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 Ų 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 dŲ/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)cosŲ
2x dx/dt = 24sinŲ dŲ/dt
when the time is 9:00, Ų = 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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