cakc problem help

[kroutons92]

The hands of a clock in some tower are approximately 3 meters and 2 meters in length. How fast is the distance between the tips of the hands changing at 9:00? (Hint: Use the law of cosines.)

 

[burakaltr]

The hands of a clock in some tower are approximately 3 meters and 2 meters in length. How fast is the distance between the tips of the hands changing at 9:00? (Hint: Use the law of cosines.)

Okay.

dt seconds later

(3+(60/dt))

and

(0+dt)

 

[srmichael]

So at 9:00 the angle between the two hands is 90 degrees. If you call the distance connecting the tips of the hands z, then you need to find \(\displaystyle \frac{dz}{dt}\) when \(\displaystyle \theta=\frac{\pi}{2}\).

The Law of Cosines would then be \(\displaystyle z^2=3^2+2^2-2(3)(2)\cos\theta\). However, you will need to calculate \(\displaystyle \frac{d\theta}{dt}\) since when you take the derivative of this equation with respect to t you will have the \(\displaystyle \frac{d\theta}{dt}\) term.

 

[burakaltr]

Okay.

dt seconds later

(3+(60/dt))

and

(0+dt)

dt dotted with (3+(60/dt)) at 90-(dt-60/dt) is 3*2*Cos(90-(dt**2-60)/dt)=6*Sin((dt**2 - 60 )/dt)=6*Sin(dt=0 - 60/dt)=-6*Sin(60/dt)

I do not know how to go further from here but I will try and post later.