#### To determine

**To find:** The rate of change of the distance between the tip of the minute hand and hour hand at one O’clock.

#### Answer

The rate of change of the distance between the tip of the minute hand and hour hand is dldt≈−18.6 mm/h.

#### Explanation

**Given:**

The length of the minute hand is 8 mm.

The length of the hour hand is 4 mm.

**Formula used:**

(1) Chain rule: dydx=dydu⋅dudx.

**Calculation:**

Let *O* be the center of the watch and at one O`clock the position of the minute hand is *A* and the position of the hour hand is *B* and θ be the angle between the minute hand at hour hand at the center of the watch and *l* be the distance between the tip of the minute and hour hand as shown in the Figure 1 given below.

Since the hour hand of the clock moves around the clock once in every 12 hours, the change in the angle of the hour hand is 2π12=π6 rad/h.

And the minute hand moves around the clock in every hour, or at the rate of 2π rad/h.

So, the angle θ between them (measuring clockwise from minutes hand to the hour hand) is changes at the rate of.

dθdt=π6−2π=−11π6 rad/h

Obtain dldt at one O`clock.

Since θ and *l* changes with the time.

Therefore, the angle θ and distance *l* are the function of the time *t*.

By using Cosine rule in the ΔOAB.

l2=82+42−(2×8×4cosθ)

Differentiate with respect to the time *t*.

ddt(l2)=ddt(82+42−(2×8×4cosθ))2ldldt=−64(−sinθ)dθdt [∵dydx=dydu⋅dudx]ldldt=32sinθdθdtdldt=32lsinθdθdt

Now at 1:00, the angle between the two hands is one-twelfth of the circle, therefore.

θ=2π12=π6 rad

Substitute θ=π6 in l2=82+42−(2×8×4cosθ).

Substitute θ=π6, l=80−23 and dθdt=−11π6 in dldt.

dldt=3280−23sin(π6)(−11π6)=−32×1180−23(12)(π6)=−8×11380−23π=−88π380−23≈−18.6 mm/h

Therefore, the rate of change of the distance between the tip of the minute hand and hour hand is dldt≈−18.6 mm/h.

That is, at 1:00, the distance between tips of the hand is decreasing at a rate of 18.6 mm/h.