enter image description here

Take this question:

"We follow the tips of the hands of an old fashioned analog clock (360 degrees is 12 hours) . We take the clock and put it into an axis system. The origin (0,0) of the axis system is the rotationpoint of the hands. The positive x-as goes through "3 hour" and the positive y-as through "12 hour" We calculate the time "t" in hours, starting from 0:00 hours.

The equation for the tip of the big clockhand is: x=3sin2πt, y=3cost2πt

The equation for the tip of the small clockhand is x=2sin(1/6)πt, y=2cos(1/6)πt

On t=0 the two hands overlap eachother. Calculate the first point in time after t=0 when this occurs.


"This is true when cos(2πt)=cost(1/6πt) and sin(2πt)=sin(1/6πt) So, t = 12/11"

I simply don't know where to start...

  • $\begingroup$ I think you want $x=3 \cos{2 \pi t}$... $\endgroup$ – Ron Gordon Mar 24 '13 at 13:32
  • $\begingroup$ please elaborate. $\endgroup$ – user1095332 Mar 24 '13 at 13:35
  • $\begingroup$ In fact, as I read through this problem, the whole thing makes no sense whatsoever. How on earth does one make an equilateral triangle from a side of $2$ and another of $3$? As for the cosine, read your statement as to the positive $x$ axis. $\endgroup$ – Ron Gordon Mar 24 '13 at 13:41
  • $\begingroup$ different question above $\endgroup$ – user1095332 Mar 24 '13 at 13:55

Let our time $t$ be $x$ hours and $y$ minutes where $0 \leq x < 12$ and $0 \leq y < 60$.

Let $\theta_1$ be the angle of the minute hand and $\theta_2$ be the angle of the hour hand where we measure angles clockwise beginning at 12 o'clock so that $0 \leq \theta_1, \theta_2 < 360^{\circ}$

It follows that $\theta_1 = 360 \times \frac{y}{60}$ and $\theta_2 = 360 \times \frac{x}{12} + 360\times \frac{y}{60\times 12} = 360(\frac{x}{12} + \frac{y}{60\times 12})$.

If the angle of the hour hand is the same as the minute hand, we have $\theta_1 = \theta_2$ and hence:

$$ \frac{y}{60} = \frac{x}{12} + \frac{y}{60\times 12}$$

We know that the hour and minute hands will have equal angles at some time shortly after 1 o'clock, so setting $x=1$ gives:

$$\frac{y}{60} - \frac{y}{60 \times 12} = \frac{1}{12} \iff \frac{12y-y}{60\times 12} = \frac{1}{12} $$

Solving for $y$:

$$11y = 60 \iff y = 60/11$$

So the two angles are equal $1$ hour and $60/11$ minutes after 12 o'clock, which is $1 + 1/11 = 12/11$ hours after 12 o'clock.

  • $\begingroup$ Thanks, but what about the formulas? How to use them in my calculator? $\endgroup$ – user1095332 Mar 24 '13 at 18:18

For the overlap thing, the next time there's an over lap is when the minute hand has already gone through a full hour (revolution). Then

$$2 \pi (t+1) = \frac{\pi}{6} t \implies t + 1 = \frac{12}{11}$$

  • $\begingroup$ How did you get: $$2 \pi (t+1) = \frac{\pi}{6} t $$? $\endgroup$ – user1095332 Mar 24 '13 at 14:18
  • $\begingroup$ $t$ is in hours, as you said. As I said, the next time there's an overlap is when the minute hand goes through a full revolution. Thus, the $t+1$. $\endgroup$ – Ron Gordon Mar 24 '13 at 14:19
  • $\begingroup$ $$t + 1 = \frac{12}{11}$$ Does that mean $$t = \frac{12}{11} - 1$$ $\endgroup$ – user1095332 Mar 24 '13 at 14:20
  • $\begingroup$ It does not seem to work when I plot it on wolframalpha nor my TI-84 What should I plot on my TI calculator? $\endgroup$ – user1095332 Mar 24 '13 at 14:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.