Two ants are taking a nap. The first one is resting at the tip of the minute hand of a cuckoo clock, which is 25 cm long. The second one is resting at the tip of the hour hand, which is half the length. At what rate is the distance between the two ants changing at 3:30?
I have drawn a diagram but do not know what to do after that point.
Let $\theta_t$ bet the angle of the minute hand in degrees. (t = minutes). $\theta_t = 6*t$. (As there are 360 degrees in a circle and 60 minutes in an hour there are 6 degree is a minute of time).
If the minute hand is of length 1. the the position of the minute hand $f(t) = (f_x(t), f_y(t)) = (\cos 6*t, \sin 6*t)$.
Let $\phi_t$ be the angle of the hour hand in degrees. (t = minutes). $\phi_t = t/2$. (In 60 minutes it will go 1/12 of a circle or 30 degrees, so in 1 minute it will go 1/2 degree ).
If the hour hand is of length 1/2. the the position of the hour hand $g(t) = (g_x(t), g_y(t))=(.5 \cos t/2, .5 \sin t/2)$.
$d(t)$ = distance between $f(t)$ and $g(t)$ = $\sqrt{(f_x(t) - g_x(t))^2 + (f_y(t) - g_y(t))^2 }$
$= \sqrt{(\cos 6t - .5 \cos t/2)^2 + (\sin 6t - .5 \sin t/2)^2 }$
$= \sqrt{\cos^2 6t + .25 \cos^ t/2 + \sin^2 6t + .25 \sin^ t/2 - \cos 6t \cos t/2 - \sin 6t \sin t/2 }$
$= \sqrt{1.25 - \cos 6t \cos t/2 - \sin 6t \sin t/2 }$
So rate the ants are moving away from each other is $d'(t)$. So figure out $d'(3:30) = d'(3*60 + 30)$.
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Or we could have taken t in hours:
cut and paste:
Let $\theta_t$ bet the angle of the minute hand in degrees. (t = hours). $\theta_t = 360*t$. (As there are 360 degrees in a circle, 1 hour = 360 degrees).
If the minute hand is of length 1. the the position of the minute hand $f(t) = (f_x(t), f_y(t)) = (\cos 360t, \sin 360t)$.
Let $\phi_t$ be the angle of the hour hand in degrees. (t = hours). $\phi_t = 30t$. (In an hour it will go 1/12 of a circle or 30 degrees).
If the hour hand is of length 1/2. the the position of the hour hand $g(t) = (g_x(t), g_y(t))=(.5 \cos 30t, .5 \sin 30t)$.
$d(t)$ = distance between $f(t)$ and $g(t)$ = $\sqrt{(f_x(t) - g_x(t))^2 + (f_y(t) - g_y(t))^2 }$
$= \sqrt{(\cos 360t - .5 \cos 30t)^2 + (\sin 360t - .5 \sin 30t)^2 }$
$= \sqrt{\cos^2 360t + .25 \cos^ 30 + \sin^2 360t + .25 \sin^ 30 - \cos 360t \cos 30t - \sin 366t \sin 30t }$
$= \sqrt{1.25 - \cos 360t \cos 30t - \sin 360t \sin 30t }$
So rate the ants are moving away from each other is $d'(t)$. So figure out $d'(3:30) = d'(3.5)$.
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$d(t)= \sqrt{1.25 - \cos 6t \cos t/2 - \sin 6t \sin t/2 }$
So $d'(t) = 1/2*\frac{1}{\sqrt{1.25 - \cos 6t \cos t/2 - \sin 6t \sin t/2 }}(\frac 1 2 \sin t/2*\cos 6t + 6\sin 6t*\cos t/2 - 6\cos 6t \sin t/2 - \frac 1 2 \cos t/2 \sin 6t)$
If t = 210, 6t = 1260 = 360*3 + 180. t/2 = 105.
$d'(210) = 1/2*\frac{1}{\sqrt{1.25 - \cos 180 \cos 105 - \sin 180 \sin 105 }}(\frac 1 2 \sin 105*\cos 180 + 6\sin 180*\cos 105 - 6\cos 180 \sin 105 - \frac 1 2 \cos 105 \sin 180)$
$= 1/2*\frac{1}{\sqrt{1.25 + \cos 105 }}(\frac 1 2 -\sin 105 + 6 \sin 105)$
$= 1/2*\frac{1}{\sqrt{1.25 + \cos 105 }}(\frac 1 2 -\sin 105 + 6 \sin 105)$
$= 1/2*\frac{1}{\sqrt{1.25 + \cos 105 }}(6.5\sin 105)$
$= \frac{3.25\sin 105}{\sqrt{1.25 + \cos 105 }}*25 \frac {cm}{minute}$