EvoN1020v

I remember this thread pretty well. It was a rather lengthy argument. I suggest this link: http://www.ctre.iastate.edu/educweb/ce353/lec05/lecture.htm. It looks really complicated, so I don't know how I can help.

Also, you should know that sine function is an odd function, while cosine is an even function. That's why they have their own special name to define the functions.

I would just assume that intelligent people have messy writing because most of them have poor communication skills.

Well, since your equation of [math](x+y)=x^2+xy+1[/math] have both x, y on the both sides of the equation, you have to use implicit differentation?

To use implicit differentation, you do this: [math](x+y)\frac{dy}{dx}=\frac{d}{dx}(x^2+xy+1)[/math].

You can find better webpages on implicit differentation than the one you found. If you still don't understand, then I will try to explain it better next time. I don't have the time right now because my girlfriend is here with me.

You'll have to use implicit differentation.

That's a strange equation you gave there. I know the numbers work out in normal situations using weight and F applied but what if you approach the top of a ferris wheel spinning fast? what forces keep you on your seat. there is no applied and your weight is going to seem different as you near the top and you begin to feel wieghtlessness.

I go with Friction force and Normal Force. friction force always being an oppiisition force and the normal force would be the force exerted up on an object keeping it from `breaking`through the ground or seat or whatever.

The tension of the seat joined to the Ferris Wheel keeps it from falling off. Also, you don't use "Force applied" when you're defining forces in circular motions. Tension is another kind of force that points toward the centre. You'll learn it soon enough.

Use google.

If you know MATLAB, another mathematical computer program.

This is assuming that MAPLE program is almost similiar to MATLAB.

You have to define datas for x-axis, let use time for seconds:

t=[0:0.01:10]

This means that the x-axis will be starting at 0 and going in an increment of 0.01 toward 10 at the end.

Then you define a function for example [math]f(x)=x^2+2x-1[/math]

You type this: f=x.^2 + 2.*x - 1

(Notice the dot? It means that the x-axis will be multiplied element-by-element. This is very important to know).

To plot the graph:

plot(t,f)

That's it.

You can google for MAPLE homepage, and there should be plenty of tutorials on it.

Without pushing someone over a horizontal surface would be a static friction. It keeps the object from slipping.

There's no way of knowing coefficient of friction without testing (i.e. pulling or pushing).

I wonder if people confuse friction and "sticktion"? By sticktion I mean the initial force (not inertia) required to initiate movement between two surfaces, and friction the ongoing resistance to continued motion. Take two rough surfaces with equal roughness. At rest, the roughness will tend to interlock. Once moving (having overcome sticktion) the rough surfaces will not have time to interlock, the surfaces roughness tending to bounce over itself. In fact, with a given combination and type of roughness, there may be an optimum "unsticking" velocity.

 

Sticktion is an engineering fact. Sometimes it is called rolling resistance as when a cartwheel requires a lot of effort to roll it out of a rut, but once moving it will tend to bounce over other ruts with no more real friction than if it were on a smooth road.

I think you're talking about static friction. It's a FACT that static friction is usally higher than kinetic friction so the object won't slip at stationary. When the object overcomes the static friction, it will experience a "stick slip" during the kinetic friction movement.

One acceleration should be positive since one block will go in the positive direction (i.e. up). That means the other block will go in the opposite direction hence negative. (i.e. down). Makes sense?

Hahaa, never thought each person would describe a tangent in many different ways!!

Ok don't laugh I realised i did have a tiny bit of time. I basically did what i said before, and i get v= 714.285714285 repeating 714285. Am i correct?

Very good Ragib. The answer is [math]714 m/s[/math]. I'll put up the complete solution for this question when I have the time, or I can wait for other people to attempt the question themselves.

EvoN1020v mind if I post next weeks question?

You're welcome to post a new question in 2 weeks' time. Blike just e-mailed me saying that he will re-create the front page with a featured physics questions thread on it. I'll post the first question, and then you can post a question after that.

Cool with you?

I would love to have the front page back, I loved having all of the top threads readily available, now i have to spend extra time searching through the forums.

You can use the "New Posts" button on the taskbar at the top. It will show you the newest threads, and functioning threads with new posts. It's almost similiar to a front page.

Anybody try the 3rd question? It's actually a high-school version type of question. Anyone should be able to answer it.

There is already a thread on this subject that was created over a year ago. bascule, you can check it out if you want to.

Not exactly...a tangent to a polynomial may intersect at more than 1 point. A tangent is best defined as the limiting Secant PQ as P approaches a stationary point Q. P and Q are points.

