A large clock has an hour hand which is 1 meter long, and a second hand which is 2 meters long.

Question

a. What is the rate of change in the distance between the ends of the two hands, with respect to time, at 2:00 PM?

     Hint: the distance between points (x1, y1) and (x2, y2) in the plane is sqrt((x1-x2)^2 + (y1-y2)^2) --> the distance formula

 

b. Define coordinate axes on the clock, and find functions for the x and y coordinates of each hand, in terms of time t seconds after 12:00.

Answer 1

hour hand = 1 foot

mnute hand = 2 feet

at 2pm, the hour hand is at 2

the minute hand is at 12

the angle between them is 360/6 = 60 degrees = pi/3

the distance between the two hands is the side opposite 60 degrees

use the Law of Cosines, a general version of the Pythagorean Theorem

c^2 = 2^2 +1^2 -2(2)(1)Cos60= 5-4(1/2) =3

c = distance between the end points of the two hands = sqr3= 1.732 feet

c^2 =5-4CosC

c=sqr(5-4CosC)

c'=[(1/2)/sqr(5-4CosC)](4SinC)

c'= 2SinC/sqr(5-4CosC)

c'(2pm) = 2Sin60/sqr(5-4Cos60) =(2sqr3/2)/sqr3 = 1 ft/hr

rate of change in distance at 2pm = 1 foot per hour