Circular oil slicks.........

The class started off with our daily quiz that will prepare us for the ap exam later on. The quiz for today consisted of 4 questions:

When both degrees on the numerator and denominator are the same we can just take the leading coefficients and divide them. Doing it the long way would not be recommended since we only have 2 minutes to do each multiple choice question.



This is a classic ap exam question on the multiple choice. Mr. K assured us it'd be on there sooo everyone make a mental note. The way it was written is just to throw us off. All it is asking for is the derivative of cos(pi/2), which is -1.

This question was made easier for us because it was already is in its factored form. So by knowing the roots of the derivative we know the critical points. Using the 1st derivative test we see that a number less than 1 will give us a negative result and a number greater than 1 will give us a positive. Therefore, by the first derivative test there is a min at 1.



I thought that this was the hardest question in the quiz. It helps a lot to just have a feel of what the graph of it would look like. This is an accumulation function so the derivative is always the underlying function. Plugging in 0 into x we get that the derivative does equal 5 at 0. By doing the 2nd derivative test we see that the result will always be negative so therefore it is concave down.

After the quiz, we proceeded to where we left off last class. By drawing a diagram we see it is like a cylindrical shell kind of question. The question is what is the mass of the oil slick. In the question, they give us density per metre squared so we'll have to multiply an area function. The area function we'll be using will be a circle since they said it was a circular oil slick. So if we err unrolled a portion of the washer we'd end up with a trapezoid. By moving one of the triangles at the side we can make a rectangle. The area of a rectangle can be found by l x w. And there is our area function.



The question wants 75 percent of the mass which is 3255. So now our integral isnt going from 0 to 1000 but to some value r since we want the smallest possible. This will give us an accumulation function that we'll equal to 3255. To find r we can use our calculator to find intersections or bring everything to one side and find roots.

Yes....another scribe post finished. "virtual memory is low" Good thing I finished before my computer died. Anywho HOMEWORK is 8.5 1-5. Next scribe will be ughh the person who said he wanted to beat up my clone...yes you Justus.