posted by Rose .
a clock has a dial face of 12 inches radius the minute hand is 9 inches while the hour hand is 6 inches the plane of rotation of the hour hand is 2 inches above the plane of rotation of the minute hand. Find the distance between the tips of the minute and hour hand at 5:40 am.
let the plane of rotation of the minute hand be z=0. So, the location of the tip of the minute hand is
Similarly, the location of the tip of the hour hand is
(6cos θ/12,6sin θ/12,2)
at 5:40 am, θ = 2π/3, so θ/12 = π/18
So, the distance between the tips is
d^2 = (9cos 2π/3 - 6cos π/18)^2
+ (9sin 2π/3 - 6sin π/18)^2
now just evaluate for d
oops. Forgot all those hours
θ = 2π*5 + 2π/3 = 10π + 2π/3 = 32π/3
For the minute hand, that's the same value as 2π/3
but, θ/12 is now 32π/36 = 8π/9
which affects its trig function values immensely.