Solved by Dave Newquist

Question:

The hands of an accurate clock are 3 inches and 4 inches long. When it's 8 o'clock, at what rate is the distance between the tips of the hands changing?

Answer:

First we need a function to describe the distance between the tips of each hand on this clock.

The law of cosines (a^2 = b^2 +c^2 - 2*b*c*cos A) describes the square of the distance between the tips. So Let

f(x) = squareroot(3^2 +4^2 - 3*4*cos(x))

f is a function from radians to inches, and we would like a function from hours to inches.

We need another function, g(x):hours->radians. In other words, given the hour, we need a function to give us the angle between the two hands. The big hand is moving at a rate of 2*PI/hr. The small hand is moving at (PI/6)/hr. So the angle between them after x hours is 2*PI -(PI/6), which is 11*PI/6. So g(x) = 11*PI*x/6.

f(g(x)) is the function we are looking for.

f(g(x)) = squareroot(25- 24*cos(11*PI*x/6))

The derivative of this function is 24*(11/6)*PI*sin(11*PI*x/6)/2*squareroot(25- 24*cos(11*PI*x/6)) Evaluated at 8, this is 11*PI*squareroot(3)/squareroot(37).

So the distance between the tips of the hands is changing at a rate of about 9.84 inches/hr at 8:00.

-- Dave Newquist