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this should say for what value of x on the interval... Sorry!
deriving trig...aint had to do this yet :)
well the derivative is -0.54sin(0.3x)-2.16(cos(0.3x) if you need it. I know I need to set it to zero to get the critical values, but I don't know how to solve for x lol
Dx(-7.2 sin(.3x)) = -7.2(.3) cos(.3x)
lol .... I was getting there :)
lets multiply by 100 to get rid of some of those decimals
-54 sin(.3x) - 216(cos.3x) = 0
can we factor out anything from 54 and 216? I know 2 comes out...
you can take out a 9 i think
54/216 = 27/108..try 3 now 9/36 try 3 again 3/12 and 3 again 1/4
looks like 54 factors out completely... 216/54 = anything useful? 4 yay!!
-54(sin(.3x) + 4cos(.3x)) this look right to you so far?
yeah I'm following I think. lol, that's what I have when I do the same work on my calculator. :)
so, lets see what sin(.3x) + cos(.3x) gets us to equal 0
they gotta be opposites of each other, so lets throw one to the other side
and dont forget that 4 that I left out lol
yeah i was going to ask why we weren't using it, lol but nevermind
4cos(.3x) = - sin(.3x) now picture these graphs for a minute
the cos is just stretched 4 tall, so the 4 doesnt matter the sin is flipped over upside down
.3 = the frequency of the graphs
these cross points at 2 places every time they go thru one frequency...does that make sense?
once in Q2 and another in Q4
the only angle that has sin and cos equal is 45 degrees
right.. I put them in my graphing calculator. lol so that means I have more than one answer right?
yes... more than one answer, and it will have something to do with 45 degrees
-9.65 and 11.28 is our interval
.3(sqrt(2)/2) = ?
I might be wrong about the 45 tho :)
derive it again to see if we can get some useful information...
derive the original f(x) or take the second derivative?
second derivative will give us a better picture....
64.8 sin(0.3 x)-16.2 cos(0.3 x)
I got: y'' = .648 sin(.3x) - .162 cos(.3x)
oh, yeah I just had it multiplied by 100 still
it might be helpful, but im still "stuck" :)
When I asked a friend, she told me I needed to use the quadratic. However, she had her math major friend do the problem for her, and can't tell me anything farther than that lol
lets go back to the original.. sin(.3x) + 4cos(.3x)
Is there a way for us to write them both in the same terms, like a way for us to make cos to sin?
there can be a square root involved, or a phase shift....
sin = sqrt(1-cos^2)
sin^2 = 1-cos^2
If it helps, when I tried using wolfram alpha, it gives me the answer.. x = 20/3 (pi n+tan^(-1)(1/4 (1+sqrt(17))))
But I don't actually even know what that means, lol so I that wasn't very helpful either. It was kind of a a last resort as it was, because my assignment is due at midnight. :\ lol
4cos(.3x) = -sin(.3x) square both sides to get 16cos^2(.3x) = sin^2(.3x) 16cos^2(.3x) = 1 - cos^2(.3x)
then: 17cos^2(.3x) -1 = 0 right?
i believe you're correct.
cos^2(.3x) = 1/17
cos(.3x) = sqrt(17)/17 then ....if that helps :)
okay so x ends up being like 4.419392212
cos-1(sqrt(17)/17) is 1.325817664 i think
25.32 is what I get for that x :) does that make any sense?
25.32 radians of course
ohh.. so why am I getting a different number, is my calculator just in the wrong mode?
maybe; and I used 1/sqrt(17) to be simple
lets plug that in to see what we get: sin(.3(25.32)) + 4cos(.3(25.32)) :=: 0 ??
I'm not getting zero. :\
0.1321 + 3.9648 = .... ack!!! lol.
need someone smarter than me I guess :)
oh no. lol what do I do now? Do you have any ideas of where else I can get help b y midnight?
how many points it take off if it is wrong?
I'm not sure.. there are only 4 sections and this one is technically two parts.
let me review our stuff for a few minutes and see what I can deduce...
alright. If it helps, on the thread for the problem, it says to take the derivative equal to zero, and that you should get tan=#, then take the inverse.
I don't know how they're getting tangent though.
What if we divided both sides by cos(0.3x) and then had tan(0.3x)=4.. does that work?
-4x -y=0...-y=4x....-y/x = 4 -4x +y = =..... y=4x......y/x=4 this is the tan they get at.... -sin(.3x) ------- = 4 cos(.3x) -sin(x) = sin(-x) so thats the same
tan(a) = 4 tan-1(4) = what?
i was getting to it.... :)
75.96 degrees is a divide that by.3
so we have 253.212=x ?
that sounds large. lol
remember that tan-1 spits out an angle between -90 and 90
(75.96 + 180) /.3 ?? maybe
(75.96 - 90)/.3?
that one gives us -46.8
i see that..... maybe our interval itself gives us a max or min at the end points?
if you put those in.. for -9.65, we get 0.0119783705 and for 11.28, we get -0.190832377
.3x gives us one cycle every 20pi/3 does that helps us out?
well in the wolfram alpha, it said something about 20/3 pi * n, but n is supposed to mean any integer, so I didn't really understand it.
n = the angle measurement a normal cycle is 2pi but we disturb that with .3.... so it increases the frequency of the cycle to 20pi/3
6 and 2/3 pi
x = 20/3 (pi n+tan-1(1/4 (1+sqrt(17))))
What is the value of our interval in terms of pi?
x(pi) = -9.65 x = pi/-9.65
-.325 pi to .279 pi right?
that aint gonna make a difference, becasue our interval is telling us the value of "x" which is the angle involved.... not the period....
okay.. I'm just lost at this point. lol I really don't even know how to get any further in this.
im at a loss for the moment as well :) its all those decimals that are messing me up
I've got it!!!! lol I graphed it and just followed the graph and put in the numbers :P thanks though hahah