Related Rates: Clock

Question

Please answer with work too. The hour hand of a clock is 5 inches long and the minute hand is 7 inches long. At the moment the hour hand is positioned at 4 and the minute hand is positioned at 5, what is the instantaneous rate of change of the distance between the tip of the hour hand and the tip of the minute hand? Express the rate of change in inches per second, and round your answer to four decimal places.

Answer 1

hour hand is 5 inches, minute hand is 7 inches

hour hand is at 4 o'clock, minute hand is at 5 o'clock

what is the instantaneous rate of change of the distance from the tip of hour hand to the tip of the minute hand?

hour hand at an angle with the horizontal of - 30 degrees =-pi/6

minute hand angle with horizontal = -60 degrees = -pi/3

distance at the moment = d where using the Law of Cosines

d^2 =a^2+b^2 -2abCosC

d^2 = 5^2 +7^2 -2(5)(7)Cos30 =25+49-70(sqr3/2)=about74-.866(70)= 13.3782

d= sqr13.3782 =about 3.65763

take the derivative with respect to time

2dd'= - 2ab(sinC)C' since a'=b' =0 the hands don't change length

d' =2(5)(7)sin30C'/3.65763

=70(.5)C'/3.65763

=(35/3.65763)C'

= 9.56905C' where C' is the rate of change of the angle between the 2 hands

C' = rate of change of the minute hand minus rate of change of the hour hand

= 90/15 - 90/180 = 4-1/2= 3.5 degrees per minute

d' = 9.56905(3.5)

=33.4917 inches per minute

=33.4917/60 in/sec

=about -5582 inches per second

d' = 0.5582 inches per second = instantaneous rate of change of the distance between the hand tips

Answer 2

Look at

d²=(7cosθ₁-5cosθ₂)²+(7sinθ₁-5sinθ₂)²

at θ₁=-60 degrees and θ₂=-30 degrees

d=3.658 in.

taking derivative with respect to t

2dd'=2(7cosθ₁-5cosθ₂)(5sinθ₂(θ₂')-7sinθ₁(θ₁'))+2(7sinθ₁-5sinθ₂)(7cosθ₁(θ₁')-5cosθ₂(θ₂')

θ₁' is constant = -2πradian/hour-=-.00175 radian/s.

θ₂' is constant = -2π/radian/12hour=-.000145 radian/s.

plugging in given values and calculated values

d'=.007678 in./s.