C can take any constant value, C is the same thing as C/5

Masque described the classical calculus solution to this where you integrate by parts twice, and the second time you will get an integral that is the same as the original integral, so you will be able to say

integral = uv - [UV - c*integral]

where the uv and the UV come from the 2 different integration by parts, and c is some constant (you will find it's -2 here). So you just solve for integral as

integral = (uv - UV)/(1-c) + C

Watch out for signs here as there is a lot of subtracting negatives.


But another way is to remember the trick I taught you about differentiating instead of integrating. If we differentiate sin(2t)e^t we get

sin(2t)e^t + 2cos(2t)*e^t.

That's what we want except for the second term. But if we differentiate cos(2t)e^t we get

cos(2t)e^t - 2sin(2t)e^t.

Now it isn't immediately apparent how to cancel the cos(2t)e^t and still end up with just sin(2t)e^t, but we can do it by multiplying the first expression by some constant a and adding that to the the second expression times some constant b, such that

a - 2b = 1
2a + b = 0

where the first expression gives us our sine term, and the second cancels the cosine term. Solving this gives

a = 1/5, b = -2/5

so our solution is

1/5*sin(2t)e^t - 2/5*cos(2t)e^t + C.

You don't have to do it that way, but it's an alternative.


As a preview of coming attractions, when you learn about complex numbers, you will learn that a sine is really an exponential, so integrating this is no harder than integrating an exponential.

cos(2t) = [exp(2jt) + exp(-2jt)]/2

sin(2t) = [e^(2jt) - e^(-2jt)]/(2j)

sin(2t)e^t = [e^(1+2j) - e^(1-2j)]/(2j).

There's no telling how long it will take the math department to get around to teaching you enough to do it this way, if ever, but as a EE this is basically the first day of kindergarten. Exponentials times sines are super important.

Plus with complex analysis they will find out a ton of other integrals (possibly unable to do with their methods so far) sums and products that can be done with Residue theory etc that they will simply see complex numbers as an amazing gift to mankind lol.

http://en.wikipedia.org/wiki/Residue_theory

Right, the ones that are now the biggest pain in the ass, with complicated trig substitutions and so forth, become things you can practically do in you head. But for that you probably will need a math class in complex analysis (or learn it on your own from the wiki page).

Lost my t:

sin(2t)e^t = [e^((1+2j)t) - e^((1-2j)t)]/(2j)

I have a rectangle of H = 2, L = 10, call it A.

I have a circle of R = 100, call it Cb

I have a smaller circle of r = 20, at the center of the bigger circle, call it Cs

Given that I know the rectangle is within the big circle, what is the probability it is touching the little circle? Rectangle has same chances of being anywhere within the big circle, in addition to being in any orientation.

Here's my attempt, though I'm confused on this one for sure so I'm not confident in my answer.

Basically I drew a picture and placed the rectangle A on top of Cs, and argued as long as center of rectangle was within H/2 of top point of Cs, then it would be "inside. So I put a point there [0,21]. Using same logic for the bottom, I put a point at [0,-21]. I used the same logic for the sides and put points at [-25,0] and [25,0]. Then I drew an ellipse connecting those 4 points (I'm not sure if this was correct, but I wasn't sure how else to determine my rectangle contact area).

Area of ellipse = 21*25*pi = 525pi
Area of Cb = 100*100*pi = 10000pi

Probability = 525/10000 = 0.0525

The multiple orientations thing confused me, my basis for ignoring it was: if rectangle can take infinite orientations all with equal probability, and these orientations would all result in me drawing an ellipse of the same size around Cs, then I only needed to calculate it for one orientation, because, say i did it for 2 orientations:

P(hit) = P(orientation 1) * 0.0525 + P(orientation 2) * 0. 0525 = 0.0525. I guess I am arguing for using law of total probability here.

Spitballing: find the area of the region that could contain the upper left corner, assuming the rectangle is placed horizontally. Compute the fraction of that are that results in rectangle overlapping circle. Good enough i think because every orientation rotates to one of these positions.

