I am studying calculus on my own. Using old text by Varberg and Purcell. Not sure if my solution is correct. (Cannot find online.) Differentiated using law of cosines but my rate of change seems quite fast. I proceeded as follows:

To find an overall $\frac{\mathrm{d}\theta}{\mathrm{d}t}$, I know that the angular rate of change of the minute-hand is $2\pi$ radians/hr., and that of the hour-hand is $\frac{\pi}{6}$ radians per hr. I subtracted the slower from the quicker to get $\frac{11\pi}{6}$ radians per hour, and allowed it to be negative, -$\frac{11\pi}{6}$, as it is in a clockwise direction.

I labelled the minute hand length in the triangle as $a$, the hour hand as $b$, and the variable distance between the tips of the hands as $c$. By the law of cosines:

$$c^2 = 5^2 + 4^2 - 2 (5\cdot 4) \cos\theta. $$

(I know that theta will be $90^{\circ}$, i.e. $\frac{\pi}{2}$ radians, at 3:00. Also, as it will be a right triangle, the distance $c$ at 3:00 will be the square root of $4^2 + 5^2$, i.e $\sqrt{41}$.

$$c^2 = 41 - 40 \cos\theta.$$

Then I differentiated with respect to time:

\begin{gather} 2c \frac{\mathrm{d}c}{\mathrm{d}t} = 0 - 40 \left[- \sin \frac{\pi}{2}\right] \frac{\mathrm{d}\theta}{dt} \\ \sqrt{41} \frac{\mathrm{d}c}{\mathrm{d}t} = 20 (1) \left(-\frac{11\pi}{6}\right) \\ \frac{\mathrm{d}c}{\mathrm{d}t} = -17.99 \text{ inches per hour}. \end{gather}

But this seems way too fast! If, so, where did I go off the tracks?

Many Thanks. Victor Jaroslaw, a beginning calculus student.

  • $\begingroup$ Do you mean the distance between the tips of the hands, or the angle between the hands? $\endgroup$ – MPW Dec 18 '17 at 22:33
  • $\begingroup$ I think it is perfectly right. $\endgroup$ – Abhiram Natarajan Dec 18 '17 at 22:51
  • $\begingroup$ If you convert to inches per minute, you get about $-0.3$. Doesn't that feel reasonable? $\endgroup$ – rogerl Dec 18 '17 at 22:58

It is right. One sanity test is this - 17.99 inches per hour is 0.3 inches per minute. The time at which the hour and minute hand are on top of each other is around 3:16.36. Assuming constant speed of 0.3 inches per minute, we can say that the distance reduced is around 4.9. The initial distance was $\sqrt{41}$ as you mentioned. So the new distance should be $\approx \sqrt{41} - 4.9 \approx 1.5$. The right answer is of course 1 (because the lengths are 5 and 4), so this answer is not too off right?

P.S. Obviously it is terribly wrong to assume constant speed, but it is fine as a heuristic given the amount of time elapsed is not too much. Anyway, the assumption was not to get an accurate answer anyway. It was, as I said, for a sanity test.


Three o'clock is a special time because the tip of the minute hand has all of the change in $x$, and the tip of the hour hand has all of the change in $y$.

So, let's try this. The distance $D=\sqrt{x^2+y^2}$, and the change can be expressed as

$$\frac{dD}{dt} = \frac{\partial D}{\partial x}\frac{dx}{dt} + \frac{\partial D}{\partial y}\frac{dy}{dt}.$$

The partials are

$$\frac{\partial D}{\partial x} = \frac{x}{D};\frac{\partial D}{\partial y} = \frac{y}{D}$$

The change in the distance $D$ with respect to $x$ is negative at three o'clock, because of the motion and position of the hands. Likewise, the change in $D$ with respect to $y$ is positive. At three o'clock, $x=4, y=5, D=\sqrt{41},$ so $\frac{\partial D}{\partial x} = -4/\sqrt{41}$ and $\frac{\partial D}{\partial y} = 5/\sqrt{41}$.

The velocities of the hands are $10\pi$ and $2\pi/3$ inches per hour for the minute and hour hands, respectively. At three o'clock, the minute hand velocity is aligned exactly on the $x$ direction, and the hour hand velocity exactly on the $y$ direction.


$$\frac{dD}{dt} = \frac{-4\cdot 10\pi + 5 \cdot 2\pi/3}{\sqrt{41}} \doteq -17.99 \text{ inches/hr}.$$

Looks like we agree!


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