****, sorry guys. It was clear to me in my head so I guess I just assumed it would be clear to you. I really thought it was obvious but now it is quite obvious that it was not obvious xD

How did they do this? (this is a practice problem, i just wanted to see the solution is why i put in "1").

Its pretty straightforward, one word hint is Jacobian.

what about without jacobian? my book didnt tell us how to do this part... i know to make x = rcos(theta) and y = rsin(theta) and dA = r dr dTheta

Yeah, you know what x and y are in terms of r and theta. So you can just write that in for the 3x. What dxdy is in terms of dr and dtheta is given by the Jacobian determinant (and why I gave that hint), which you also got. So make those substitutions. Then we can talk about what happens with the limits of integration.

So I'm trying it for this problem:



So I plotted the points it gave me:
(well, I plotted the limits of integration to see what's going on in the question)


So it wants the volume of the shape made when this triangle is the base and the height is 2x (or, the height is 2*rcos(theta)). So the shape is like a pyramid I guess, or at least a triangle base with a height given to it.

I know my limits for theta, it needs to go from pi/4 to pi/2
edit: (I can tell this b/c if I go from 45 deg to 90 deg it will cover the area of the triangle. Obv, the problem is it will cover more than teh area of the triangle which is why i need the limits for r)

I dont know my limits for r. I would say that my limits for r are 0 to sqrt(x^2+y^2) but when I turn that Into polar coordinates all I get is r = r which isnt very helpful... (i get this by using the fact that sin^2 t + cos^2 t = 1)

You know that y = r sin theta and that the line you're trying to trace is y = 5. Therefore, r = 5/sin theta. Notice that r is a function of theta, so you're going to need to do your theta integral second.

makes perfect sense. thanks a lot.

Got an A in my calc 3 class. Thanks a lot for the help guys.



At 4:20 he talks about experiments for evolution, but I had this idea:

What about doing artificial selection first to go from "standard" -> "artificial" state, then introduce them back into the population and see if it goes from "artificial" -> "standard" again.

For instance, take a bunch of elk, like 2000 from the wild, and put 1k in a sanctuary and 1k in another sanctuary. For the first one, shoot the bulls with the largest antlers at the start of the rut season. Do this until antlers are 1/4th the size in your sanctuary. For the second one, do nothing.

Once they are at 1/4th in sanc 1 as in sanc 2, shoot all the bulls in sanc 2 and replace them with the bulls in sanc 1. Then observe sanc 2 and see if the antlers of the elks return to their original "large" shapes (as they were before you started the experiment). or just leave sanc 1 alone and see if it goes back to sanc 2 lvls i guess.

Has anyone done this experiment before? I would very much like to read about it.

Hello guys,

Quick math question. The poker site I play only gives me HH at the end of a session and they disappear after I close the client. They give me number of hands played and BB/100. I want to track this manually but I am unsure how to properly track BB/100.

For example, if I had three sessions where I had:

100 hands 5bb/100
600 hands 3bb/100
50 hands 10bb/100

How would I find my total bb/100 over all hands?

To get the value of total bb/100, you can:

1. For each line of data, multiply the total number of hands for that line by the number of bb won (for example, for the first line this would be 100 * 5, note that the /100 isn't included)
2. Add up the results from each line
3. Divide the answer from step 2 by the total number of hands played
4. The result of this will give you the bb value for the total bb/100

So, in your example:

100*5 + 600*3 + 50*10 = 500 + 1800 + 500 = 2800
2800 / (100 + 600 + 50) = 2800 / 750 = 3.733

So your total bb/100 for all hands in your example sessions were 3.733 / 100

Thanks a lot. Appreciate it.

Back in high school, different Geometry topics were covered in my algebra, pre-calculus, and calculus books, but I never completed a formal Geometry course, so right now for fun I'm working through a Geometry textbook that I picked up. I've ran across one of those questions where I'm certain I'm right and the book is wrong, so I'm looking for input:

Question:

"On a scale map of a state, each square represents 5 square miles. The water works company needs to connect two purification stations that are located at the points (3, 11) and (14, 26) on the grid. About how many miles apart are these two stations?"