Can you re-word this, having hard time understanding what you are suggesting. I drew a figure using AutoCAD of theproblem, and it looks like my ellipse idea isn't correct.

Edit: I think I was trying to do what you were saying, but I'm having problems finding the shape of the area that contains a corner or midpoint of the rectangle

Make that c = -4.

First: Thanks Bruce/Wyman/Masque for the integration lessons during my break; we just got to integration by parts in class and, compared to my comrades, I am an integration jock.

Physics:

Assuming no air resistance: Ball is thrown straight up into the air and returns to the ground after 5 seconds. How high did the ball go?

I'm getting two different answers using two equations that I thought would give me the same answer.

We know that if it took 2.5 seconds to go from Velocity_initial to Velocity=0 that the balls initial velocity is 24.5m/s. So using the following equation where V_i=24.5, a=-9.8 and delta t=2.5 seconds I get 29.375m.



If I use the next equation where V_i=24.5 V_f=0 and delta t=2.5 I get 30.625m



As it turns out, 30.625m is what the trailing term evaluates to in our first equation (60 being the first term), so if the first term was 0 then I would get the negative version of the answer I get in the second equation.

The obvious first question is where did I make a mistake, the second question is, if I didn't make a mistake, are these equations for two different things and I am simply confusing them for an equation for the same thing?

I calculated wrong. Instead of entering in V_i=24.5, I entered V_i=24.

Anyone got any hints or ideas on my circle rectangle problem? Kinda stumped on it. Will check out office hours on Monday otherwise.

It's aggravating because I'm pretty sure I understand the concept and their trick w.r.t. to "infinite orientations" and using total probability, but the geometry of the problem is throwing me for a loop.

Fix your frame of reference to be the rectangle.

Still lost. The problem isn't as simple as just adding 5 to the radius of the small circle is it? (in the case where rectangle is tilted at angle theta measured from the circle). The hint says to find conditional probability for one rectangle orientation first, so I guess that's why I was focused on the area a horizontal rectangle touching the circle would make

Consider a rectangle E with center O, the origin in the plane, and the sides of E parallel to the x/y axes.

1. What's the set of points X in the plane are there so that a circle with center X and radius R cover E?




2. What's the set of points X so that a circle with center X and radius r (little r) touch E?

Sorry, I didn't check my work because I don't have a pen and paper handy (to you know, draw a picture and connect a bunch of lines and write down lengths), that's also why I didn't arrive at a final answer (because I'd have to do algebra)

Ideas I feel are correct though

How about if you just consider taking the rectangle and rotating it around the small circle so that the midpoint of a short side is tangent to the circle. Then you can do that for smaller and smaller circles with radius 0 to 20, and the rectangle will always touch the small circle. But if you do that for circles of larger radii than 20, it will not touch, so increase the radius until it touches the outer circle. Then just take 20 divided by the the total radius from 0 until it touches the outer circle which should be sqrt(100^2 - 1) - 10. That would give an answer of 22.2%. I'd simulate that with Monte Carlo to be sure.

Consider the center of the rectangle call it M the 4 vertices are ABCD. The center has coordinates Xm,Ym maybe call them in polar coordinates Rm,fm.

Ask yourself what is the probability a rectangle with center M is inside the big circle. Obviously the possible vector MA can be rotated 2*Pi to cover all orientations. Some angles will force vertices out for some M. Some other M will never have such problem and all orientations are inside etc. (some M will have vertices outside and no possible orientation to avoid this and have 0 chance eg if M far enough from center of the small circle but outside a minimum radius that can be easily calculated etc).

What you probably need to do is find the function that gives you the probability a rectangle with center M is inside the big circle (find that rotation angle range /2pi ratio.). Then for each one that is inside find the probability its not touching the small circle (another ratio smaller of course in general or equal to the above).

