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I might be missing something obvious, but...
1/2 + 1/3 + 1/4 = 13/12? Doesn't one of the partners in k+78 own a 1/12 imaginary share?
I think that this problem could only be solved if we knew who was holding the fake share.
siva.eas, your answer works if there is a final term in the sequence and that it is indeed 1. But exactly how one would show that a final term exists at infinity is beyond me for this equation.
There is no sum since the series diverges. Conceptually it approaches infinity.
If you divide your equation by x-1 you will get x² - 2x - 15
This means that (x-1)( x² - 2x - 15) = x³ + 3x² +13x - 15
Factoring the second part of the product gives (x-5)(x+3)
So (x-1)(x-5)(x+3) = x³ + 3x² +13x - 15
This is completely factored because there are no powers of x above one.
Integrating a line equation gives the area below the curve (or line). The equation for the first quadrant is:
y = √r²-x²
Integrating this equation from zero to r gives:
x/2 (√r²-x²) + r²/2 (arcsin x/r) = πr²/4
Since this is only the area of one quadrant, you must multiply this by 4 which gives πr²
Probably from the lack of sugar. Easily disassociated compounds like sugar and salt tend to lower the freezing point of water.
You are getting a little confused with the notation. f(x)=x^2 is the same thing as y=x^2. f'(x) is the same as dy/dx=2x. Taking an integral is the exact opposite of taking an integral. An integral is also called an antiderivative. ∫dy = ∫2x dx is the same as y=x^2.
Too many people get hung up on the different notation. In my opinion, you should become more comfortable with the dy/dx type than the f'(x) kind. It makes the higher maths a little easier to grasp.
Those increments are the most important piece of calculus to understand. Without those increments you really can't understand limits. And limits are used as the basis for almost all of calculus' proofs.
By the way if you want to remember to divide by the "class width" in a general way, just divide your function by (b-a). They are constants and will not affect your integral. But that is more general notation. a being the upper and b the lower bound of a definite integral.
I've been away. I am not disputing that it is generally accepted that the line is vertical when the slope is undefined, but you can not prove it. Say that a wall is has a slope of 3meters/1micrometer. Well, unless you have a very sensitive measuring device you will say that the slope is undefined and that the wall is perfectly vertical. But if the top of the wall were smooth a ball bearing would roll right off. Would you then say that gravity were wrong?
There are a lot of discussions here about infinity and how it applies to mathematics. You can say that an undefined slope approaches vertical but you can not say that it is absolutely vertical. And in this case it cannot be proven that this angle is 90°. (or -90°)
I too taught myself algebra, trig, and calculus, and I commend you for doing so. I dropped out of school in eighth grade, but went to college for engineering some 15 years later. The university wouldn't even let me matriculate until I had earned 20 credits because they thought that I had no shot. Actually I tested out of 3 of my calculus classes without ever stepping foot in a classroom. I asked the head of the math department if I could simply go over the texts on my own and be tested by him directly when I felt comfortable with the material. Thankfully he agreed to allow me to do so.
It was hard at times, especially with calculus. Some concepts are grasped right away and others must be gone over and over until it finally sinks in. If you had the motivation to do this much on your own, you will be just fine. Some people just enjoy learning. Good luck on your continuing studies.
P.S. When you become proficient in differential equations maybe you could help me out!
Ricky was correct in that there are no solutions to your problem.
76800/147600 = 64/123 ≈ 52%
I saw that you posted in another thread about the arc length integral, very good. (useful too) Try to remember that one it comes in handy ofter. I actually solved the integral that I was having so much trouble with.
As funny as it sounds, in the end it was nothing more than 2π (√u) evaluated from cos²0° to cos²32.77053659° which provided the answer of 1 square unit. The u came from a substitution for y² after much simplification.
Anyway, I still haven't found a relationship between this interior angle (65.54107318° which is twice the half angle used for the surface of revolution) and the 3282.80635°² in one steradian. I think it is because there is no relationship to be found. By definition a 4π steradians = 4πr².
Sorry that is the best that I can do on this one for now.
Besides, even if it were a vertical line how would you know if it were positive or negative in slope? Would the line be 90° above the x axis or -90° below it? These are questions that the best math minds are still trying to grasp. Perhaps you should contact the publisher of the book that you are using to prove that he/she is correct.
Uh.....sorry.
y = √x
dy/dx = 1÷2√x
dy/dx = 1/0 at x = 0
Either way it approaches ∞ and that is not a slope it is a concept!
Again, approaching infinity is not vertical, close but no cigar.
I would ask your teacher for such a proof if nobody here wants to prove it.
I don't know how your book made the leap of faith to say 90° was definitely a solution. Maybe someone here can show such a proof.
I however see the intersection at zero as an indeterminant slope for the square root function. Just because the slope approaches infinity does not prove that the slope would indeed be completely vertical. I couldn't see how to apply L'Hopital's rule to this situation. Anything less than perfectly vertical would result in less than a 90° angle between the two lines.
As for the other intersection, well that is obvious. Simply take the derivative of both equations at x = 1. I would like to point out that the slope of √x at 1 is ½ and not -½.
The difference in angles from the origin is the angle between them:
arctan 3 - arctan ½ = 45°
but...
arctan 3 - arctan -½ = 98.13°
By the way, that is all you need to do for this problem. By definition m = tan dy.
Yes John, that is precisely what I was trying to write. As far as my efforts to figure this thing out this is what I have come up with.
The problem that I was having with the integration was silly after realizing that I was making substitutions too early complicating the matter more than was necessary.
f(y) = √1-y²
When I first saw this I immediately start using the trigonomic substitution which made this completely unworkable, at least for me.
