 12, 2012
The assigned problems for 18.1 are 1,3,5,7,9,15,17,19,21,32,33,34,35,39,41.
The assigned problems for 18.2 are 3,7,9,13,17,23,25,29,31,33,35.
Monday, July 9, 2012
Here is the Mathematica file that I used today. It is called Surfaces.nb. Right-click on the underlined word "Here"; in the pop-up menu that appears, your browser will offer you to save the file in your directory. Make sure that you save it with the exactly same name. After saving the file you can open it with Mathematica 5.2. You will find Mathematica in
Start -> All Programs -> Math Applications -> Mathematica.
Open Mathematica first, then open Surfaces.nb from Mathematica. You can execute the entire file by the following manu sequence (in Mathematica):
Kernel -> Evaluation -> Evaluate Notebook.
You can view animations by double-clicking on a cell that contains an animation. You can recognize such a cell by noticing a small arrow at the bottom of its cell bracket; which is on the far right side of the cell.
Thursday, July 5, 2012
The assigned problems for 17.5 are 1, 3, 5, 15, 17, 19, 25, 28, 29, 30, 34, 37, 39.
Here is the Mathematica file that I used today. It is called 20120705.nb. As before, right-click on the underlined word "Here"; in the pop-up menu that appears, your browser will offer you to save the file in your directory. Make sure that you save it with the exactly same name. After saving the file you can open it with Mathematica 5.2.
The assigned problems for Review of Chapter 17 are 7, 9, 11, 12, 25, 26, 27-29, 30, 31, 32, 34, 35, 37, 38, 42, 48, 50, 55, 57, 58.
Tuesday, July 3, 2012
The assigned problems for 17.3 are 1-10, 11, 12, 13, 15, 16, 17, 20, 22, 23, 25, 26, 27, 28, 31.
The assigned problems for 17.4 are 4, 5, 7, 8, 9, 15, 16, 17, 18, 20, 21, 22, 23.
Here is the Mathematica file that I used today. It is called VectorFields.nb. Right-click on the underlined word "Here"; in the pop-up menu that appears, your browser will offer you to save the file in your directory. Make sure that you save it with the exactly same name. After saving the file you can open it with Mathematica 5.2. You will find Mathematica in
Start -> All Programs -> Math Applications -> Mathematica.
Open Mathematica first, then open VectorFields.nb from Mathematica. You can execute the entire file by the following manu sequence (in Mathematica):
Kernel -> Evaluation -> Evaluate Notebook.
Friday, June 29, 2012
Yesterday I pointed out that there is a typo in Example 6 in Section 17.2 page 891 in the textbook. The typo that I had in mind is the use of the equality sign at the last step in the solution. At the beginning of the problem the authors state that we will encounter "an integral which must be calculated numerically". (This is the first clue.) It is surprising that at the end of the calculation it turns out that the last integral equals $9.69$. What they mean is that the last integral approximately equals $9.69$. The common mathematical notation for this is $\approx 9.69$. The second clue is in the last sentence of the solution. The authors do not write $2\pi = 6.28$, but correctly $2\pi \approx 6.28$.
Example 6 is an excellent opportunity to learn about a mathematical function that is rarely mentioned in calculus classes. It is called the complete elliptic integral of the second kind. For $0 \leq k \leq 1$ it is defined as
\[
E(k) = \int_0^{\pi/2} \sqrt{1- k^2 (\sin \theta)^2 } \, d \theta
\]
Clearly, $E(0) = \pi/2$ and
\[
E(1) = \int_0^{\pi/2} \sqrt{1- (\sin \theta)^2 } \, d \theta
= \int_0^{\pi/2} \cos \theta \, d \theta = 1.
