CSSA Trials Question (1 Viewer)

[Dreamerish*~]

A is a fixed point on a circle centre O, radius 1 cm.

P is a variable point which moves around the circle with a constant speed of one revolution per second.

∟AOP = θ

(i) Show that ^dθ/_dt = 2π radians per second.
(ii)Find the rate at which the area of ΔAOP is changing when θ = ^2π/₃.

Thanks.

 

[physician]

Is this further applications of calculus...?

 

[Dreamerish*~]

Is this further applications of calculus...?

Yes, yes it is.

Relative rates.

 

[physician]

Yes, yes it is.

Relative rates.

so sorry... thats the only topic we havn't covered...

by the way which year is this question from?

 

[Dreamerish*~]

so sorry... thats the only topic we havn't covered...

by the way which year is this question from?

That's ok.

1997.

EDIT: LOL, it's ok! I found the answers!

 

[KFunk]

A is a fixed point on a circle centre O, radius 1 cm.

P is a variable point which moves around the circle with a constant speed of one revolution per second.

∟AOP = θ

(i) Show that ^dθ/_dt = 2π radians per second.
(ii)Find the rate at which the area of ΔAOP is changing when θ = ^2π/₃.

Thanks.

I can't think of a really solid algebraic way to do i) but basically since the variable point goes through one revolution every second it's rate of change is 2π radians per second i.e. dθ/dt = 2π radians s^-1

ii) Area of triangle = 1/2.a.b.sinC
where a = 1 , b = 1 , C = θ so
A = 1/2.sinθ
d(A)/dθ = 1/2.cosθ

dA/dt = dA/dθ . dθ/dt = π cosθ

hence when θ = ^2π/₃ , dθ/dt = -π/2 cm²s^-1

Edit: you have the answers so this probably doesn't matter

 

[Dreamerish*~]

I can't think of a really solid algebraic way to do i) but basically since the variable point goes through one revolution every second it's rate of change is 2π radians per second i.e. dθ/dt = 2π radians s^-1

ii) Area of triangle = 1/2.a.b.sinC
where a = 1 , b = 1 , C = θ so
A = 1/2.sinθ
d(A)/dθ = 1/2.cosθ

dA/dt = dA/dθ . dθ/dt = π cosθ

hence when θ = ^2π/₃ , dθ/dt = -π/2 cm²s^-1

Yeah that's what I was trying to find out for (i). I knew how it works, but I can't prove it. The answeris just that since dP/dt is 1 revolution per second, theta changes at 2 pi per second.

 

[Jago]

hey i did that exact same question a few days ago!