Question

On a certain clock the minute hand is 4 inches long and the hour hand is 3 inches long. How fast is the distance between the tips of the hands changing at 4 pm?

Answer 1

Is 4.39 rad/hour right?

Answer 2

great question, I used to give this as a bonus question on my initial "rate of change" test.

the minute hand is rotating at 2pi rad/hour
the hour hand is rotating at 2pi/12 or pi/6 rad/hour
then dα/dt = pi/6 - 2pi = -11pi/6 rad/hour, where α is the angle between the two arms

let x be the distance between the arms
then
x^2 = 4^2 + 3^2 - 2(3)(4)cosα
x^2 = 25 - 24cosα

2x(dx/dt) = -24(-sinα)dα/dt
dx/dt = 12(sinα)(dα/dt)/x

at 4 pm, α = 120º, sinx = 1/2 and cosα = √/3

x^2 = 25 - 24(√3/2)
x = 2.05314
and
dx/dt = -12(1/2)(-11pi/6)/2.05314
= -70.95

so at 4 pm the distance between the two hands is decreasing at 70.95 inches/hour or 1.18 inches/minute.

check my arithmetic

Answer 3

you asked for "how fast is the distance between the hands" changing, but you gave your answer in radians which is an angle measurement.

how did you get that answer?
Show me your work.

Answer 4

Thanks. I didn't think my answer made any sense. I tried to do too much, instead of taking the derivative of the law of cosines equation, which I had used initially...

Answer 5

For cosa, do you mean -1/2?

Answer 6

Thanks for catching my error, no excuse for that.

of course sin 120 = √3/2 and cos 120 = =-1/2

then

x^2 = 25 - 24cos120
x = 6.0828 and

dx/dt = 12(sinα)(dα/dt)/x
= 12(√3/2)(-11pi/6)/6.0828
= -9.84 inches/hour, the negative shows the distance is decreasing

or -.164 inches per minute

(keep checking my arithmetic)

Answer 7

Thank You Sir

Answer 8

@Reiny I think you multiplied the final result instead of dividing cause I got 16.832 inch/hour