Trigonometry and differentials type question?

Question

The law of cosines states that for triangle ABC, a^2 = b^2+c^2−2bc cos(∠A), where a, b and c denote the lengths of the sides opposite from the vertices A, B and C respectively.
Suppose that a large clock has a 2 foot hour hand and a 3 foot minute hand. How fast is the distance between the tips of the hands changing when it is 4:05? (Note: both hands are moving.) Give your answer in feet per hour.

dawonzcher · Accepted Answer

Let:
A= vertex between hour and minute hands oh the clock
b= the minute hand = 3 ft
c= the hour hand = 2 ft
a = the changing distance between the tips of the "hour and minute hands"

Now we solve for a and vertex A at time 4:05:
At exactly 4:00 o'clock, vertex A is 120 degrees
At 4:05, vertex A is 90.5 deg. Then solve for a:

a^2 = b^2 + c^2 - 2bcCos A
a^2 = 3^2 + 2^2 - 2(3)(2)Cos(90.5)
a^2 = 9 + 4 - 12(-0.0087)
a^2 = 13 + 0.1044)
a = sqrt(13.1044)
a= 3.62 ft ...<=====length of "a" at time 4:05

a^2 = b^2 + c^2 - 2bcCosA ...==>The Law of Cosine
2a(da/dt) = 0 + 0 - 2bc(-SinA)(dA/dt)...<===Taking the derivative.

Derivative of b^2 and c^2 are both zero since they are constant. Meaning, their respective length are not changing with respect to time.

2a da/dt  = 2bcSinA (dA/dt)......<=====Solve for da/dt
da/dt = ( 2bcSinA)(dA/dt) / 2a

Then Plug-in values: 
dA/dt = 390 deg / hr = 2pi + pi/6 = 13/6 pi
A = 90.5 deg
b = 3 ft
c = 2 ft

da/dt = ( 2*3*2(sin 90.5)(13/6 pi)] / 2(3.62)
da/dt = ( 12(0.9999)(13/6 pi) / 7.24
da/dt = ( 81.68) / 7.24
da/dt = 11.28 ft/ hr ....<=====Answer (The distance between the tips of the hour and the minute hands is changing at a rate of 11.28 ft / hr at time 4:05.

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Trigonometry and differentials type question?

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