Monday, February 25, 2008

Post-Work Scribe...

Sorry to all, this scribe is very late due to the fact that I worked till 11 both nights and I just woke up from falling asleep at the computer... o_O

Okay, so Monday's class started for most people with walking into room 66 and getting ready to learn. Mine however started in the library as I was getting my Grad Pictures taken.

So, when I walked into class, a couple minutes had already gone by and Mr. K. was quickly going over the Exercises on Volumes of Revolution from Thursday's class. Now, over the weekend Grey-M did a pretty decent job posting the answers in his Scribe, but happened to make a couple of mistakes.

To make sure you don't make the same ones:

In depth corrections may be seen in the slides from today .

We then returned to the "Oil Leak Question" from Tuesday's class...

If you don't remember, the question was:
Now, to explain it briefly... the function given is in Gallons/Hour and the value we wish to obtain is in Gallons. So, you do the math... (we have to multiply by Hours). Now the amount of hours it takes to spill can also be represented as ∂t (change in time) which, when multiplied by the function, results in a neat looking integral which can then be solved from 0 to 10.

The next question we looked at was the "Density Question".
For this question, first visualize that there is a circular oil slick with a changing density (it is more dense in the center than it is towards the circumference). We are then told the density at any radius is given by a function. Once again looking at units, we see the function is in KG/m^2 and we need a value with units KG. So we need to multiply the function by the area of the slick and the area in terms of r is 2πr•∂r. So multiplying that onto the function once again provides an integral (thanks to the ∂r) and may be evaluated from 0 to 1000 to get the answer.

Part B of this question asks to find the smallest radius that contains 75% of the slick's mass.
So for this one, we take the integral to find mass used in the previous question and use "o to r" as the interval. As well, set it equal to 75% of the total mass (3255). By doing this, we have set the end of the interval as the variable and therefor the first value of "r" from 0 that equals 75% of the mass will be the result.
Once this is set up, integrate by parts with u=1+r^2 to solve it... the rest is grunt work =P

Then we moved on to a new question dealing with density of cars along a length of street.
This is very similar to the first question in terms of how we solve it. It says the function gives value with the units Number of Cars/KM and we need just the Number of Cars. So, we somehow need to multiply by KM. It says that x=KM so the distance from zero (∂x) can be multiplied by the function to once again CREATE AN INTEGRAL!!! (Surprise!)
To find how many cars are along a stretch of street of given length, just plug the length into the equation (x=length) and solve.

I believe that is all for this scribe... don't forget to do some homework!

Next Scribe is.........

ETHAN!!! Because I think he's the only one who hasn't done 7... I may be wrong, but yeah. You're scribe

1 comment:

etimz said...

hey I was sick. Il be in class, next class. :) sorry for the delay and making you do my work for me. If its any help, I really like your scribe.