Leaving f(y) alone until I had the integrand simplified as much as possible was the solution for me. Sadly it took a lot of crumpled pieces of paper and a lot of time pouring over the integration tables to figure this out.
I solved it easily after that point. Just setting u = y². (smacking myself!) Then I was reproached as the surface area for the angle that I proposed earlier was only .019883871 units²!
So then I solved the integral backwards to find the true angle within the sphere at the apex. Basically I just solved for when the integral equalled one. (That was what was stated in your original post)
This solution is only half of the interior angle because of the way the surface formula works. The angle that solved the integral was about 32.77°. So the whole angle would be about 65.541°. It turns out my theory of 360° multiplied by the interior angle was further off than your original guess from the beginning.
I could not find a relationship between this interior angle and the square degrees, and I tried very hard to believe me. That 65.541° is reliable though because that integral is used to calculate just the type of thing that we are doing here.
By the way you should note that if you remove the 2πf(y) from the integrand it will now solve for the length of the arc! You can use either f(x) or f(y) depending on which is more convenient for the length of arc.
I hope all of this jabbering that I've done on this thread has helped you somewhat, I know that I have learned a thing or two since I first ventured into this topic.
Oh! Is there somewhere that I can learn about that [math] syntax?
It has happened to me on several occasions, but it seems more prevalent when using the "Post reply" screen than the small quick post area.
It has happened enough that I now copy the contents of my post before attempting to send it in case that it disappears. I think that it has something to do with the amount of time spent composing a post. Like some sort of session times out before posting actually takes place.
John, I was trying to verify my suggestion that the interior angle was indeed correct at 90/π²°. This was because if you multiply this by 360° you will get the amount of square degrees in a steradian.
As it turns out I am having a very difficult time integrating the formula to prove it. I have to integrate ∫2πf(y)√1+(dx/dy)² dy.
I tried integration by parts and trigonomic substitution to no avail.
The integration is the general form for a surface of revolution about the y axis. I think that that is what you need for this proof.
f(y) = √1-y² is what needs to be applied to the formula above. Without going much further at this point the upper and lower bounds are no problem here but do depend on how the integration is solved.
So if anyone can solve the integration above for the general case, we could all finally put an end to the mystery of how exactly square degrees are related to steradians within three dimensional space. Because once the integration above is completed and evaluated over the correct range it should simply equal 1.
If you are looking for the distance along the surface of the earth.
Since 80° N above the equator and 70° S below the equator total 150° for the interior angle. We can use:
s = rθ in rads or s = r π θ ÷ 180 in degrees (s = arc length)
s = 6370(150π):div180 = 16676.6 km
If you want straight line distance between the two points then:
Since the interior angle of the apex is 150° and the radius is estimated to be 6370 we can use the law of sines. The radius is constant and 180° - 150° = 30° tells us that the other angles are both 15°
x/sin 150° = 6370/sin 15°
So x = 6370 sin 150° / sin 15° ≈ 12305.9 km
Hope that helps.
Let me know it there is anything that I can do. I will be happy to submit anything that you (collectively) think could be beneficial. It is great to hear a little excitement in this coversation because I think that it has great potential.
This is where the logic fails.
For what it is worth, I would be willing to do a few myself. After all I am the one that suggested the idea in the first place. Your job would be mostly in finding a way to edit and maybe organize them in some meaningful way. I personally think that the hardest project will be designing an efficient search mechanism for such a large library that will surely ensue.
Thanks MathIsFun, I learn something new everyday here.
Now I can write 32400 ÷ π² instead of 32400/pi^2
I am quite new here, however the depth of the worksheets is laughable. I do not mean to be offensive in any way as it seems that many of you have surpassed my level of education. It is just that the level of the worksheets fails to exceed all but the most elimentary level of mathematics.
While the basics of mathematics are necessary and justified here the worksheets page seems truncated to the point where it is useful only to very few. My short stint here has left me with the feeling that most of the problems dealt with in the Help Me! forum are at the calculus or pre-calculus level. Would it not be reasonable, even beneficial, to bridge the gap between what now exists there up until at least that point?
The collective knowledge of all of the people who visit this site must be immense. Why not tap that potential to create a kind of knowledge base that is not found on very many websites if any? Just the moderators could have a significant impact upon the contents of this page. If each moderator were to create a small set of problems with worked out solutions just once a week within his/her given specialties for a year, it would create over 260 lessons for those who visit! What kind of prestige would that bring to this place within two or three years time?
That is not the limit however. Perhaps you could create a depository for worksheets that could be created by all of those who visit here. You never know who might drop by and want to make a contribution. A radio engineer could drop by and leave a page of formulas that would have taken one of us days or even weeks to derive. Everyone has an area that is particularly highly developed within them because of our unique personalities and interests. It is from extracting this specialized knowledge that we can all most benefit.
I know that there is a place to create worksheets, but who is willing to drop a page or two of related rates within such an elimentary environment? It is your job as moderators to encourage and reward such activity. I suspect that many here and who are visiting regularly are in or are heading toward an establishment of higher learning. Why not allow others to place their solved homework problems in the worksheets page?
Perhaps I am rambling but I see a vast waste of potential in this area. Little sites with grand ideas can become Googles. The more knowledge contained here, the more knowledge that it draws. Can you see an exponetial function emerging?
Sincerely,
Tom
The first one is 1.165561185 and then my little solar calculator couldn't go any further. It took six iterations using Newton's Method.
x - f(x)/f '(x)
Oh, and be sure to use radians because degrees will not give a solution.
You can use degrees but then your equation will be: y = 2x - tan(180x/pi)