Posted by **Chelsea** on Tuesday, November 23, 2010 at 8:57pm.

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?

- calculus -
**Reiny**, Tuesday, November 23, 2010 at 10:09pm
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

- calculus -
**Connor**, Thursday, March 22, 2012 at 1:24pm
this is wrong

- calculus -
**Connor**, Saturday, October 27, 2012 at 4:28pm
This is right

- calculus -
**john**, Thursday, October 17, 2013 at 7:37pm
this is wrong, right answer is 4.4pi in/h

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