# Calc.RR1

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First consider the distance between the tip of the second hand and the 9 o'clock position; let us call this $r$. Further, let $\theta$ be the angle between the 9 o'clock position and the second hand, and let $t$ be the time, in seconds. Then we have ${\frac {d\theta }{dt}}={\frac {2\pi \ {{\rm {radians}}}}{60\ {{\rm {seconds}}}}}={\frac {\pi }{30}}{{\rm {rad/sec}}}$. If we use the Law of Cosines, we can get a formula for $r$ in terms of $\theta$. That is, $r^{2}=6^{2}+5.5^{2}-2\cdot 6\cdot 5.5\cos \theta$. (Draw a picture to help visualize this if you need to. The two smaller legs of the triangle will always be 6 and 5.5 inches, for the 9 o'clock position and second-hand tip, respectively. As $\theta$ changes, so does $r$.) Using implicit differentiation, we get ${\frac {dr}{d\theta }}={\frac {6\cdot 5.5\sin \theta }{r}}$. Thus, the desired related rate is calculated using the chain rule: ${\frac {dr}{dt}}={\frac {dr}{d\theta }}\cdot {\frac {d\theta }{dt}}={\frac {11\pi \sin \theta }{10r}}$. At 3:30, we have $\theta ={\frac {\pi }{2}}$ and $r={\sqrt {6^{2}+5.5^{2}}}\approx 8.1394$. Then ${\frac {dr}{dt}}\approx 1.39\,{{\rm {in/sec}}}$.
The desired rate for the minute and hour hands is calculated in a nearly identical fashion. For the minute hand, we have ${\frac {d\theta }{dt}}={\frac {2\pi \ {{\rm {radians}}}}{3600\ {{\rm {seconds}}}}}={\frac {\pi }{1800}}{{\rm {rad/sec}}}$ and ${\frac {dr}{d\theta }}={\frac {6\cdot 5\sin \theta }{r}}$, so ${\frac {dr}{dt}}={\frac {\pi \sin \theta }{60r}}$. At 3:30, $\theta =-{\frac {\pi }{2}}$ and $r={\sqrt {6^{2}+5^{2}}}\approx 7.8102$, so ${\frac {dr}{dt}}\approx -0.0067\,{{\rm {in/sec}}}$, or $\approx -0.4022\,{{\rm {in/min}}}$.
For the hour hand, we get ${\frac {d\theta }{dt}}={\frac {\pi }{21600}}{{\rm {rad/sec}}}$ and ${\frac {dr}{d\theta }}={\frac {6\cdot 3\sin \theta }{r}}$. Then ${\frac {dr}{dt}}={\frac {\pi \sin \theta }{1200r}}$. At 3:30, if we assume that $\theta =\pi$, we see that this rate is 0. This is rather intuitional, as we may think of this as the peak in midair when we throw a ball up into the air; at this peak, the ball's velocity is 0. We can get a bit fancier though, as at 3:30 we could assume that the hour hand is halfway between the 3 and 4 o'clock positions. That is, at 3:30, $\theta =-{\frac {11\pi }{12}}$, so $r={\sqrt {6^{2}+3^{2}-2\cdot 6\cdot 3\cos(-11\pi /12)}}\approx 8.9316$. Then, ${\frac {dr}{dt}}\approx -7.5864e-4\,{{\rm {in/sec}}}$, or $\approx -0.0046\,{{\rm {in/min}}}$, or $\approx -0.2731\,{{\rm {in/hr}}}$.