The way I worked it out is this:

Since each square (1 unit by 1 unit on the map) is equivalent to 5 mi^2, then that means that the length of each side of these squares is equal 5^.5 miles or ~ 2.236 miles

The distance in units from point A to point B is ((14-3)^2 + (26-11)^2)^.5 = (11^2 + 15^2)^.5 = (121+225)^.5 = 346^.5 or ~ 18.6 units in length.

So the distance in miles is 18.6 units * 2.236 miles / unit, or ~ 41.6 miles.

The book answer is "around 93 miles", which is equal to 18.6 units * 5 miles / unit, but that can't be right to me, since 5 is the area of each square, not the length of the side of each square.

Right??

Can this question be answered as is? My first thought is that I cant answer this until I know the speed of the transmission line. Should I assume that the speed of the transmission line is c?

It's given, 150 Mbps.

It can send 150 Mb of info from here to edge of galaxy in 1 second? I guess this is what I was asking.

Hint: You're probably setting up your problem wrong if you're concerned about that.

Here are some sticking points I notice in this type of problem:

When measuring disk space, I believe 1 GB is equal to 2^30 bytes, which is larger than 1 billion bytes (1,073,741,824 bytes)

However, when measuring bandwidth, 1 Mbps is precisely 1,000,000 bits per second.

Operating under those assumptions, I've got an answer for the first part of the question, the range where the dog has a higher data rate.

I'll outline my process fairly generically.

1. Calculate the total amount of data stored in the tapes
2. Divide this amount by the data line bandwidth (watching units carefully) to determine how long it would take the data line to transmit an amount of data equal to what is stored in the tapes
3. Using that time, determine how far the dog can travel in that time, any distance less than that means the dog has a higher data rate

So the problem is assuming that it can send info at 150Mb/s to anywhere in the universe, is this correct? This is why I had a problem with the question. Or maybe I just dont understand....

If I told you I could speak at 100 words per minute, would it matter how far away from me an observer was standing for this to be true?

For the record, using my above assumptions about storage and transmission nuances, I get an answer of

How exact of an answer is being sought? Nearest tenth? Hundredth?

Dog: 3x7 GB tapes = 21 GB = 21*8 bits = 168Gb
Speed: 18km/hr / 60 / 60 = 18/60^2 Km/s = 0.005 Km/s
Data needing transferring: 168 Gb = 168,000 Mb
Data transfer rate: 150 Mb/s
Data transfer time: 168,000 Mb / 150 Mb/s = 1120 s
Dog travels in 1120s: 0.005 Km/s * 1120s = 5.6 Km
If distance is > 5.6 Km, the dog will be slower at transferring the data.

this seem right?

edit:
I) Bernie has a higher data rate than a transmission line if the distance is < 5.6 Km.
II) If Bernie's speed is doubled, the distance over which he is faster also doubles (11.2Km).
III) If the tape capacity is doubled, the distance over which he is faster also doubles (11.2Km).
IV) If the data rate of the line doubles, the distance over which he is faster is halved (2.8 Km).

This is my last question for the day =P



My problem with this one is that I don't understand what they are asking. There are many details left out of the problem that I think would be relevant in coming to any given solution.

For instance, does checking A----B take 100ms, or does checking all connections going out of A take 100ms, and if you check A---B connection do you need to check B---A connection? Do you need to check the "no connection" and if so, does that also take 100ms?

I think I would have to go with the answer being 5 p 4 * 100ms = 120*100 ms?

Edit again: Ok, I think I found this exact question solved online, and my assumptions on what they are looking for is not even close to being anywhere near right. The answer is much, much more complicated than I thought, and more complicated than your initial attempt as well, unfortunately.