calculus
posted by nico on .
If the hands on a clocktower's clock are 2ft and 3ft long, how fast is the distance between the tips of the hands of the clock changing at 9:20?
I know that the derivative of the minute hand is pi/30 rad/min and derivative of the hour hand is pi/360 rad/min.
what is the equation i have to take the derivative of to find dtheta/dt?

starting at t=0 at noon, measured in minutes, the position (with (0,0) at the center of the clock) of the
hour hand is 2cos(2pi/720 t)
min. hand is 3cos(2pi/60 t)
so, the distance d between the hand tips at time t is
d^2 = 2^2 + 3^2  2*2*3 cos(pi*t/30 * 119/120)
2d dd/dt = 12(pi/30 * 119/120)sin(pi/30 * 119/120 t)
at 9:20, t=560, so
d = 3.663, so
dd/dt = 1/3.663 * 12(pi/30 * 119/120)sin(pi/30 * 119/120 560)
= 0.34 rad/min
Better check my math. The distance should be shrinking.
I'll try to revisit it later; maybe the above will give you a starting point, and you can fix my error. 
I do see one error. The sign of the angle when figuring d:
pi/30 t(1/120  1) = pi/30 t * 119/120, so that will change the sign of the dd/dt, so it is in fact shrinking. 
in the first line of equation, where did "cos(pi*t/30 * 119/120)" come from?

the law of cosines uses the angle between the sides, which in our case is the angle between the hour hand and the minute hand.
since the equations are
hour hand is 2cos(2pi/720 t)
min. hand is 3cos(2pi/60 t)
the angle between the hands is
2pi/720 t  2pi/60 t
= pi/360 t  pi/30 t
= pi/30 t * (1/120  1)
= pi/30 t * (119/120) 
oops, that would be pi/30 t (1/12  1) = pi/30 t * 11/12
that should change things a bit... 
If you are 4\7mile from your house and you can walk 4 5\7 miles per hour how long will it take you to walk home?
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