Get a helping hand
American River College has an extensive array of support services that you should not hesitate to take advantage of. In addition to the direct support you get from your instructor during class time and office hours, ARC offers drop-in tutoring in the Learning Resource Center, academic counseling in the administration building's Counseling Center, and a broad collection of services in the Student Services building. Most of these support services can be quickly found on the college's webpage titled ARC Support Services. There are currently 48 separate entries on that site.
Some of the programs, like the aformentioned Learning Resource Center, have their own website. The hours for drop-in tutoring are posted there. It is also the location for an open computer lab, the Reading Center, and the Writing Across the Curriculum program.
Check out ARC's student services and learn about them now rather than later.
Monday, January 21, 2013
Saturday, August 25, 2012
Math Student Survival Guide
Although there is no magical secret for succeeding in a math class, there are some things that are universally true. Here's one:
What does “enough time” mean? There is no one-size-fits-all answer. Some students breeze through math with a minimum of effort and others have to struggle. If you are taking prealgebra or algebra in college, you have probably struggled with math on occasion. If you plan to succeed, you must plan to find the necessary time for reading math, studying math, and doing your math homework. One important tip:
There will be days when you just don't have a lot of time. This happens. But do some math anyway, even if it's just 15 minutes of review or solving a homework problem. Doing some math every day is a recipe for success and maintains forward progress. “Saving up” your math for weekends is almost always a mistake.
Here's another important truth:
It's amazing how many students quietly fail their classes instead of getting help. Community colleges are all about helping students. Even when state funding is tight and budgets are cut, we scramble to provide assistance opportunities to our students. Here's a list:
Be an active learner and take charge of your learning. The teacher can't teach you without your consent and your active participation. Take responsibility for your education.
You will not pass your math class
unless you spend enough time on it.
unless you spend enough time on it.
What does “enough time” mean? There is no one-size-fits-all answer. Some students breeze through math with a minimum of effort and others have to struggle. If you are taking prealgebra or algebra in college, you have probably struggled with math on occasion. If you plan to succeed, you must plan to find the necessary time for reading math, studying math, and doing your math homework. One important tip:
Do some math every day.
There will be days when you just don't have a lot of time. This happens. But do some math anyway, even if it's just 15 minutes of review or solving a homework problem. Doing some math every day is a recipe for success and maintains forward progress. “Saving up” your math for weekends is almost always a mistake.
Here's another important truth:
Students who get stuck need to get help.
It's amazing how many students quietly fail their classes instead of getting help. Community colleges are all about helping students. Even when state funding is tight and budgets are cut, we scramble to provide assistance opportunities to our students. Here's a list:
- Questions during class (ask them!)
- Questions during office hours (know them!)
- Tutorial sessions (attend them!)
- Learning Resource Center (go there!)
- Student Support Services (there's an entire building!)
- Study groups (team up with classmates!)
Be an active learner and take charge of your learning. The teacher can't teach you without your consent and your active participation. Take responsibility for your education.
Course syllabi
If you're looking for our course syllabus, it's available in pdf form from the Box.net widget in the left-hand sidebar. Just click on the one you want and you will be able to download it.
In the sidebar you'll also find some potentially useful links that you may want to try out. The Internet has many kinds of math aids and tutorials, including videos, that might be helpful. For example, click here to see a list of Khan Academy's algebra videos and here for videos on calculus.
Note: The whole point of Quiz 0, which you receive on the first day of class, is to give you some points for showing you know how to follow directions.
In the sidebar you'll also find some potentially useful links that you may want to try out. The Internet has many kinds of math aids and tutorials, including videos, that might be helpful. For example, click here to see a list of Khan Academy's algebra videos and here for videos on calculus.
Note: The whole point of Quiz 0, which you receive on the first day of class, is to give you some points for showing you know how to follow directions.
Monday, May 07, 2012
Wednesday, April 18, 2012
Tuesday, April 03, 2012
Tuesday, March 27, 2012
Tuesday, February 28, 2012
Monday, February 20, 2012
Monday, November 22, 2010
The intersecting sphere and cylinder
The bonus task I gave you last week involved looking at the curve created by the intersection of a sphere and cylinder. For purposes of this demonstration, I will assume that the sphere has radius 4 and the cylinder correspondingly has diameter 4. Thus the sphere is x2 + y2 + z2 = 16 and the cylinder is (x − 2)2 + y2 = 4.
