Linear Approximation — horizontal change in position of tip of a clock hand.

Question

The problem is:

A clock's minute hand has a length of $r$ meters, linearly approximate the change in the horizontal distance of its tip as the arm moves from 6.01 am to 6.02 am.

I've worked it out as follows:

$\quad \quad f(x) \approx f'(a)(x-a)+f(a)$

$\quad \quad f(x) - f(a) \approx f'(a)(x-a)$

$\quad \quad \Delta f \approx f'(a)(x-a)$

Since the horizontal position in a circle is given by the cosine value of the angle, and since at $\quad \quad$ 6.00 am the change is $0$, and also since a $15$ minutes duration corresponds to turning by $\frac{\pi}{2}$ radians, a one minute duration corresponds to turning by $\frac{\pi}{30}$ radians.

Then at $6.01 \, am$, the change in the horizontal position of the tip of the minute hand is:

$\quad \quad \quad \quad \Delta f \approx f'(a)(x-a)$

$\quad \quad \quad \quad \quad \,\,\,\, \approx r \cdot \frac{d}{d\theta}\Bigg(cos\theta \bigg\rvert_{\theta=\frac{\pi}{2}}\Bigg)\Bigg(\Big(\frac{\pi}{2}-\frac{\pi}{30}\Big)-\frac{\pi}{2}\Bigg)$

$\quad \quad \quad \quad \quad \,\,\,\, \approx r \cdot -sin(\frac{\pi}{2})(-\frac{\pi}{30})$

$\quad \quad \quad \quad \quad \,\,\,\, \approx r \cdot \frac{\pi}{30}$

By the same reasoning, the change in the horizontal position of the tip between $6.01 \, am$ and $6.02 \, am$ is:

$\quad \quad \quad \quad \quad \,\,\,\, \approx r \cdot \frac{d}{d\theta}\Bigg(cos\theta \bigg\rvert_{\theta=\frac{14}{30}\pi}\Bigg)\Bigg(\Big(\frac{14}{30}\pi-\frac{\pi}{30}\Big)-\frac{14}{30}\pi\Bigg)$

$\quad \quad \quad \quad \quad \,\,\,\, \approx r \cdot -sin(\frac{14}{30}\pi)(-\frac{\pi}{30})$

$\quad \quad \quad \quad \quad \,\,\,\, \approx 0.9945 \cdot \frac{\pi r}{30}$

Problem is, that's not the answer. I don't get the answer until I reach a number of failed attempts. The answer is required in terms of $\pi$. Please help.

Answer 1

The angle of rotation of the clock's minute hand is $2\pi$ for $60$ minutes.

For one minute the angle of rotation is $\frac{2\pi}{60}=\frac{\pi}{30}$. So the tip of the hand travels $\frac{\pi}{30}r$ on the circle.

This is a short distance, compared to the circumference of the circle. So the small arc of circle can be confused with a straight line of same length, which is quite horizontal at 6am.

Thus in a first approximate the answer is $\simeq\frac{\pi}{30}r$ .

This is consistent with your more accurate result $\quad 0.9945\frac{\pi}{30}r\quad$ since $\quad 0.9945\simeq 1.$

Answer 2

If we measure from the center of the clock, at $6:01$ the $x$ position of the hand is $r\cos \frac \pi {30}$ and at $6:02$ it is $r \cos \frac \pi{15}$. If the time is $t$ minutes past $6:01$ the linear approximation to the horizontal position is $r\cos \frac \pi {30}+t(r \cos \frac \pi{15}-r \cos \frac \pi{30})$. It sounds like you are asked for one value, but then I don't understand what the question is asking for.