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R' = 3 -(30t)/(t^2-3) i think
3t^2 -30t -9 ------------- t^2- 3
3(t^2 -10 -3) would be the top if I did it right....
I Got R'=3-15/(t^2-3)(2t)
yep, as long as thats just screen clutter and not a glaring error :)
add the parts like fractions, 3(t^2 -3) -30t ------------ t^2 -3
Sorry how did you get 3(t^2-3) on top?
the top becomes a quadratic.... a fancy name for "highest power is a 2"
ok... you pretty much have the form: 30 3 - ---- now how do we add fractions togther? b
You have to have a common denominator?
But if R'(t)=3- 15 (2t) ----- (t^2 -3)
That would be 3 - 30t ------- (t^2 - 3)
30t 3 - ------- is what we got right? (t^2 -3)
That's what I got
then to add the "fractions" mulitply the left part by: 3 t^2-3 -- x ----- right? 1 t^2-3
:) deriving...no problem.... fractions? ehhhh :)
3t^2 -30t -9 ----------- t^2 -3
quadratic formula for the top to find the critical points...
5 +- [sqrt(900 - (4)(3)(-9))]/6
Where is the 5 from?
5 +- [sqrt(1008)]/6
you familiar with the quadratic formula?
good heres what I get: t = 5 +- 2sqrt(63)/3
Did you do quadratic formula of 3t^2-30t-9?
yep..... as long as I kept track of myself
30 sqrt(900-(4)(3)(-9)) --- +- -------------- 6 6
So 30 +- sqrt(900+108) ---------------------- 6
we can simplify that by factoring out a 3 first.... sneaky little bugger it is :)
3(t^2 -10t -3) now quad the left side
10 +- sqrt(100 -(4)(-3)(1)) ----------------------- 2
10 +- sqrt(112) --- -------- same result, just easier to see 2 2
112 = 16*7 4sqrt(7)
So 5 +- 2sqrt(7)
thats better :)
you know how to test for min and max?
So those are the critical points? The neg and pos value from that?
yep... those are the critical points. to see how they are behaving, you take a second derivative....derive the derivative.
I don't think I have to do that.
you sure? its fun :) if you have to determine if they are MIN MAX or inflections, then we would, but if you just gotta find criticals, thats it.... dont forget to include the end points of the graph as well, they might be important
Wow, thanks! Can you help me with one more?
i dunno....... i feel a bout of stupidy brewing amongst me brain cells :) ........sure why not :)
Find all critical points of P(t)=-cos(4t)+39-2t
Holy cow sorry my computer spazzed out
thats just the cosine function moved about.... should be rather simple to determine
I have found P'(t) if that helps
P' = 4sin(4t) - 2 right?
Yes that's what I got
good, then we are either both genuises or idiots :) Ill go with genuisesss
0 = 4sins(4t) -2 solve for sin(4t)
what does 4t have to equal for sin to be 1/2?
there are 2 angles, that will satisfy this
I have no idea.
what? you gotta have SOME idea.... its just a circle :)
sin(30) = 1/2 and sin(150) = 1/2
How do I figure the two angles that will satisfy that?
you can either remember back to trig that a 30-60-90 triangle has sides of 1-2-sqrt(3) and interpolate from that, OR use the sin inverse function on a calculator
what from that tells me that 30 and 150 are what I'm looking for?
Sorry I want to understand
sin^-1 will only give you one angle tho...30. You have to remember to cross over the yaxis to get to the other one...150
4t = 30 and 4t = 150 solve for t :)
might be better to use radians instead of degrees, half dozen of one or 6 of the other
Oh! Okay thanks. Moving on
4t = pi/6 4t = 5pi/6
whoa where did that come from?
what!!! you see a bee??
No I mean the 4t=pi/6 and 4t=5pi/6
Wait is that 30 degrees and 150 degrees in radians?
:) there are a few ways to express angles. you can use degrees: 30 and 150 are degrees. OR you can use "radians" which measures the angle using the radius itself as a measuring stick. a circle is 360 degrees around OR 2pi radians pi/6 = 30 degrees 5pi/6 = 150 degrees
To get a number for "t" we should use radians :)
t = pi/24 t = 5pi/24 is what I come up with
30/4 or 150/4 depending on what your answers are looking for
so t=.1309 + pin/2
t = pi/24 = .1309 or t = 5pi/24 = .6545
for degrees... t = 7.5 degrees or t = 37.5 degrees
Okay. I have a question.
go ahead and ask :)
usually before solving for t my professor asks "which quadrants is sin negative in" why did we not have to look for that in this problem?
sin(angle) = 1/2 sin is only positive in Q1 and Q2 .... no need to look any where else, unless you can think of a reason to..
Okay. That makes sense. Thank you sooo much!
when sin = -1/2, it puts a different value into the equation that is not a critical point
you are soooo welcome :)
if you have any more questions, just start a new posting :)