\]
The function complete elliptic integral of the second kind
The complete elliptic integral of the second kind function can be used to express the circumference of an ellipse exactly. Consider the ellipse in polar coordinates: $x(\theta) = a \cos \theta, y(\theta) = b \sin \theta$. Assume that $0 \leq a \leq b$. Then the circumference is given as
\begin{align*}
\int_0^{2 \pi} \sqrt{a^2 (-\sin \theta)^2 + b^2 (\cos \theta)^2 } \, d\theta & = 4 \int_0^{\pi/2} \sqrt{a^2 (\sin \theta)^2 + b^2 \bigl(1- (\sin \theta)^2 \bigr) } \, d\theta \\
& = 4 \int_0^{\pi/2} \sqrt{ b^2 - \bigl(b^2 - a^2\bigr) (\sin \theta)^2 } \, d\theta \\
& = 4 b \int_0^{\pi/2} \sqrt{ 1 - \biggl(1 - \frac{a^2}{b^2}\biggr) (\sin \theta)^2 } \, d\theta \\
& = 4 b \, E\,\Biggl(\sqrt{1 - \frac{a^2}{b^2}}\Biggr)
\end{align*}
The quantity $\sqrt{1 - \frac{a^2}{b^2}}$ is called the eccentricity of an ellipse. It measures how much ellipse deviates from a circle. If eccentricity of a circle is 0. If the eccentricity is 1, then ellipse degenerates to a line segment of length $2b$.
In Example 6 we have $a=1$ and $b=2$, so the eccentricity is $\sqrt{3}/2$. Thus the exact expression for the circumference is $8 E\bigl(\sqrt{3}/2\bigr)$. Mathematica gives $E\bigl(\sqrt{3}/2\bigr) \approx 1.211$. Hence, we have that the circumference of the ellipse in Example 6 is approximately equal to $8\times 1.211 = 9.688$.
One more comment about the solution of Example 6. At the end of the solution the authors use two circles to obtain rough estimates for the circumference of the ellipse. In general it is an excellent idea to think of simple ways to find rough estimates for numbers that you are seeking in a problem. In this particular problem there are simpler and better rough estimates given by a rectangle in which the ellipse is inscribed and a parallelogram about which the ellipse is circumscribed, see the picture below.
$2\pi \lt 8 E\bigl(\sqrt{3}/2\bigr) \lt 4\pi$
$4\sqrt{5} \lt 8 E\bigl(\sqrt{3}/2\bigr) \lt 12$
Thursday, June 28, 2012
The assigned problems for 17.2 are 2, 3, 6, 8, 10, 12, 13, 14, 17, 19, 23, 25, 26, 31, 33, 34, 36, 37, 38, 45.
There is a typo in Example 6 in Section 17.2 page 891 in the textbook. Do you see a typo? The authors give you two clues for this typo, one before it one after it.
The assigned problems for 16.7 are 2, 3, 5, 9, 11, 12, 15, 16, 21, 22, 23, 24, 25, 26.
Thursday, June 21, 2012
First about Mathematica.
Today in class I used Mathematica version 5.2 to verify my calculations. Mathematica 5.2 is quite widely available on campus. The current list of the labs which have Mathematica 5.2 you can find here. On some campus computers we also have Mathematica version 8. These versions are not compatible. For this class I will only post version 5.2 notebooks since there are more computers with version 5.2.
Here is the Mathematica file that I used today. It is called 20120621.nb. Right-click on the underlined word "Here"; in the pop-up menu that appears, your browser will offer you to save the file in your directory. Make sure that you save it with the exactly same name. After saving the file you can open it with Mathematica 5.2. You will find Mathematica in
Start -> All Programs -> Math Applications -> Mathematica.
Open Mathematica first, then open 20120621.nb from Mathematica. You can execute the entire file by the following manu sequence (in Mathematica):
Kernel -> Evaluation -> Evaluate Notebook.
If you have problems running files that I post please let me know. If you spend some time learning how to use Mathematica you will enhance the understanding of math that we are studying.
Second about Problem 25 in Section 16.3. It is an interesting problem.
It is hugely disappointing that the book does not mention at all that in this problem we are dealing with a pyramid with a triangular base. The volume of any pyramid can be calculated as one-third of the volume of the prism with the same base and the same hight. That is $V = \dfrac{bh}{3}$, where $b$ is the area of the base, $h$ is the height and $V$ is the volume of the pyramid, see Wikipedia's page Pyramid.
The formula $V = \dfrac{bh}{3}$ can be proved using integral calculus of functions with one variable. The fraction $1/3$ originates from the integral $\displaystyle \int_0^1 x^2 dx = \dfrac{1}{3}$.
Without calculus one can make paper models of three pyramids that put together make a cube. This is the simplest case for the above formula. To make this task easier I provide blueprints for the model. An illustration how model works is given in the animation below.
Here is the model: Print two copies of this
pdf file
on heavy yellow paper.
Print this
pdf file on a transparency. Follow the instructions provided.
If you make the model above you will have a hands-on demonstration of the following animation: