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?

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)

Is 4.39 rad/hour right?

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

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.

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...

For cosa, do you mean -1/2?

Thank You Sir

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

To find the rate at which the distance between the tips of the hour and minute hands is changing, we can use the concept of derivatives. Let's denote the distance between the tips of the hands as 'd', the length of the minute hand as 'm', and the length of the hour hand as 'h'.

We know that the minute hand rotates 360 degrees in 60 minutes, while the hour hand completes 360 degrees in 12 hours. Converting the hour hand's rotation to minutes, it rotates 360 degrees in 720 minutes.

At any given time, the angle θ between the minute hand and 12 o'clock can be found using the equation θ = (360/60) * t, where 't' represents the number of minutes past the hour. Similarly, the angle φ between the hour hand and 12 o'clock is given by φ = (360/720) * t.

Using trigonometry, we can express the length of the minute hand projection (x) and the length of the hour hand projection (y) as:

x = m * cos(θ)
y = h * cos(φ)

The distance 'd' between the tips of the hands can be found using the Pythagorean theorem:

d^2 = x^2 + y^2

Taking the derivative of both sides of the equation with respect to time (t), we get:

2d * dd/dt = 2x * dx/dt + 2y * dy/dt

Simplifying the equation, we have:

dd/dt = (x * dx/dt + y * dy/dt) / d

Now, let's substitute the given values at 4 PM. At 4 PM, the minute hand is at 0 degrees (θ = 0), and the hour hand is at (360/720) * 240 = 120 degrees (φ = 120 degrees). Additionally, the lengths of the hands given are m = 4 inches and h = 3 inches.

Substituting these values into the equations, we have:

x = 4 * cos(0) = 4
y = 3 * cos(120) = -1.5

To find the values of dx/dt and dy/dt at 4 PM, we need to know the rotational speed or the rate at which the clock hands are moving. Unfortunately, the question does not provide this information. Hence, without the rotational speed, we cannot determine the exact rate at which the distance between the tips of the hands is changing at 4 PM.