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Unformatted text preview: 18.014 Problem Set 6 Solutions Total: 24 points Problem 1: A water tank has the shape of a right-circular cone with its vertex down. Its alititude is 10 feet and the radius of the base is 15 feet. Water leaks out of the bottom at a constant rate of 1 cubic foot per second. Water is poured into the tank at a constant rate of c cubic feet per second. Compute c so that the water level will be rising at a rate of 4 feet per second at the instant when the water is 2 feet deep. Solution (4 points) The relationship of interest in this problem is V = r 2 h/ 3, where V,r,h are, respectively, volume of water in the cone, radius of the cone at the maximum water height, and height of the water in the cone. Notice that all known quantities are related to volume and height, so determining V in terms of h will make things a bit simpler. Using similar triangles, we see that h/ 10 = r/ 15 or 3 h/ 2 = r . Substituting we get, V = 3 h 3 / 4. Now, as both h and V depend on t we compute dV 9 h 2 dh = . dt 4 dt Now, h = 2 ft , dh/dt = 4 ft/s , and dV/dt = ( c 1) ft 3 /s . That is, 9 4 c 1 = 4 = 36 . 4 Thus we need c = (36 + 1) ft 3 /s . Problem 2: Let f ( x ) = 1 x 2 / 3 . Show that f (1) = f ( 1) = but that f ( x ) is never zero in the interval [ 1 , 1]. Explain how this is possible in view of Rolles theorem. Solution (4 points) As ( 1) 2 / 3 = (( 1) 1 / 3 ) 2 = 1, the equalities f (1) = f ( 1) = follow trivially. From Example 6 (page 163), we can set n = 3 and consider the limit for all x = 0. (Recall we have defined the n-th root for x < when n is odd.) Then, following the argument of Example 3 (page 166), we see f ( x ) = 2 / 3 x 1 / 3 for all x = 0. In particular, wherever f is defined, f = 0. Rolles theorem states that for a continuous function f on [ a,b ] that is differen- tiable on ( a,b ), if f ( a ) = f ( b ) then there exists c ( a,b ) such that f ( c ) = 0. The 1 reason that the conclusion of Rolles theorem fails here is that one...
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