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So, can't we express it in the first derivates of y(t) ? Thats what i'm looking for.
some kind of chainrule or smthing?
I'm not convinced... i dont get it really. Do you mead x = a(-cos(yt/y)) ?
How do i differentiate that one and get the same thing in the brickets sin(y).
Ganesh... maybe i didn't make my self clear that y is a function of t.?? Or is it
me that have misunderstood you? Coud you show me how you diferentiate the
expression?
Hi guys, I'm stuck again!
And it's mathematical, like always (regret those lazy days back in analytic math courses, haha).
I'm in this situation:
dx=asin(y)dt
(a is a constant value)
where y=y(t)
I want to solve this for x.
smthing like x = smthing + C (constant of integration, or what ever you want to call it)
I can recall it had something to do about, maybe substitution, but i can't see how thats going to happen...
But it's a mechanics problem.
That's the problem, man! The answer should be 0.75, but I can't see it. I need to know if I can solve it,
or if I have to find another way of attacking the original problem. I would also like to read it like you did,
but unfortunately I have drawn it correct. The side 0.3 is not the hyp. of the little triangle. I have gone
forth and back on if there is some kind of connection between the ratios, or (what you call the relationship
between the sides that's not the hyp.)
Hi, check this out:
http://engman.bravehost.com/mechanics/side.JPG
I've been sitting with it for 2 hours now, it just makes me crazy. I mean, it is
a possibility that the mechanical problem that this trig. is derived from is ment
to be solved in an other way, but i wouldn't bet my money on that.
OK! THIS IS ALL SOLVED NOW! No more problems!
Hi.
How do I solve sin(θ) = 0.38cos(θ) ???
CORRECTION: eq(2) should be: cos(y) - xsin(y)=0!
Of course. I'm just happy that you give it a try.
This suddenly felt impossible:
How do I transform these three equations...
(1): xz=bc
(2): cos(y) + xsin(y)=0
(3): zsin(y) + xzcos(y) - c(a+b)=0
(three unknown x,y,z with three equations, always solvable right!? But i only want to solve this for x!!!)
(I've looked it through carefully, so it shouldn't be any errors in there)
...to either of these too:
answer, version1: x=1/√((a/b+1)²-1)
, version2: x=tan(arcsin(b/(a+b))
This is how far I've got:
(1) can be rewritten to: x=bc/z , new(4)
(2) -"- : x=cos(y)/sin(y) , new(5)
(3) -"- : z=c(a+b)/(sin(y)+xcos(y)) ,new(6)
(6) in (4) --> x=b(sin(y)+xcos(y))/(a+b) , new(7)
(5) in (7) --> cos(y)/sin(y)=b(sin(y)+cos²(y)/sin(y))/(a+b) --> [multiplying both sides with sin(y)] cos(y)=b/(b+c)
So, finally: I'm getting y=arccos(b/(a+b)), BUT HOW DO I PROCEED... It's a dead end. I don't know what more to do...
Maybe you do?
Uhum, I was just hoping that there's someone knowing something about
mechanics, or just can flick an eye on my calculations for this problem.
MathIsFun and mathysperson I'm sure will recognize one of these two problems.
http://engman.bravehost.com/mechanics/2probs.JPG
On 7.14 you can have a look of how far I've got, but I'm stuck.
http://engman.bravehost.com/mechanics/incomplete_solution714.JPG
7.39 I haven't a clue how to do it. The answer is: angle between 32.8 and 45 degrees. The 45 degrees
seems logic, and I guess that should calculate the intuition way. I'll buy that.
IF this is the wrong forum, if you wanna strictly do math around here, I would really appreciate a similar
forum with mechanics.
/Mr. G
You have both been of great help, and I'm so greatful. To MathIsFun: That was the
solution I knew was there. Mathsyperson: I'm impressed! And regarding the "i want this angle
so bad"; It is written by me, and have been more than true for the last 72 hours. But not any more...
hihi... it's like a burden has been lifted from my shoulders, and the salvator:
"www.mathisfun.com". You guys rock!
Heyy Mathsyperson! That is such a cool answer, it's a real pleasure to read it!
But, Im convinced that the problem that this algebra subproblem is conected with,
is not ment to be this advanced... so you said that the 1.5m measurement is useless. It seems so
to me to... but it should be involved somehow. So, you are absolute positive that that measurement
cannot unlock an easier solution to this?
Ahh! An easy solution, just what i was hoping to be buried somewhere
under. And the mechanical excercise is now solved, thank
to your algebraic consultancy. A big thanks to you!
Hi! Thanks so much for the fast intrest. I mean, this is so great.
Hate having to sit by my self with this kind of lousy problems.
Actually this is a part of a quite simple mechanical problem, and
I'm almost 100 percent sure of that the way they mean, is to be
able to calculate that angle. But, it doesn't seem like that should
be a big problem. I mean, calculating that angle shouldn't be the
head task of the exercise. So, I hope that your answer is not too
advanced, even thought I will appreciate it equally much, my friend!
/Gustav
Hi, thought that i might just put up all my problems in here, to hope
for the best. Well, i have other problems too, but to get rid of one or
two of them would make my day!
Just check this link out, for the problem:
http://engman.bravehost.com/Jobb/algebra2.JPG
/Gustav
Hello! I have a problem with an angle. I just cant calculate it.
But I'm pretty sure that this is possible. What do you say!?
I would be more than happy if some of you guys could see how to do this.
The problem is depicted on this adress.
http://engman.bravehost.com/Jobb/algebra.JPG
/Gustav from Sweden
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