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Now I have never heard of these steradians before, but after a little research it appears that:
sr = A/4pi
And more importantly: 4pi sr = 129600/pi degrees
So in our case 1 sr = 32400/pi^2 degrees ≈3282.8
These are the relationships from the experts, although I am still having difficulty grasping the jump from regular units to square units.
The reason that the number is so large is because steradians are literally square radians.
So they say that a sphere contains 4pi sr (square radians) = 4pi rad^2
Since each rad = 180/pi degrees:
4pi(180/pi)^2
Hence, 32400/pi^2 degrees per sr, and our answer of about 3282.8
The only logic that I can grasp at here is that this number represents the interior angle of the apex at the center of the sphere multiplied by the degrees of revolution.
If my logic holds then the angle at the apex would be 3282.8/360 = 9.1189..revolved 360.
I'll see if this makes sense in a while.
I don't know if a surface of revolution will help in this problem as it calculates a volume. Anyway it works like this.
Make a function that models the outer edge of an object that is three dimentional for a cartesian coordinate system. After you have the function it is simply a matter of integration.
If y=f(x) the Volume would be ∫2pixf(x) dx] evaluated from 0 to the radius. (for a solid object) Perhaps an example is in order.
Take a cone. If you look at the right half of it the side will match a decreasing line with the y intercept being the apex.
The equation for that line will be y = H - (H/R)x, this is true for all cones. This line intercepts the x axis at x = R.
Now multiply your line function by (2 pi x) and then integrate from zero to R. You will get:
2pix(H - (H/R)x) = 2piH ∫x - x^2 / R dx (remember to factor out constants?)
2piH [ x^2 / 2 - x^3 / 3R ] from 0 to R
2piH (R^2 / 2 - R^3 / 3R)
2piH ( 3R^2 - 2R^2 ) / 6
This all equals piHR^2 / 3
This is the common formula for the area of a cone. This is the same way that the volume of a sphere was calculated using the curve of one quadrant and multiplying the integration by two to add the bottom half. You can use this method for any object with a little imagination.
Note that we used y = f(x) because we were rotating around the y axis. If you rotate around the x axis you will need a function like x = f(y).
You can also calculate surface area if at the end of your computation you divide by the thickness or radius of the object in question. If you want a volume of an object that is not solid you simply subtract the integration of the inner boundary from that of the outer boundary.
I hope this helps and would love if you post any uncommon volumes that you figure out in the future.
edit: I am sure that you understand that an integration can be interpreted as the area below the curve or line. What we have done above is that we have integrated that area for every value of the radius also. That is why we multiply the function by 2pix (circumference formula) or 360 degrees.
Hi, Thinkdesigns. Calculus is basically just the study of rates of change. If you always keep that basic definition in the back of your head you should never become overwhelmed by the breadth of the subject.
Anyway, you asked for some simple questions related to maximums and minimums.
1) If y = 4x - x^2 then find the maximum value of y and for what value of x does this value occur?
2) If y = x^2 - 6x + 10, determine if there is a local extrema and at what point it occurs.
3) If y = 10x + 1, determine if there is a local extrema and at what point it occurs.
Those are a couple of the most basic types of minimum/maximum problems. If you want tougher examples or want a question answered go to the "Help Me!" thread.
Sorry about that. I had [3a(3a^2)^(1/2)] / 4 before I blew in on the last step of simplification which equals the same thing that you posted. darn!
Thanks a lot ryos and justlookingforthemoment. I truly appreciate the time you spent explaining to an old man that new fangled computer stuff....really thanks!
Poor Danster, not only do you have to create seperate functions for each axis but you also have to have a variable within all of those functions which changes them depending on the location of the source of the wave.
To make matters even worse the data suggest that you have to model a dampening force within these functions. This data will never correspond to the general wave equation that we have been talking about. My poor fellow you are going to have to do a lot of careful studying of the interaction between the forces interacting here. I wish you all of the best and I tip my hat to you for giving it a go.
Good luck and drop a line if you can think of anything else. I have an inkling that you will be working on this for a while unless you are extremely gifted in physics.
I believe that this is the answer. If we model the roofline as a curve and then treat the roof as a surface of revolution, I believe that it produces the answer needed.
Using points where x = -4.25, 0, 4.25 we get the curve:
y = -32x^2/61 + 1092x/1037 + 5
y = (-544x^2 + 1092x + 5185) / 1037
Sorry for the large constants, but it was specified an exact answer was needed.
V = ∫2 pi x f(x) dx evaluated from zero to r
V = (pi / 1037) (-272r^4 + 728r^3 + 5185r^2)
I spared you the simplification process, but I am sure that you all know how to do this.
V/thickness = A , in our case V/r = A
A = (pi / 1037) (-272r^3 + 728r^2 + 5185r)
Since our radius is 4.25 our area is 1683pi / 122 (exact)
Good catch John, I missed the 2x^2 term equaling the r^2 and not the radius. Do I get partial credit? It should have read:
2x^2 = 16 + 8 + 8^(1/2)
x^2 = 12 + 4*8^(1/2)
x^2 = 3 + 8^(1/2)
Since x^2 = Area, this is your answer.
Sorry for any inconvenience joojoo.
I have the answer to the square question too, but I can't figure out how to use the bleeping hide tag thing. Old guys like me have issues with computers you know?
Could someone please tell me step by step how to use the hide tag, the description in this thread didn't get through my thick skull. I tried what I thought it said to no avail.
Using ESP, I posted the answer to this today in HELP ME!>>calculus help: maximum/minimum problem. Well for a circle with a radius of four anyway. My ESP isn't perfect. I thought that it was funny that you asked the same question as someone else.
