Last-minute encouragement

Here’s a soundtrack for your last-minute exam preparation. To be clear: you are Rocky and tomorrow’s exam is every opponent Rocky ever faced (in Rocky I through IV at least; I forget what happens in Rocky V). It might knock you around a little bit, but stay motivated and you’ll come out on top!

Also, I’ll probably forget to say this tomorrow, but it has been a real pleasure teaching this class this semester. So many of you have been on the ball since August. Your consistent effort and positive attitude has made waking up for these early classes so much more pleasant for me! I’m keeping my fingers crossed for all of you tomorrow. Break a leg!

Posted in Uncategorized | Leave a comment

Extra Credit Assigment


As announced in class today, you will have the opportunity to complete an assignment for extra credit (link to PDF above). Assignments that are complete and thoughtful can earn up to an extra 1% for your overall grade. This might not sound like a lot, but it has the potential to boost your grade from an F to a D…or a B+ to an A-, for example.

The assignment shouldn’t take you more than 5 or 10 minutes….don’t think too hard about it; just be honest and write what comes naturally.

Print out the PDF and fill in your responses by hand. Hard copies are due either in my hands or in my mailbox in N711 (the mailbox with my name on it will be on your left when enter the room) by noon next Monday, December 22. (Keep in mind that N711 is the math department office, and so it’s closed outside of usual business hours.) No late assignments will be accepted. I’ll bring some printouts to your exam on Thursday for people who want to complete the assignment then and there.

My idea is that I can scan and share your responses with my MAT 1575 students next semester (I’ll remove your names), so please include anything you think they’ll find helpful. If you don’t want your assignment shared with strangers, just write somewhere on it that you’d prefer it be kept private.

Posted in Uncategorized | Leave a comment

Test #3 Solutions


Posted in Uncategorized | Leave a comment

WebWork and Tuesday’s class

Just a reminder…all WebWork sets have been reopened. They are now all due next Wednesday night, right before your final exam on Thursday morning.

In class on Tuesday, I will show one or two examples computing arclengths or surface areas, but the rest of class time will be devoted to…whatever it is you want to talk about. I expect you’ll come prepared with questions you had trouble with from WebWork, textbook homework, or the final exam review sheet.

Posted in WebWork | Leave a comment

Cylindrical shells – last example from today’s class

As soon as class ended today, I realized that we had forgotten something important when setting up the integral that computed the volume of the solid obtained by rotating the region between y = \sqrt{x} and the x-axis, between x=0 and x=4, about the x-axis using the method of cylindrical shells. I’ll review the set-up here, and add the part I forgot.

Since we’re rotating about a horizontal line (the x-axis) and we’re going to use the method of cylindrical shells, that means we’ll integrate with respect to the vertical coordinate y. So the first part of the set-up is to write:

\int \dots dy.

This means that our integrand has to be given in terms of the variable y. The integrand is the area of the infinitely thin cylindrical shell that you get from rotating a horizontal segment at height y about the x-axis:

\int (area of cylindrical shell) dy.

When you cut open this infinitely thin cylindrical shell, you just get a rectangle whose area is its length times its width. Its length is the circumference of the cylinder; its width is the length of that horizontal segment at height y. The radius of the cylindrical shell is just the height of the horizontal segment at height y, so the circumference of the cylindrical shell is 2 \pi y. We can add this to our integral above:

\int 2 \pi y*(length of horizontal segment) dy.

Now we just have to figure out the length of that horizontal segment at height y. Since it’s horizontal, its length will just be the difference of the $x$-coordinates of its endpoints. The endpoint at the far right will be the same no matter what y is. Here, the right-hand boundary of the region is the line x=4, so the right endpoint of the horizontal segment has x-coordinate 4. The left endpoint of the horizontal segment will depend on what y is, but it will always be a point on the graph of y = \sqrt{x}. So the left endpoint of the horizontal segment at height y is just x. This means that the length of the horizontal segment at height y is 4-x…but our integrand can’t have the variable x in it…we’ve set it up so that we’re integrating with respect to y. Since y = \sqrt{x}, y^2 = x. We just replace the x we see in the formula for the length of the segment with y^2. So the length of the horizontal segment at height y is 4-y^2:

\int 2 \pi y * (4-y^2) dy.

This is the integral we set up in class. It will not compute the volume of the solid because it’s an indefinite integral and not a definite one. We forgot to add limits of integration. Remember that we’re integrating with respect to y, which plays the role of the radius of the cylindrical shell. We know that x goes from 0 to 2, and we know that y = \sqrt{x}, so this means that the radius y goes from \sqrt{0} to \sqrt{4}. This means that the definite integral that computes the volume is:

V = \int_0^2 2 \pi y * (4-y^2) dy.

Here’s what I’d do to evaluate this integral:

\int_0^2 2 \pi y * (4-y^2) dy = 2\pi \int_0^2(4y - y^3)dy = 2 \pi (2y^2-\frac{1}{4}y^4)]_0^2 = 2 \pi((2*2^2 - \frac{1}{4}2^4) - (2*0^2 - \frac{1}{4}0^4))

= 8 \pi.

Therefore the volume is 8\pi.

Posted in Uncategorized | Leave a comment

Last quiz – Thursday, December 11

Just a reminder: this week’s quiz will cover sections 5.1 and/or 6.1 of your text.

Posted in Quizzes | Leave a comment

WebWork due dates

I’ve updated the due dates for your remaining WebWork sets. Don’t worry…they won’t change again! All sets are now due on Wednesday, December 17; the night before your final exam. Please don’t wait until then to complete these sets! If you find any of the exercises particularly tricky, we can talk about them in class or during office hours.

You should be able to complete the sets Setting Up Integrals – Volume and Volumes of Revolution now.

The remaining topics are 6.4: The Method of Cylindrical Shells and 8.4: Length and Surface Area. We’ll go over these in class on Thursday. (The method of cylindrical shells is a second method for computing volumes of revolution. Formulas for computing length and surface area are different, but they operate on the same principles as the computations of areas and of volumes that we’ve seen already.) The corresponding WebWork sets are pretty short, so you’ll want to make sure to practice exercises from your textbook as well.

Posted in WebWork | Leave a comment

Test #3 Grades – Posted

Test #3 grades are now posted in Blackboard’s gradebook. It will still be a few days before solutions are available.

Posted in Uncategorized | Leave a comment

Links for 5.1, 6.3, and 6.4

Here are some helpful links for this Thursday’s and next week’s lectures:

Riemann sums left endpoint, right endpoint, midpoint

Riemann sum calculator

Solids of revolution (disk/washer method and shell method)

Posted in Uncategorized | 3 Comments

Taylor Polynomials and Taylor Series – Links

I’ll show these two graphs in class on Tuesday. (It’s convenient to store the links here; you can click on them before Tuesday’s class if you like.)

f(x) = e^x

f(x) = \ln(1+x)

Posted in Uncategorized | Leave a comment