Axzce's bad questions

[axzce]

(0) what's the implied domain and range of y=5arcsin(2cos(3x))?
no idea how to approach this.
-calc free

can someone please help me? it's really appreciated

thanks in advance

[axzce]

(1) given the vectors (4,-3,-1), v=(-2,0,5) and w=(2,2,1), what is the vector component of v(subscript:that plus sign with the bottom cut off) of v perpendicular to w?

(2) Does arcsin(cot(3pi/4)) = -pi/2 OR -pi/2 +-2pi?

I think it's the latter but teacher says it's the first.
Who is right and why?

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(3) given cos(x)=-3/4 , x is an element of [pi,3pi/2]
then how do i calculate tan(2x)

I worked out what sin(x) =, then worked out tan, then I applied the compound angle formula for tan(x+y) with y=x in that formula....and my answer for 120/sqrt(7)

my solution book says the answer is 3sqrt(7),

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oh yeh I'm creating this thread because I have so many questions.

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4) u=(4, -3, -1)
v=(-2,0,5)
The angle between u and v?
I went:
cos(angle)=(u*v)/(uv)
so:
cos(angle)=9/(sqrt(26)*sqrt(29))
therefore angle = arcos(sqrt(754))
However my solution booklet says: arccos( - (sqrt (13)) / sqrt (58) )

[RobM8]

(0) is going to be a really ugly graph with lots of intervals for the domain due to the periodic cos(x).

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(2) cot(3pi/4) = 1/(tan(3pi/4)) = 1/(-tan(pi/4)) = -1
therefore, arcsin(cot(3pi/4)) = arcsin(-1) = -pi/2
there is only one solution due to the range of the arcsin(x) function [-pi/2,pi/2].
your other solutions are outside of this range.
teacher is right.

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(3) cos(x) = adj/hyp = -3/4                                                                                    
draw a triangle where the hypotenuse is 4 and the adjacent side is 3.                
you can find the length of the opposite side using Pythagoras (a^2=b^2+c^2)
opp^2 = hyp^2 - adj^2 = 16 - 9 = 7    
so opp = sqrt(7)      
sin(x) = -sqrt(7)/4 (As x is in the 3rd quadrant)
tan(x) = sin(x)/cos(x) = sqrt(7)/3
tan(2x) = 2tan(x)/(1-tan(x)^2) = [2sqrt(7)/3]/[1-7/9] = [2sqrt(7)/3]/[2/9] = (9)(2sqrt(7))/((3)(2)) = 18sqrt(7)/6 = 3sqrt(7)

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(4) u.v = cos(theta)|u||v| where theta is the angle between the vectors.
u.v = (4)(-2) + (-3)(0) + (-1)(5) = -8 + 0 -5 = -13
|u|= sqrt(16 + 9 + 1)= sqrt(26)
|v|= sqrt(4 + 25) = sqrt(29)    
therefore, -13 = sqrt(26*29)cos(theta) = sqrt(754)cos(theta)
rearrange to get theta = arcos(-13/sqrt(754)) degrees
For some nonsensical reason your solutions have taken out a common factor of sqrt(13) from -13/sqrt(754) to get -sqrt(13)/sqrt(58).
                                                                                                    
EDIT: your first question is usually question 1 not question 0 ?

[axzce]

thanks RobM8!!!!!!!  the (0) was kinda odd but I  doesn't matter eh?!

(5) *attached*

[axzce]

does anyone know how to do this attached question?

[RobM8]

I have done the question however I am not sure of my answer.
Post solutions so I can verify or wait for someone else to have a crack.

Also, what textbook and chapter/topic is this from?

[axzce]

Could I please ask you if rather than show me the working out,, you show me where to go from where I'm at, or what I'm doing wrong up to there?

: ) !! thanks

so cause it's inversly proportional we know it's : dh/dt = k/sqrt(h)

then we just sub in the values and rearrange right?

so it's given that:

dh/dt = k/sqrt(4)  and we should try and we also know that t = 2

I tried some integration stuff from here to try and sub in t=2 but it didn't work out...

Solution is attached.

[moekamo]

yea got got the first bit sort of...



the second part is just subbing in h=25 and solving for t, should be easy, use the formula on the second last line so you're not unecessarily resolving something

the last part, you sub in , solve for h, chuck it back into the same equation you used for part 2 to solve for t

[brightsky]

i) dh/dt = k/sqrt(h), sqrt(h) dh = k dt, int sqrt(h) dh = int k dt, 2h^(3/2)/3 = kt + C, which means h(t) = (3/2)^(2/3) (kt + C)^(2/3)
we know h(2) = 4, sub that in:
(3/2)^(2/3) (2k + C)^(2/3) = 4
we also know that h(0) = 0:
(3/2)^(2/3) (C)^(2/3) = 0, C = 0, sub back into above, k = 8/3
so the equation for h(t) is
h(t) = (3/2)^(2/3)(8/3 t)^(2/3) = (4t)^(2/3)

ii) 25 = (3/2)^(2/3)(8/3 t)^(2/3)
t = 125/4 years

iii) dh/dt = k/sqrt(h) = 8/3(sqrt(h)) = 1
h = 9/64 metres

hope there aren't any errors

EDIT: beaten and moekamo's solution is way neater for i)

[RobM8]

Confirmed, same method and answers as Brightsky.

[axzce]

yea got got the first bit sort of...



the second part is just subbing in h=25 and solving for t, should be easy, use the formula on the second last line so you're not unecessarily resolving something

the last part, you sub in , solve for h, chuck it back into the same equation you used for part 2 to solve for t

Farkk!!!!! that's so fricken clever!! thank you! you're working out was perfect and it made me understand so easily!!! thanks!!! Moekamo you're a legend man thanks so much!

thanks for your help Brightsky too and again kindest regards to Robm8, up reputation all round!