Rate of change problem

The circle below has a radius of 6 cm and point C is moving clockwise around the circle. Assume that point B is moving away from point A at a rate of 2 cm/sec. i.e. x is increasing at a rate of 2 cm/sec. At this moment, assume θ is pi/3 radians.

a. What is the instantaneous rate of change of θ at this moment. i.e. how fast is θ decreasing at this moment?

b. Also, at this moment, is θ decreasing faster or slower?  By how much?

Student X

At any particular instant, let's focus on the right angled triangle ABC where AB= x units and angle CAB = θ, where both x and θ vary with time. 

A relationship can be formed between x and theta, which is x = 6 cos θ 

Differentiating both sides wrt y gives 

dx/dt = -6 sin θ*dθ/dt -------(1) 

Substituting dx/dt =2 cm/s, θ = π/3,  

instantaneous rate of change of θ at this moment = dθ/dt = 2 ÷ [ -6 *sqrt(3) /2 ] =-0.385 rad/s (shown)

Since x increases at a constant rate of 2cm/s, its second order derivative wrt time is zero, ie d^2 x/dt^2 = 0 cm/s^2 

Differentiating (1) on both sides wrt t once again,  

d^2 x/dt^2 = -6 [ cosθ *( dθ/dt)^2 + sinθ * (d^2 θ/dt^2 )]  (Be mindful of how the product rule is employed in the RHS) 

Substituting d^2 x/dt^2 = 0 cm/s^2, θ = π/3  and  dθ/dt = -0.385 rad/s,  

-6 [ cos(π/3) *( -0.385)^2 + sin(π/3) * (d^2 θ/dt^2 )] =0

Solving gives d^2 θ/dt^2 = -0.0856 rad/s^2  

Hence, at this moment, θ is decreasing slower (because of the negative sign) by 0.0856 rad/s. (shown) 

 Hope this helps. Peace.

Best Regards,

Mr Koh