Calc.RR1

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A clock face has a 12 inch diameter, a 5.5-inch second hand, a 5 inch minute hand and a 3 inch hour hand. When it is exactly 3:30, calculate the rate at which the distance between the tip of any one of these hands and the 9 o'clock position is changing.


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}}}.