Page 247 - Calculus Demystified

CHAPTER 8
Applications of the Integral
234
2
distance moved is x feet and the force applied is about F(x) = 3x −x +10.
See Fig. 8.23. Thus work performed in that little bit of the move is w(x) =
2
(3x − x + 10) · x. The aggregate of the work is obtained by summation. In
this instance, the integral is the appropriate device:
10 2
10
2 3 x
W = 3x − x + 10 dx = x − + 10x = 1050 foot-pounds.
0 2 0
EXAMPLE 8.14
A man is carrying a 100 lb sack of sand up a 20-foot ladder at the rate of
5 feet per minute.The sack has a hole in it and sand leaks out continuously
at a rate of 4 lb per minute. How much work doesthe man do in carrying
the sack?

Fig. 8.24

SOLUTION
It takes four minutes for the man to climb the ladder. At time t, the sack has
100 − 4t pounds of sand in it. From time t to time t + t, the man moves
5· t feet up the ladder. He therefore performs about w(t) = (100−4t)·5 t
foot-pounds of work. See Fig. 8.24. The total work is then the integral
4 4

W = (100 − 4t) 5 dt = 500t − 10t 2 = 1840 foot-pounds.

0 0