# Gert K. Pedersen's Analysis Now PDF

By Gert K. Pedersen

ISBN-10: 0387967885

ISBN-13: 9780387967882

ISBN-10: 3540967885

ISBN-13: 9783540967880

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Sample text

The side of the bottom is p feet in length, and water flows in the reservoir at the rate of c cubic feet per minute. Find an expression for the rate at which the surface of the water is rising at the instant its depth is h feet. Calculate this rate when p = 17, h = 4 and c = 35. The volume of a frustum of pyramid of height H, and of bases A and a, is V =~(A +a+ J Aa). It is easily seen that, the slope being 45°, for a depth of h, the length of the side of DIFFERENTIATION 39 the upper square surface of water is (p + 2h) feet ; thw01 A = p 2 , a=(p+2h) 2 and the volume of the water is lh{p 2 +p(p+2h)+ (p+2h) 2} cubic feet =p 2h+2ph2 +th3 cubic feet.

57) ; but we can nevertheless manage it now without any difficulty. Developing the cube, we get v = 27t6 - 32·4t5 + 39·96t4 - 23·328t3 + 13·32t2 - 3·6t + 1 ; dv hence dt = 162t5 -162t4 + 159·84t3 - 69·984t2 + 26·64t- 3·6. (5) Differentiate y=(2x-3)(x+1) 2. dy =( 2 x _ 3 ) d[(x+ 1)(x+ 1)] + (x+ 1)2 d(2x -3) dx dx dx =(2x-3)[(x+1)d(x+l) +(x+1)d(x+ 1 dx dx d(2x-3) 1)2 ( dx + x+ )J =2(x + 1)[(2x -3) + (x + 1)] =2(x + 1)(3x -2); or, more simply, multiply out and then differentiate. DIFFERENTIATION 37 y=0·5x 3(x-3).

57) ; but we can nevertheless manage it now without any difficulty. Developing the cube, we get v = 27t6 - 32·4t5 + 39·96t4 - 23·328t3 + 13·32t2 - 3·6t + 1 ; dv hence dt = 162t5 -162t4 + 159·84t3 - 69·984t2 + 26·64t- 3·6. (5) Differentiate y=(2x-3)(x+1) 2. dy =( 2 x _ 3 ) d[(x+ 1)(x+ 1)] + (x+ 1)2 d(2x -3) dx dx dx =(2x-3)[(x+1)d(x+l) +(x+1)d(x+ 1 dx dx d(2x-3) 1)2 ( dx + x+ )J =2(x + 1)[(2x -3) + (x + 1)] =2(x + 1)(3x -2); or, more simply, multiply out and then differentiate. DIFFERENTIATION 37 y=0·5x 3(x-3).