A certain clock has a minute hand that is 4 inches long and an hour hand is 3 inches long. How fast is the distance between the tips of the hands changing at 9:00?

Answer 1

let Ø be the angle between them

the angular velocity of the minute hand = 2π/60 rad/min = π/30 rad/min
the angular velicity of the hour hand = 2π/(12(60)) or π/360 rad/min

so dØ/dt = (π/30 - π.360) rad/min
= 11π/360 rad/min

let the distance between the tips be x inches
by Cosine Law
x^2 = 16 + 9 - 2(4)(3)cosØ
2x dx/dt = 24sinØ dØ/dt

when the time is 9:00, Ø = 90° and x = 5
10dx/dt = 24sin90° (11π/360)
x = 24(1)(11π/360)/10
= .2304 inches/min

Answer 2

this is wrong

Answer 3

This is right

Answer 4

this is wrong, right answer is 4.4pi in/h