I was just browsing for interesting posts in the past and I noticed that this one was never really answered. The question asked how far the boy had walked.
let v be the velocity of the boy
D/v = t1
D/(4v/3) = 3D/4v = t2
delta t = t1 - t2
we know that delta t = 4/3 and it also equals D/v - 3D/4v
4/3 = D/v - 3D/4v
4/3 = D/4v
D = 16v/3 = 48km
Great!!! LOL!
The things that we punch into these keyboards in a frenzy.
Nice explanation mathsyperson, I use Newton's method all of the time. It is very useful for difficult problems. I must be showing my age by calling it Newton's Method.
By the way, I didn't compute an extra digit for nothing, I wanted to be sure that the fourth digit did not change if he was going to keep it as accurate.
I don't think that anyone can tell you precisely how to do it. You must have a decent imagination and a good understanding of geometry. With that said:
The top circle is in the corner of the square, therfore the radius of the circle is one inch away from each side. Using the Pythagorean Theorem (x^2 + y^2 = r^2) tells you that the center of the first circle is 2^(1/2) away from the corner. This relationship will hold in the opposite corner as well.
So the total diagonal distance will be those two distances from the opposite corners plus the entire diameter of the center circle plus the two half circle or radius remaiders from the corner circles.
The diagonal distance would then be: 2(2)^(1/2) + 2 + 2
The sides of the square will also be solved using the Pythagorean Theorem with the diagonal distance being the hypotenuse.
x^2 + y^2 = r^2, since x = y in this case, 2x^2 = r^2
Our radius from earlier was 4 + 2(2)^(1/2) = 4 + 8^(1/2)
2x^2 = 4 + 8^(1/2)
x^2 = (4 + 8^(1/2)) / 2
x = (2 + 2^(1/2))^(1/2)
Since the area of a square is simply a side squared the area is:
A = 2 + 2^(1/2)
That's the exact answer. It is approximately 3.41421356237.................
I figured out number three for you.
I am sorry that I can't draw this for you, but maybe you can draw it yourself and follow along. Any triangle will meet the circle in three places. Using the top vertice as a bisector of the circle you notice that it also bisects the base of the triangle also. Now if you draw a line from the center of the circle to one of the triangle's lower vertices you will see another right triangle appear with the hypotenuse being the radius of the circle and the base being that of half of the triangle's base. The other side will be the amount of distance greater than the radius that the height of the triangle will occupy. I hope that you can see this.
So, h = (r^2 - b^2/4)^(1/2) + 4
A = bh/2
A = b/2 [r + (r^2 - b^2/4)^(1/2)
A = (b^2 r^2/ 4 - b^4/ 16)^(1/2) + br/ 2
Substituting in our value of r = 4 and taking the derivative gives:
dA/db = [(32b - b^3) / 8 (4b^2 - .25b^4)^(1/2)] + 2
This derivative equals 0 when b = 48^(1/2) = 6.9282
h = 6
Maximum area is 3 (48)^(1/2) = 20.78
I hope that helped, I will try and work on the trapazoid later.
Wow John, I never looked at it like that. Good observation. Now I suppose that this type of situation really demands vector notation to be truly correct. Well at least safra's program is working correctly now.
All "whole" numbers have a denominator of 1 not 0.
After only 20 cuts you would be left with a ballot .2900390625mm high by .205078125mm wide!! (That's 1/87 X 1/125 inches) Since this far less than the normal diameter of mechanical pencil led I doubt that anyone could vote with it because even a "dot" would not fit on the ballot.
After 30 cuts it gets even more ridiculous. The piece of paper (?) would be only 41 nanometers wide by 58 nanometers high. That means you could place approximately 33000 of these ballots atop a human hair!!!
I will come late on this one also. Different strokes for different folks. I like to do these types of things algebraically.
7 hours = 420 minutes
special questionaires = .2 q (q=any questionaire)
simple questionaires = .8 q
total time = .8q(10 minutes) + .2q(30 minutes)
420 = .8q(10) + .2q(30)
420 = q(8 + 6)
q = 420/14 = 30
The average concept used by mathsyperson never even occured to me. Go figure.
Your first problem was because of how I structured the equations. By definition point A will have a smaller x value then point B. A simple "if" statement can easily cure that.
The second problem I hadn't thought about at all. There is no slope in such a situation. This would also seem a good place for an "if" statement. Then you can just pick a location like half of delta y or whatever you feel is a good place for that instance.
Anyway, I hope your project is coming along now. Good luck.
I am thinking that you will not be able to use a single equation to give you all of those motions in three-dimentional space. For example, the formula that mathsyperson gave you is generally used to find the position of a particle within a wave. However, this relates to only movement along the y axis for a specific point in time.
My thinking is that all of your movements will correspond somehow to the frequency of the wave in question. I am leaning toward the necessity of creating a different wave function for each individual movement. This is because the amplitude (Or A of the original equation) will be different in magnitude as well as direction for these different yet connected parameters.
Also, just wondering, is the origin of these waves stationary or moving? If the origin moves all of your functions will become much more complicated also. Each function will have to change as the source and direction change.
I hope that you are getting paid for this endevour because it sounds like you will have to make a lot of relationships between a lot of variables to pull this off. Good luck pouring over that data.
I have not been on this site for long, but there may be a place to post such information. If there is I would like to try to model this motion. It seems like a worthy challenge. Sorry I could not help you more, but I would try to make a function that works along each axis first and then try to find a relationship among them. This way it won't be so daunting. In short, I do not believe that a particular phase shift will work to model this movement.
I have been checking this thread every day waiting for the answer. It really is bothering me that I can't figure it out yet. (I am sick, I know.)