Chapter 3.9, Problem 50E

### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

Chapter
Section

### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# The minute hand on a watch is 8 mm long and the hour hand is 4 mm long. How fast is the distance between the tips of the hands changing at one o' clock?

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.

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=dydududx.

Calculation:

Let O be the center of the watch and at one Oclock 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.

Obtain dldt at one Oclock.

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

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