# The Minute Hand Of A Certain Clock Is 4 In Long. S

 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 :) BigNewton From: USPosts: 5Votes: 15 A = pr² * t/60 dA/dt = pr²/60 = p*4²/60 = 4p/15  Anonymous From: -Posts: -Votes: - 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: - Area swept in one min = pr^2/60 = p*16/60=0.84 in^2  Enter 