Any of you do any work with vibration?
http://desmond.imageshack.us/Himg12/scaled.php?server=12&filename=calcw.jpg&res=landing
Can anyone explain how to evaluate this? It’s an integration by parts problem. You can send me a pm if it’s too complicated and long to post here. I evaluated it to -7/4cos(2x)-7/8sin(2x) but it was obviously wrong. I reworked it multiple times and I can’t get the correct answer. I know you have to use trig identities but I’m not seeing where the 2x is coming from. Any help would be appreciated, if not I’ll just ask my teacher on Monday.Thanks.
Edit:
I’ve managed to work it to 7/8 sin(2x) - 14x/4 cos(2x). Still can’t figure out where the 8 is coming from in the denominator part of the cosine portion.
Nevermind. I figured it out…and it only took me 3 hours lol. For anyone who actually cares, here is my solution. And yeah, I know I have bad hand-writing. Peace!
I’ve been meaning for a while to link to these videos. I know there are a ton of video lectures available online at this point, but I think these are better than most. Worth bookmarking for any math, physics, or engineering students. He also has single and multivariable calc videos up.
So I have been studying for a Pre-Calculus test all week and I came across this:
(2x-4)^1/2 - (x-1)^1/2 -1 = 0
A cuadratic equation hidden within radicals, I think. I would appreciate any help on how to solve this.
Simple question, but I don’t understand how to graph this properly if anyone can get this.
My Question
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Wants us to use the y - y = m ( x + x ) formula and what not.
But I don’t understand how the answers are:
a) 33.65
b) 38.6
- Add 1 to both sides
- Square both sides (use (a-b)^2 ), which will leave you with one radical
- Get that radical by itself on one side with the non-radical parts on the other side of the equation
- Square both sides again
- Use the quadratic formula to get the roots. You will get two roots. However, only one is valid here. Why?
You have two points, so you find the average increase in population (aka the slope) of the data you are given.
You then use the point-slope form of the line. They used the 1990 population value to find the population for 1997 and 2010.
You also misspelled the equation. It’s y - y1 = m(x - x1)
Just like you said:
x1 = 1990
y1 = 31
x2 = 2004
y2 = 36.3
y = (y2 - y1) / (x2 - x1) * (x - x1) + y1
33.65 = (36.3 - 31) / (2004 - 1990) * (1997 - 1990) + 31
However, assuming that the trend is linear is completely ridiculous and you should point out that these questions cannot be properly answered without without factoring in the age distribution of the population and the life expectancy. In fact if they told you that in 2000 the population over 65 was 32.6 million it would be evident already.
Yea, misspelled.
Thanks for the feed back!!! I’ll get to work, now 
Also, if the given answer was actually 33.65 million, that’s false accuracy. It should only be given to 2 sig figs. Problems like these are why people think math is boring and useless. Who thinks students get motivated by “real-world” problems of interpolating the number of old people?
I found the answers to be 10 and 2; both were valid. What I did was after adding 1 both sides, I added (x-1)^1/2 on both sides as well. After that, I squared both sides of the equation and basically followed the rest of your instructions. Thanks for the help!
Did you plug in 2 in the original equation? -2 most certainly does not equal zero.
circumference = 2 * pi * r
divide circumference by 365
yeah i stressed out from studying and over thought it waay to much lol
You’re welcome.
The big tip off was when it said to assume the orbit is CIRCULAR. You were also given the radius. You want distance around the circle, so it’s a perimeter, but its of a circle, so that kind of “perimeter” is the circumference.
Learn to think about 1) the kinds of info you are given and 2) the relations between the information (ie, the formulas)
Make a concentrated effort to NOT think of math in terms of “oh its this kind of problem” … unless you’re doing differential equations … then you’re limited to that kind of mindset.
In the education world we call this rote learning. +1 for pointing out necessity of critical thinking. 
some of the next few problems i need to work on ask me to evaluate the expression without using a calculator
sin (pi/6)+cos(pi/6)
then it goes on to tell me to evaluate
sin 30 csc30
just trying to make sure i fully understand what is it is i’m doing here
unit circle,
I don’t have time right now for a full post, but the short version of it is that the expressions sin(pi/6), cos(pi/6), sin(30), and csc(30) can all be associated with numbers… so for instance, rather than saying
1/2 + sqrt(3)/2
They say
sin (pi/6)+cos(pi/6)
Your job is to know that certain angles evaluate to certain numbers when put into trig functions. All these things are just numbers. You just find the expression and substitute the number it evaluates to. Then it boils down to basic arithmetic.
And as far as the beginnings of trig, knowing the ratios of a 30-60-90 triangle is like 90% of the way to knowing all trig functions at all the “standard” (ie appearing on tests) angles. You don’t have to memorize the ratios either, just know that cutting an equilateral triangle down the middle gives a 30-60-90 triangle with one leg half the hypotenuse, and now Pythagoras’ theorem does the rest. All the rest are multiples of pi/2, which are easy if you know what the unit circle definitions are, and multiples of pi/4. Basically, if you don’t know sin(3pi/2) or cos(pi) almost immediately, then you don’t understand the definitions at all.
pi/4 is easy as well because by symmetry sin(pi/4)=cos(pi/4) and sin^2+cos^2=1. Knowing just this almost gets you through the first month or so of a trig course.
Just don’t let the unfamiliar names like sin, cos, csc intimidate you. When you get down to it, the idea is pretty basic.