So every point that is a conceivable center of the rectangle has associated with it 2 probabilities. The chance it belongs to a rectangle that is inside the big one and the chance that its outside the small one. So each point in the plane has a probability to belong to a rectangle (as its center) that is inside the big and outside the small circle call it P(A and B) and a probability P(B) to be inside the big circle regardless what it does with A. The conditional probability is;

P(A|B)=P(A and B)/P(B).


So for all set of points that satisfy B find P(A|B). Integrate over all of those points and find the avg conditional probability that way.


So for each point M (xm,ym) in the plane you have a conditional probability P(A,B) and you can find it by P(AandB) and P(B) as suggested.

Now all you need to do is express those as function of the coordinates to create an effective density function so to speak in x,y or r,f etc. I suppose the hard way is to find the 4 coordinates of ABCD as function of the coordinates of M and the rotation angle F (basically find for what F range in 0 to 2Pi your conditions are satisfied for all 4 vertices and that will give you the above probabilities for each point).

Study also the figures, there may be some easy geometrical argument with additional circles to find those ratios more easily.

Due to spherical symmetry of course you expect those density functions to be functions of r strictly for any point M (r,f). Basically for some small rmin and less its 0 then it starts growing and it goes to 1 stays to 1 for a while and then stars dropping again and goes to 0 at some rmax as your r grows from 0 to infinity. You need to express those as functions of r,a,b where a and b are the side lengths of the rectangle.

The problem with my solution is that if I use the long side of the rectangle, I get a different answer 1.8% lower. Still may be worth considering though.

Actually there is a similar problem where you actually can get different answers. That's the one where compute the probability that a random chord is longer than a side of the inscribed equilateral triangle. It comes down to what "pick a random chord" means.

Probability has to be withing 0.04 and 0.0625 right?

We know that it's inside the big circle. I'm pretty sure this is just introduced to give a bound on the problem (center of rectangle is located within a region of A = pi*100^2/4 Probability that center of rectangle lies within small circle is

20^2/100^2 = 0.04

Now imagine case where rectangle is rotating around the edge of the circle such that the midpoint of the rectangle is 5 units away from the circle edge, but it is pointing towards the center of the circle such that the border of the circle and border of the rectangle touch (call that a hit). So now we have the maximum possible hit area, of

25^2/100^2 = 0.0625

But the center of the rectangle being inside the big circle isn't enough to ensure that the rectangle is entirely inside the big circle. Doesn't it have to be entirely inside?

I see the problem with mine is that while the midpoint of a side must always lie on a circle of some radius, it doesn't have to be tangent at that point, so I'm not covering all the orientations, just ones with 2 sides parallel to a radius. That is one orientation though, like your hint suggests.

I think the answer is near 91% from simple approximation arguments along the lines of thoughts i described above eg if r>30.2 or r<89.8 all those points are 100% both inside the big and outside the small circle so conditional is 100% then from 89.8 to a bit less than 99 the chance rectangle is inside the big circle is less than 1, eventually going to 0 near 99 but the conditional is still 100%. Then of course you have additional small corrections as you go below 30.2 where the conditional probability i described is less than 1 for the small r but remains 1 for the large r, but those areas are small so the avg is probably near 91% (see next). (the faction of area between the 2 critical radii i described, although the real critical ones are a bit lower and a bit higher respectively lets call them primary critical vs secondary critical eg around 21+ where the conditional drops to 0 and 99- where the conditional remains 1 but the chance its inside drops below 1 ).

So i guess the conditional density grows from 0 starting a bit near 21+ to 1 at about 30.2 keeping 1 all the way to 99- (a bit below 99).

So i suppose we can roughly call it (its not really 99 a bit less) ~(99^2-30.2^2)/99^2 or about 91%

No, it is given that the center of the rectangle is inside the big circle, and that's all that matters for the big circle.

The big circle has nothing to do with the problem other than to give bounds. Another way of wording it would be that the midpoint of a rectangle can land anywhere within a R=100 circle. What is the probability that it overlaps a smaller circle of R=20 centered inside the bigger circle

Sorry for the confusion

I took that to mean entirely inside, not just the center inside.