We will now show that the self-intersection curve crosses itself at right angles at the point (4, 0, 0).
Here is one way to do it:
We parameterize the path by r = xi + yj + zk, where we need to find appropriate functions for x, y, and z. Let's rewrite everything in cylindrical coordinates, because that form is more useful in this situation. The sphere's equation is r2 + z2 = 16, the cylinder is r = 4 cos θ, and the intersection curve we are looking for is r = (r cos θ)i + (r sin θ)j + zk.
Since the cylinder requires that r = 4 cos θ, we can substitute for r in the expression for the curve r. We obtain r = (4 cos θ cos θ)i + (4 cos θ sin θ)j + zk.
What can we do about that z? According to the sphere, we must have z = (16 − r2)1/2. Substituting r = 4 cos θ into the equation for z, we get z = (16 − (4 cos θ)2)1/2 = (16 − 16 cos2 θ)1/2 = 4 sin θ. We can now write our curve with θ as our parameter:
r(θ) = (4 cos θ cos θ)i + (4 cos θ sin θ)j + (4 sin θ)k.
Observe that r(0) = 4i and r(Ï€) = 4i, indicating that the curve intersects itself for θ = 0 and θ = Ï€. If we find the tangent vectors at these two points, the angle between them is supposed to be a right angle. Of course, the easiest way to find tangent vectors is to use r′(θ), the velocity vector:
r′(θ) = (−8 cos θ sin θ)i + 4( cos2 θ − sin2 θ)j + (4 cos θ)k.
Plug in θ = 0 and θ = π to find the pertinent tangent vectors:
r′(0) = 4j + 4k
r′(Ï€) = 4j − 4k.
If we dot these vectors we get zero. The vectors are perpendicular, so the intersection curve crosses itself at ninety degrees. QED.
We will now show that the self-intersection curve crosses itself at right angles at the point (4, 0, 0).
Here is one way to do it:
We parameterize the path by r = xi + yj + zk, where we need to find appropriate functions for x, y, and z. Let's rewrite everything in cylindrical coordinates, because that form is more useful in this situation. The sphere's equation is r2 + z2 = 16, the cylinder is r = 4 cos θ, and the intersection curve we are looking for is r = (r cos θ)i + (r sin θ)j + zk.Since the cylinder requires that r = 4 cos θ, we can substitute for r in the expression for the curve r. We obtain r = (4 cos θ cos θ)i + (4 cos θ sin θ)j + zk.
What can we do about that z? According to the sphere, we must have z = (16 − r2)1/2. Substituting r = 4 cos θ into the equation for z, we get z = (16 − (4 cos θ)2)1/2 = (16 − 16 cos2 θ)1/2 = 4 sin θ. We can now write our curve with θ as our parameter:
r(θ) = (4 cos θ cos θ)i + (4 cos θ sin θ)j + (4 sin θ)k.
Observe that r(0) = 4i and r(Ï€) = 4i, indicating that the curve intersects itself for θ = 0 and θ = Ï€. If we find the tangent vectors at these two points, the angle between them is supposed to be a right angle. Of course, the easiest way to find tangent vectors is to use r′(θ), the velocity vector:
r′(θ) = (−8 cos θ sin θ)i + 4( cos2 θ − sin2 θ)j + (4 cos θ)k.
Plug in θ = 0 and θ = π to find the pertinent tangent vectors:
r′(0) = 4j + 4k
r′(Ï€) = 4j − 4k.
If we dot these vectors we get zero. The vectors are perpendicular, so the intersection curve crosses itself at ninety degrees. QED.
Sunday, August 22, 2010
Computing a weighted average
It's how your grades are computed
It's not too difficult to turn the points you earn on an assignment into a percentage. You just divide them by the total number of points possible and move the decimal point two places. For example, if you get 15 points on a 20-point quiz, your percentage score is 15/20 = 0.75 = 75%. No problem.
However, I do not use just a point system to figure grades for a class. I use a weighted average of the scores you earn in three categories: HWQ (homework & quizzes, lumped together), Exams (the chapter tests), and Final (the comprehensive final exam at the end of the semester). For example suppose that HWQ is 15% of your grade, Exams are 70% of your grade, and the Final is 15%. (If you check, you'll see that the percentages add up to 100%, as they should. I may vary these weights depending on the number of exams I give.)
