Random Question (1 Viewer)

[currysauce]

Since i am working through randoms, i'll post any i can't do


1. Evaluate cot 2a and tan 2a in exact form, if cot²a - cota = 1

0< a < 90degrees answer given is 1/2, 2

 

[Slidey]

1. Evaluate cot 2a and tan 2a in exact form, if cot²a - cota = 1

0< a < 90degrees answer given is 1/2, 2

let tana=k
(1/k^2)-1/k=1
1-k=k^2
1-k^2=k
now, cot2a=(1-k^2)/(2k) (double angle)
Thus cot2a=k/(2k)=1/2
And obviously tan2a=2

Hey. Awesome. New there was an elegant way.

 

[currysauce]

New question, thanks slide

Find the general solution of sin2x = tanx


answer = n(pi) , n(pi) +- pi/4, n(pi) +- 3pi/4

 

[Slidey]

Sister's 19th now, but I imagine you do something like:
2cos@.sin@-sin@/cos@=0
2cos^2@.sin@-sin@
sin@(2cos^2(@)-1)=0
sin@=0
cos@=+1/sqrt(2)

Bye.

 

[Trebla]

1. Evaluate cot 2a and tan 2a in exact form, if cot²a - cota = 1

0< a < 90degrees answer given is 1/2, 2

cot²a - cot a = 1

Change to cot a into 1/tan a:

_1__- _1__ = 1
tan²a tan a

Put every thing on common denominator tan²a

.: 1 - tan a = tan²a

1 - tan²a = tan a

_tan a__ = 1
1 - tan²a

Multiply both sides by 2:

_2tan a_ = 2
1 - tan²a

Use the formula/identity:
_2tan a_ = tan 2a
1 - tan²a

.: tan 2a = 2

__1__ = 1/2
tan 2a

.: cot 2a = 1/2

 

[currysauce]

A cylinder of radius x cm is inscribed in a cone of radius r and height h cm.

a) show that the volume of the cylinder is V = [(pi)hx²/r][(r-x)]

b) SHow that x = 2r/3 for the maximum value of V and the corresponding ratio of the volumes of the cylinder and the cone is 4:9.

 

[currysauce]

Sketch y = sin ^-1 x . Hence or otherwise evaluate INT from 0 to 1/2 sin ^-1 x dx.

 

[FinalFantasy]

A cylinder of radius x cm is inscribed in a cone of radius r and height h cm.

a) show that the volume of the cylinder is V = [(pi)hx²/r][(r-x)]

b) SHow that x = 2r/3 for the maximum value of V and the corresponding ratio of the volumes of the cylinder and the cone is 4:9.

a) just draw it and use similar triangles to obtain: height of cylinder=h-(hx\r)
.: volume of cylinder V=pi x²(h-hx\r)=pi x²(hr-hx)\r=pi x² h (r-x) \r

b)V=pix²h-pix³h\r
dV\dx=2pi h x- 3pi h x²\r
put dV\dx=0
2pi hx=3pi hx²\r
2pi r=3pi x
.: x=2pir\3pi=2r\3
blah blah blah to show it's max..

juz put it in den get volume of V den get volume of cone den u get ratio...

 

[FinalFantasy]

Sketch y = sin ^-1 x . Hence or otherwise evaluate INT from 0 to 1/2 sin ^-1 x dx.

"Sketch y = sin ^-1 x "
*sketches it*
"Hence or otherwise evaluate INT from 0 to 1/2 sin ^-1 x dx"

i dun like graphs so i'll choose "otherwise"

let I=int. sin^-1 x dx from 0 to 1\2
let u=sin^-1 x and dv\dx=1
du\dx=1\sqrt(1-x²) and v=x
.: I=[xsin^-1 x] from 0 to 1\2 -int. x\sqrt(1-x²) dx from 0 to 1\2
consider int. x\sqrt(1-x²) dx from 0 to 1\2
let u=1-x²
du\dx=-2x
.: int. x\sqrt(1-x²) dx=-1\2 int. du\u^(1\2) [from 1 to 3\4]
=-u^(1\2) from 1 to 3\4
=-[(3\4)^(1\2)-1]=1-sqrt3 \2

.: I=pi\12-(1-sqrt3 \2)

 

[currysauce]

ermm is that integration crap 3unit?

 

[FinalFantasy]

not quite........

 

[Slidey]

Rectangle method is what they are looking for. Do a search of my posts for an explanation.

http://www.boredofstudies.org/community/showthread.php?t=76720

 

[Trev]

Slide, you seem to have some obsession with that rectangle method lol!

 

[currysauce]

new one

Solve 2x^3 +9x² -27x -54 = 0 if the roots are in geometric progrression

 

[FinalFantasy]

new one

Solve 2x^3 +9x² -27x -54 = 0 if the roots are in geometric progrression

roots are -3\2, 3, -6 by inspection
common ratio is -2

 

[currysauce]

book says gp

thats good!@ how the fuck u do it

 

[maths > english]

let the roots equal a/d, a, ad

product = a^3 = 27

a = 3

3/d + 3 + 3d = -9/2

2d^2 + 5d + 2 = 0

d = -1/2 or -2

choose either

roots = -3/2, 3, -6

 

[FinalFantasy]

hmm.. wonder if i'll get the full marks in an exam if i juz do it by inspection like dat lol

 

[maths > english]

hmm.. wonder if i'll get the full marks in an exam if i juz do it by inspection like dat

ull prob get full marks cause they rarely say using this method show or prove