help with HSC question 10 (1 Viewer)

[FOB all the way]

hi could some one plz help with this question

 

[Forbidden.]

is this a Trial paper ? I have not seen this question before ? Which year is it, because that picture is way too small.

EDIT: This does not appear in Q10 from 1986-2006 HSC Maths papers.

 

[FOB all the way]

it is a trial question

 

[FOB all the way]

sorry about the picture

the question reads

A vertical wall in danger of collapse is to braced by a beam which must pass over a second lower wall b metres high and located a meters from the first wall. Let the lenght of the beam be y meters the angle the beam makes with the horizintal be @ and x the distance from the foot of the beam to the smaller wall


i) show that y=asec@+bcosec@

ii) by finding the stationary points on the curve y=asec@+bcosec@ prove tan @=cubedroot(b/a)


iii) hence show that the shortest beam that can be used is given by
y=a squareroot (1+(b/a)^2/3) + b squareroot(1+(a/b)^2/3)



hope this picture is a bit better

 

[ellen.louise]

A vertical wall in danger of collapse is to be braced by a beam, which must pass over a second, lower wall b metres high and located a metres from the first wall. Let the length of the beam be y metres, the angle the beam makes with the horizontal be @ and x the distance from the foot of the beam to the smaller wall.

i) show that y = asec@ + bcosec@
ii) by finding the stationary points on the curve y = asec@ + bcosec@
prove tan@ = (b/a)^(3/2)
iii) hence show that the shortest beam that can be used is given by
y= ???


what are the questions plz?

amirite so far?

 

[ellen.louise]

A vertical wall in danger of collapse is to braced by a beam which must pass over a second lower wall b metres high and located a meters from the first wall. Let the lenght of the beam be y meters the angle the beam makes with the horizintal be @ and x the distance from the foot of the beam to the smaller wall


i) show that y=asec@+bcosec@

ii) by finding the stationary points on the curve y=asec@+bcosec@ prove tan @=cubedroot(b/a)


iii) hence show that the shortest beam that can be used is given by
y=a squareroot (1+(b/a)^2/3) + b squareroot(1+(a/b)^2/3)

the two triangles this diagram makes are similar

Im sorry i dont have time to do this question right now.

get back to you later if noone gets it

 

[FOB all the way]

this picture might even be better

 

[Mattamz]

for (i):
In the big triangle:

cosα = (x + a)/y
secα = y/(x+a) ------(1)

from the small triangle:

tanα = b/x
x= bcotα

y = (x+a)secα
= asecα + xsecα
= asecα + secα*bcotα
= a secα + bcosecα

 

[Mattamz]

for (ii):
y = a/cosα + b/sinα

dy/dx = -a(cosα)^-2 . -sinα - b(sinα)^-2 . cosα
stat. pt occur when dy.dx = 0
0 = a(sinα)^3 - b(cosα)^3
a(sinα)^3 = b(cosα)^3
(tanα)^3 = b/a
tanα = cubed root of (b/a)

 

[Mattamz]

for(iii):
consider a trangle with sides:
b^(1/3), a^(1/3); from tanα = (b/a)^(1/3)
and from pyth. theorem:
hypotenuse of sqrt [a^(2/3) + b^(2/3)]

From this:
sinα = b^(1/3) / sqrt [a^(2/3) + b^(2/3)]
cosecα = sqrt [a^(2/3) + b^(2/3)] / b^(1/3)
= sqrt {[a^(2/3) + b^(2/3)] / b^(2/3)}
= sqrt {(a/b)^(2/3) + 1}

cosα = a^(1/3) / sqrt [a^(2/3) + b^(2/3)]
secα = sqrt [a^(2/3) + b^(2/3)] / a^(1/3)
= sqrt {[a^(2/3) + b^(2/3)] / a^(2/3)}
= sqrt {(b/a)^(2/3) + 1}

sub these in to the equation for y;
and bingo..lol

 

[FOB all the way]

thanks for ya help

but sorry i dont follow how you got to this line

dy/dx = -a(cosα)^-2 . -sinα - b(sinα)^-2 . cosα
stat. pt occur when dy.dx = 0
0 = a(sinα)^3 - b(cosα)^3

how did you simplify that and go from -2 to 3

and is tan sin/cos

so why does b/a give tan and not cot

 

[Mattamz]

thanks for ya help

but sorry i dont follow how you got to this line

dy/dx = -a(cosα)^-2 . -sinα - b(sinα)^-2 . cosα
stat. pt occur when dy.dx = 0
0 = a(sinα)^3 - b(cosα)^3

how did you simplify that and go from -2 to 3

and is tan sin/cos

so why does b/a give tan and not cot

not a problem

dy/dx = -a(cosα)^-2 . -sinα - b(sinα)^-2 . cosα
stat. pt occur when dy.dx = 0
0= -a(cosα)^-2 . -sinα - b(sinα)^2 . cosα

then times all terms by [(cosα)^2 . (sinα)^-2]

0 = a(sinα)^3 - b(cosα)^3

EDIT: remove minus sign

 

[FOB all the way]

cool

but how did you no to times all terms by
[(cosα)^2 . (sinα)^-2]

 

[Forbidden.]

thanks for ya help

but sorry i dont follow how you got to this line

dy/dx = -a(cosα)^-2 . -sinα - b(sinα)^-2 . cosα
stat. pt occur when dy.dx = 0
0 = a(sinα)^3 - b(cosα)^3

how did you simplify that and go from -2 to 3

and is tan sin/cos

so why does b/a give tan and not cot

y = a sec α + b cosec α

y = a / cos α + b / sin α

y = a (cos α)^-1 + b (cos α)^-1

dy/dx = a x -1 (cos α)^-2 x - sin α + b x -1 (sin α)^-2 x cos α

dy/dx = a sin α (cos α)^-2 - b cos α (sin α)^-2

dy/dx = 0

a sin α (cos α)^-2 - b cos α (sin α)^-2 = 0

a sin α (cos α)^-2 = b cos α (sin α)^-2

a sin α / cos² α = b cos α / sin² α

a sin³ α = b cos³ α

a sin³ α / cos³ α = b

sin³ α / cos³ α = b / a

tan³ α = b / a

tan α = ³√(b/a)



Notes:

sinⁿ α = (sin α)ⁿ e.g sin² α = (sin α)²
The "x" I've used means multiply
I hope you know that a^-n = 1 / aⁿ

 

[FOB all the way]

thanks so much

you are my new hero