HELP! Stuck with this Question from my Trials (1 Viewer)

[andy21lau]

Please Help!

a) A vertical tower CD of height 15 metres stands with its base C on horizontal ground. A is a point on the ground due South of C such that the angle of elevation of the top D of the tower from A is (π/4) radians. B is a variable point on the ground due East of C such that the angle of elevation of the top D of the tower from B is α radians, where 0<α<(π/2). The value of α is increasing at a constant rate of 0.01 radians per second.

i) Show that AB = 15cosecα.

ii) Find the rate at which the length AB is changing when α = (π/3)
____________________________________________________
My failed attempt:

AC = 15 cos (π/4) = 15/√2
BC = 15 cos α

AB = √(AC^2 + BC^2)
= √[(15/√2)^2 + (15 cos α)^2]
= √[(225/2) + 225(cos α)^2]
= √225√(1/2 + (cos α)^2)
= 15√(1/2 + (cos α)^2)

how to get 15cosecα!!!!?? :mad1:

i think its got something to do with mixing up radians and degrees... i dont know.

please help if you can.

Thanks in Advance!

 

[untouchablecuz]

Please Help!

a) A vertical tower CD of height 15 metres stands with its base C on horizontal ground. A is a point on the ground due South of C such that the angle of elevation of the top D of the tower from A is (π/4) radians. B is a variable point on the ground due East of C such that the angle of elevation of the top D of the tower from B is α radians, where 0<α<(π/2). The value of α is increasing at a constant rate of 0.01 radians per second.

i) Show that AB = 15cosecα.

ii) Find the rate at which the length AB is changing when α = (π/3)
____________________________________________________
My failed attempt:

AC = 15 cos (π/4) = 15/√2
BC = 15 cos α

AB = √(AC^2 + BC^2)
= √[(15/√2)^2 + (15 cos α)^2]
= √[(225/2) + 225(cos α)^2]
= √225√(1/2 + (cos α)^2)
= 15√(1/2 + (cos α)^2)

how to get 15cosecα!!!!?? :mad1:

i think its got something to do with mixing up radians and degrees... i dont know.

please help if you can.

Thanks in Advance!

independent trials ay? did the same one

AB^2 = AC^2 + BC^2

tan(pi/4) = 15/AC => AC = 15

tan(α) = 15/BC => BC =15cot(α)

.'. AB^2 = (15)^2 + (15cotα)^2 = (15^2)(1+cot^2 α) = (15^2)(cosec^2α)

AB = 15cosec(α) since AB>0

 

[andy21lau]

independent trials ay? did the same one

AB^2 = AC^2 + BC^2

tan(pi/4) = 15/AC => AC = 15

tan(α) = 15/BC => BC =15cot(α)

.'. AB^2 = (15)^2 + (15cotα)^2 = (15^2)(1+cot^2 α) = (15^2)(cosec^2α)

AB = 15cosec(α) since AB>0

OH sh1t!! cant believe i did cos..

cant believe it!


Thanks alot man

 

[andy21lau]

independent trials ay? did the same one

AB^2 = AC^2 + BC^2

tan(pi/4) = 15/AC => AC = 15

tan(α) = 15/BC => BC =15cot(α)

.'. AB^2 = (15)^2 + (15cotα)^2 = (15^2)(1+cot^2 α) = (15^2)(cosec^2α)

AB = 15cosec(α) since AB>0

would you mind helping me with ii)?
i dont know how to differentiate 15 cosec α.

 

[Aquawhite]

would you mind helping me with ii)?
i dont know how to differentiate 15 cosec α.

15 cosec α is 15/sinα ...yes?

Then that is equal to 15(sinα)^-1 using index rules.

And I'm confident you can do that one ^_^. Note, that is not inverse sine, just sine to the power of -1 (i.e. would be 1/sin as I gave).

 

[andy21lau]

15 cosec α is 15/sinα ...yes?

Then that is equal to 15(sinα)^-1 using index rules.

And I'm confident you can do that one ^_^. Note, that is not inverse sine, just sine to the power of -1 (i.e. would be 1/sin as I gave).

thanks
but when i search on the web
"differentiate cosec"

i see it is given by:
d (cosec x) = -cosec x cot x ??
dx

is it just another formula?

 

[untouchablecuz]

thanks
but when i search on the web
"differentiate cosec"

i see it is given by:
d (cosec x) = -cosec x cot x ??
dx

is it just another formula?

d(cosec x)/dx = derivative of (sin x)^-1 = (-1)(cosx)(sinx)^-2= -cosx/sin^2 x = (-cosx/sinx)*(1/sinx)= -cot x cosec x

AB = 15cosec x

d(AB)/dx = -15cosx/sin^2 x

d(AB)/dx = d(AB)/dt * dt/dx

d(AB)/dt = d(AB)/dx * dx/dt

dx/dt = 0.01

d(AB)/dx when x = pi/3 is -10

so d(AB)/dx = 0.01*-10 = -0.1 m/s

 

[Aquawhite]

Just take it a couple more lines!

 

[andy21lau]

d(cosec x)/dx = derivative of (sin x)^-1 = (-1)(cosx)(sinx)^-2= -cosx/sin^2 x = (-cosx/sinx)*(1/sinx)= -cot x cosec x

AB = 15cosec x

d(AB)/dx = -15cosx/sin^2 x

d(AB)/dx = d(AB)/dt * dt/dx

d(AB)/dt = d(AB)/dx * dx/dt

dx/dt = 0.01

d(AB)/dx when x = pi/3 is -10

so d(AB)/dx = 0.01*-10 = -0.1 m/s

thank you!