Anonymous From: - Posts: - Votes: - |
The minute hand of a certain clock is 4 in long. Starting from the moment when the hand is pointing straight up, how fast is the area of the sector that is swept out by the hand increasing at any instant during the next revolution of the hand? Please be VERY CLEAR and show ALL STEPS :) |

ZenDerived25 From: US Posts: 4 Votes: 12 |
We need to "relate" area of a circle to angular speed (of the minute hand). It doesn't even have to be a calculus problem because we are dealing only with constant rates. Since we know that the area of the entire circle is pi*r^2, then pi*(16) is the area. We cover 1/60th of this area in one minute, and if you divide the area by 60 you get 0.84 in^2/min. That's the amount of the area swept out in one minute, and that is our answer. It's the "rate of change" of area, in terms of in^2/min. |

FabulousBob From: US Posts: 4 Votes: 8 |
A circle with a 4 inch radius will have an area of: A = 2 p r^2 = 2 p (4 inches)^2 = 8 p inches^2 That is swept in 60 seconds, at a rate of 8 p inches^2 / 60 seconds = 2 p inches^2/15 seconds Multiply the time by the rate to get the area. A(t) = t * 2 p inches^2/15 seconds Using the methods of a solid of revolution is overkill in this situation, as that is aimed as 3-dimensional objects, not 2-dimensional figures. |

Anonymous From: - Posts: - Votes: - |
A = prē * t/60 dA/dt = prē/60 = p*4ē/60 = 4p/15 |

Anonymous From: - Posts: - Votes: - |
Area of the whole face under 4 in hand in one revolution = pi*r^2 since there are sixty minutes thus the area swept in on min = pi*r^2/60 = 3.142 * 4^2 /60 = 0.837 in^2 |

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