Tuesday, January 20, 2009

Student notes January 2009

7 comments:

ranand said...

Today in class, we started by establishing the fact that, in integrals, variables are the same, but may have a different meaning depending upon what the variable is. From then on, we started a new topic: Accumulating area functions (Indefinate Integrals). These are integrals that have a starting value, but have an "indefinate" ending value, or a variable as an ending value. So today in class, we went through a group exploration to further our understanding of the indefinate integrals. From this exploration, we discovered that the relationship between the graph of the integral function and te graph of the slopes are the same. We also discovered that the accumulating area function is one order above the orginal function. We will go into more detail about this exploration in tomorrow's class.

Charley Field said...

On Friday, January 23rd, we first completed a handout re-iterating the net value theorem which we read about the night before. This states that the integral of f'(x) is equal to the net change of f(x) on the interval "a" to "b", where "a" is the lower bound of the integral and "b" is the upper bound of the integral. After this, we worked in pairs to practice this theorem and learn a new and useful aspect of it. The integral of the absolute value of f'(x) is equal to the total amount f(x) on the same interval mentioned before.

Unknown said...

Today Mr. Hansen breifly checked over last night's homework, and Cassandra did problem 5 up on the board. Then, Mr. Hansen did a short demonstration on how a certain function on the xy coordinate plane can be "wrapped around" to form a 3 dimensional solid. We then proceeded to explore how to determine the area of disks/cylinders using Riemann Sums and then definite integrals.

Unknown said...

Today, Thursday Feb 5th, we took a 20 minute quiz.
After which we derived the volume formulas for a cone and a circle. The trick for the cone formula is to put everything in terms of R=radius and H=height. Most of us were able to figure out that a linear function with a positive slope can be rotated about the x-axis to make a cone. We can use cylindrical approximation by setting our r value to = r/h in the function V=pi r^2 h. When we took the integral of this function we ended up with (pi r^2 h)/3.

-alex

Insomnia EDM said...

Today in class, we got our quizzes back and started working on the AP Question "Accumulators", which used indefinite integrals to calculate the accumulation of sand on a beach over time. Key to this question was the fact that if one does not use their calculator, the question will take hours, literally (With a calc, it takes approx 15 min).

The first part of the question was a warm-up/review : it involved evaluating integrals. Next, we had to create a function that incorporated both the removal and addition function. The third part was evaluating the amount of sand removed at a certain point.

The fourth part was more tricky, however. This involved finding where the slope of the function was 0 and switches from negative to positive. We then took this x value and plugged it back into the original equation. Finding the x-value was easiest, we found, by using the graph screen on the interval from 0 to 6.

Remember, when calculators must be used, they must be used.

Sun said...

On Thurs. Februar 5, 2009, we took a 20 minute quiz on integral then we had a fire drill before we were able to begin working.

crouse said...

On Monday, February 9, we talked more about how to find the volume of a solid of revolution. This time, the region that we were revolving around the axis was bounded by two different functions. We found that the volume could be expressed as the integral of [π(greater function)^2-π(lesser function)^2 ]dx.