You're just saying it in a different way, rather much too fancier for a person who never hears "tangent" before. Keep it simple for newbies please.

are you kidding lol? i didnt think either of my answers were even close to being right but the way i did it was:

50,000 x 5 = 250,000 (momentum of submarine)

100 x 10 = 1000 (mementum of torpedo)

250,000-1000=249,000

249,000/50,000=4.98 m/s

then for the torpedo i did:

7/9(10)+5=12.78

7/9 was there because there is 180 degrees that the torpedo could be launched. However 0-90 degrees and 90-180 degrees mimic the same effects on the velocitity of the torpedo. Therefore since the torpedo was facing 20 degrees up or down the velocity of the submarine would only be about 7/9 effective. I then added 5 for the original difference between the submarine's velocity and the torpedo's.

 

..I gotta learn how to right my work down when I do it, I had to redo the problem to find my work.

With the after collision of the momentum, you have to be careful. The mass of the submarine before the injection of the torpedo was 50,000 kg. Since the torpedo is 100 kg, the mass of the submarine will be 49,900 kg after the injection.

Try the 3rd question yourself. It's not that hard.

After some hours of biting my fingernails trying to solve [math] \int cos^4x[/math], I finally got it. What can I tell you? It's far from your enterprise in what you already know. (You told me you are in high school, right?) In my example of [math]cos^3x[/math], it worked perfectly, because it has an odd power, so you can have a factor of one function. With even power, it's a different story. You have to use the half angle trigonometric identity.

[math]\int cos^4x dx = \int (cos^2x)^2[/math]

The half angle trig identity for [math]cos^2x[/math] is: [math]\frac{1}{2}(1+cos2x)[/math].

You input it in the above equation:

[math] \int (\frac{1}{2}(1+cos2x)^2)^2[/math]

It will expands to:

[math]\frac{1}{4} \int (1+ 2cos2x + cos^22x)dx[/math]

You should notice that there is another even power for cosx. ([math]cos^22x[/math]) so you have to use the half angle identity again.

[math]cos^22x = \frac{1}{2}(1+cos4x)[/math]

Input it in the equation:

=[math]\frac{1}{4} \int (1+ 2cos2x + \frac{1}{2}(1+cos^24x))[/math]

[math]\frac{1}{4} \int (\frac{3}{2} + 2cos2x + \frac{1}{2}cos4x)dx[/math]

Now you can use the Power Rule for Integration:

=[math]\frac{1}{4}(\frac{3}{2}x + sin2x + \frac{1}{8}sin4x)dx[/math]

So you have the answer of:

[math]\frac{3}{8}x + \frac{1}{4}sin2x + \frac{1}{32}sin4x + C[/math]

You can use the integration checkup at http://integrals.wolfram.com/index.jsp

AND IT MATCHES!!

Let me know if you don't understand any of my steps in the calculations. I might have skipped a step.

This was definitely out of my enterprise too.. I think I will learn more about this next semester.

Anyways, let me know if you have any further questions.

Tangent means a slope that touches a curve at one point.

You would need to get Education degree, namely B.Ed to become a teacher.

YEAH, I'm interested!! I never thought I am really bad at integrating trigonometric functions.

I appreciate the help but I'm a bit lost on what Evo posted. I thought you could only add integrals when something was being added in what was being integrated, rather than multiplied. If you can do it that way I wouldn't be aware of it, I'm only in cal II. Another thing I'm confused on is if you let u=sin(x) then du=cos(x) dx, and there isn't any cos(x) dx to substitute for. Anyway the assignment due tomorrow morning so if anyone wants to work out the problems and test their skills rather than doing the work for me, feel free to do so. I doubt my professor will go over it in class and I'm interested in how to work the problems.

I must apologize to you, but I'm not so good in Calculus either, as I'm only in my 1st year in university.

But I KNOW this: You need to know how to use chain rule to be able to integrate the powers. Let use the example of [math]\int cos^3x dx[/math], ok?

[math]\int cos^3x dx = \int cos^2x \cdot cosx [/math]

You know that [math]cos^2x+sin^2x = 1[/math]

So you have:

[math]\int (1-sin^2x)cosx[/math]

Then you use the Chain Rule, so you let [math]u=sinx[/math] and [math]du=cosxdx[/math].

[math] \int (1-u^2)du[/math]

Use the Integration Rule, you get:

[math]u - \frac{1}{3}u^3 + C[/math]

Replace the u and you have the answer.

[math]sinx - \frac{1}{3}sin^3x + C[/math]

For your example of [math]cos^4x[/math], you would have [math] \int cos^2x \cdot cos^2x[/math]. Then you do the rest and use [math]u=sinx[/math].

Hope this helps!!

I don't see any problem in having a CRT monitor. It might be bulky but at least it still show you the screen of the computer. The differences between LCD and CRT are quality and price.

I don't see any purpose in having an account on MySpace, Facebook, MSN Spaces, or those other blog sites. I think it's a total waste of my own time, and having strangers look in your personal life. No offence.

I was reading my university weekly newspaper, and it had an article in it about a guy named Josh who got his 16 millionith friend on MySpace. What life is that?