Suppose that you earn 72% of all the points possible on the homework and quizzes, you average 84% on the exams, and you get 79% on the final. Did you earn a B? Let's find out. The weighted average is computed by multiplying each score by its percentage weight and then adding everything together. Recall that the percentage weight of HWQ is 15% of your grade, which we can write as the decimal 0.15. For Exams we have 70% or 0.70, and the Final is 15% or 0.15, just like HWQ. The formula for grades is
Using the scores we pretended you got during the semester, your course grade would be given by
Congratulations! You earned a B for the class, with just a little bit to spare.
It's not too surprising. Most of the weight of the grade is on Exams. Homework and quizzes get only 15% because I treat them more as a learning experience. The final gets only 15% because it's only one exam. While other instructors might choose to give the final more weight, I prefer to limit the importance of any single exam.
In general
If you ever want to compute a weighted average for any reason, the main thing to keep in mind that the weights have to add up to 100%. If, for example, you wish to compute the weighted average of four numbers, X1, X2, X3, and X4, you can do it as long as the weights, w1, w2, w3, and w4, add up to 100%. Then you compute the sum of their respective products:
That's all there is to it. Of course, if you choose to use equal weights, so that w1 = w2 = w3 = w4 = 25% = 0.25, your result will just be the usual arithmetic average. (It would be simpler to add up the numbers and divide by 4.)
I'll leave it to you to figure out what to do if there are more than four numbers for which you wish to compute a weighted average.
It's not too difficult to turn the points you earn on an assignment into a percentage. You just divide them by the total number of points possible and move the decimal point two places. For example, if you get 15 points on a 20-point quiz, your percentage score is 15/20 = 0.75 = 75%. No problem.
However, I do not use just a point system to figure grades for a class. I use a weighted average of the scores you earn in three categories: HWQ (homework & quizzes, lumped together), Exams (the chapter tests), and Final (the comprehensive final exam at the end of the semester). For example suppose that HWQ is 15% of your grade, Exams are 70% of your grade, and the Final is 15%. (If you check, you'll see that the percentages add up to 100%, as they should. I may vary these weights depending on the number of exams I give.)
Suppose that you earn 72% of all the points possible on the homework and quizzes, you average 84% on the exams, and you get 79% on the final. Did you earn a B? Let's find out. The weighted average is computed by multiplying each score by its percentage weight and then adding everything together. Recall that the percentage weight of HWQ is 15% of your grade, which we can write as the decimal 0.15. For Exams we have 70% or 0.70, and the Final is 15% or 0.15, just like HWQ. The formula for grades is
Course Grade = HWQ*0.15 + Exams*0.70 + Final*0.15.
Using the scores we pretended you got during the semester, your course grade would be given by
Course Grade = 72%(0.15) + 84%(0.70) + 79%(0.15) = 81.45%.
Congratulations! You earned a B for the class, with just a little bit to spare.
It's not too surprising. Most of the weight of the grade is on Exams. Homework and quizzes get only 15% because I treat them more as a learning experience. The final gets only 15% because it's only one exam. While other instructors might choose to give the final more weight, I prefer to limit the importance of any single exam.
In general
If you ever want to compute a weighted average for any reason, the main thing to keep in mind that the weights have to add up to 100%. If, for example, you wish to compute the weighted average of four numbers, X1, X2, X3, and X4, you can do it as long as the weights, w1, w2, w3, and w4, add up to 100%. Then you compute the sum of their respective products:
Weighted average = X1w1 + X2w2 +X3w3 +X4w4.
That's all there is to it. Of course, if you choose to use equal weights, so that w1 = w2 = w3 = w4 = 25% = 0.25, your result will just be the usual arithmetic average. (It would be simpler to add up the numbers and divide by 4.)
I'll leave it to you to figure out what to do if there are more than four numbers for which you wish to compute a weighted average.
Saturday, May 08, 2010
Subscribe to:
Posts (Atom)

















































