# Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 44

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At what rate is the distance between the tips of the hands changing at one'o clock

Given that the hour hand of a watch is 4mm long and the minute hand is 8mm long. Determine how fast is the distance between the tips of the hand changing at an one o' clock?

For our strategy, we will use the law of cosines to relate the length of the sides and the distance between the tips of the hands. Also, both of the hands are moving at constant rates so the angle between them is also changing at a constant rate.

Law of cosines: $a^2 = b^2 + c^@ - 2(b)(c) \cos \theta \qquad \Longleftarrow \text{ Equation 1}$

\begin{aligned} \text{we let } a & \text{ be the distance between the tips of the hands}\\ b & \text{ be the length of the hour hand}\\ c & \text{ be the length of the minute hand}\\ \theta & \text{ be the angle between the hands} \end{aligned}

At one o' clock, the angle between the hands is $30^\circ$, $\displaystyle \frac{360}{12} = 30^\circ$

\begin{aligned} a^2 &= 4^2 + 8^2 - 2(4)(8) \cos (30)\\ a &= 4.9573 \text{ mm} && \text{, the distance between the tips precisely at one o' clock} \end{aligned}

Notice that in Equation 1, the only variables that change are $a$ and $\theta$,

Now we differentiate Equation 1 with respect to time...

$\displaystyle 2a \frac{da}{dt} = -2 bc(-\sin \theta) \frac{d \theta}{dt}$

Solving for $\displaystyle \frac{da}{dt}$

$\displaystyle \frac{da}{dt} = \frac{bc \sin \theta}{a} \left( \frac{d \theta}{dt} \right) \qquad \Longleftarrow \text{ Equation 2}$

To solve for $\displaystyle \frac{d \theta}{dt}$, now the angle is changing with respect to time, note that the hour hand makes one revolution every hour in the clockwise direction. Hence, the hour hand angle has a rate of change of $2 \pi$ radians every 12 hours and is equal to $\displaystyle \frac{2\pi}{12} = \frac{\pi}{6} \text{rph}$ (radians per hour). Also, the minute hand has an angle that changes by $2 \pi$ radians every hour and is equal to $2 \pi$ rph. Thus, the relative velocity of the two hands is then $\displaystyle \frac{\pi}{6} - 2 \pi = \frac{-11\pi}{6} \text{rph}$ the value is negative because it is always the hour hand that takes the lead from the minute hand.

So $\displaystyle \frac{d \theta}{dt} = \frac{-11 \pi}{6}$, using this to solve for the rate of change of the distance between the tips of the hands in Equation 2.

\begin{aligned} \frac{da}{dt} &= \frac{bc \sin \theta}{a} \left( \frac{d\theta}{dt} \right)\quad ; \theta = 30^\circ\\ \\ \frac{da}{dt} &= \frac{4(8) \sin (30)}{4.9573} \left( \frac{-11 \pi}{6} \right) \end{aligned}

$\boxed{\displaystyle \frac{da}{dt} = -18.59 \frac{\text{mm}}{h}}$