Evaluate the following limit without using l Hopital s Rule. x x. = lim = (1)(1) = lim. = lim. = lim = (3 1) =

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Duane Douglas
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1 5.4 1 Looking ahead. Example 1. Indeterminate Limits Evaluate the following limit without using l Hopital s Rule. Now try this one. lim x 0 sin3x tan4x lim x 3x x 2 +1 sin3x 4x = lim x 0 3x tan4x ( ) 3 = (1)(1) 4 = lim x x = lim x x = lim x x 3 4 ( ) x x ( ) x ( 3 = (3 1) = x x 2 )

2 Work In this section we define and learn how to calculate the mathematical quantity work. Work Done by a Constant Force When an object moves a distance d along a straight line as the result of a force, with constant magnitude F in the direction of motion acting on it, we define the work W done by the force on the object by W = Fd Recall that the unit of force is the pound in the British system and is called the newton in International System (SI units). Hence the units for work are newton-meter (SI units) foot-pounds (British units) Example 2. Professional bowlers regularly lift a bowling ball from a height of 18 in to a height of 60 in. How much work, in British units, is required to lift a 16 pound ball? W = Fd = = 56 foot-pounds

3 5.4 3 Work Done by a Variable Force When working with a variable force, say compressing a spring, we must replace the formula W = Fd. Now suppose that a variable force of magnitude F(x) acts along the x-axis moving an object from x = a to x = b. As usual we let P = {x 0,x 1,x 2,...,x n } be a partition of the interval [a,b]. Now if F is continuous it will not vary much over the subinterval [x j 1,x j ] so that the work done over this subinterval is approximately F(c j ) x j for some c j [x j 1,x j ]. It follows that the work done on the object over the interval [a, b] is approximated by W n F(c j ) x j j=1 Notice that this is a Riemann sum. We expect this approximation to get better and better as the norm of the partition goes to zero. Thus lim P 0 n F(c j ) x j = j=1 ˆ b a F(x)dx Definition. The work done by a variable force of magnitude F(x) directed along the x-axis from x = a to x = b is given by W = ˆ b a F(x)dx

4 5.4 4 It is well known that the force required to stretch or compress a spring from its natural length is proportional to the displacement. Let x equal this displacement. Then Hooke s Law states that F = kx with the spring constant k depending on the thickness and materials of the spring.

5 5.4 5 Example 3. A spring has a natural length of 10 in. It requires a of 600 lb force to compress the spring to 7 in. a. Find the spring constant k. k = F x = 600 3/12 = 2400 lb/ft b. How much work is required to stretch the spring from 10 in to 16 in? W = ˆ 1/ x dx = 1200x 2 1/2 0 = 300 ft-lb c. How far beyond its natural length will a 900 lb force stretch the spring? x = F k = = ft

6 5.4 6 Example 4. Pumping Liquid from Containers The 10 ft high parabolic tank (shown above) is filled with water to height of 9 ft. How much work does it take to pump the water to the top of the tank? 10 y = x 2 9 radius = y The tank s cross-section is shown above.

7 5.4 7 We proceed as follows. As before, we partition the interval (along the y-axis) [0,9] and we approximate the work required to lift a small slab from a height of y ft to the top of the tank. The force acting on this slab is its weight. (Recall that the weight of a fluid is given by its volume the weight-density ρ of the fluid.) Thus V πr 2 y = π( y) 2 y and so the weight of the slab is F(y) ρ V = 62.4 V (for water ρ = 62.4 lb/ft 3 ) = 62.4πy y Now the work required to lift this slab to the top of the tank is W (10 y) F(y) (10 y)(62.4π)y y and the work required to pump all of the water to the top of the tank is W W = (10 y)(62.4π)y y

8 5.4 8 It follows that the work required to pump the water from the tank is W = ˆ 9 0 dw ˆ 9 = 62.4π y(10 y)dy 0 ) 9 = 62.4π (5y 2 y3 3 = 62.4π(162) ft-lbs 0

9 5.4 9 Fluid Pressures and Forces Many students have experienced the increase in pressure (force per unit area) as one descends in a body of water. The increase in pressure as a function of depth is governed by the following. The Pressure-Depth Equation. In a standing fluid of weight-density ρ, the the pressure p at depth d is given by p = ρd The resulting force on a constant-depth surface is then given by F = pa = ρda For example, the pressure exerted on the floor of a 8 feet deep in swimming pool is p = 62.4 lb/ft 3 8 ft = lb/ft 2 If the swimming pool has dimensions 12 ft 20 ft, the resulting force on bottom of the pool is F = lb/ft ft 2 = 119,808 lb Suppose instead we wish to measure the force against a submerged vertical surface, e.g. the base of a levee. In this case the pressure is not constant.

10 Example 5. The vertical ends of a 15 ft long watering trough are isosceles triangles (see sketch). If the water depth is 8 ft, what is the force exerted by the water on one end of the trough? Assume that the water weighs ρ lb/ft 3. 8 ft (4,10) 15 ft Guided by our early success with work, we try a similar approach.

11 We partition the interval (along the y- axis) [0, 8] as usual and approximate the force across each the generated thin vertical surfaces of width 2x and height y (see sketch). Since the pressure is a continuous function of depth, the pressure will not vary much provided y is small. We then sum the forces on each of these vertical surfaces as we have done above. 8 ft y = 5x/2 width = 2x = 4y/5 Notice that the width and depth of the surface at y is given by 2x = 4y/5 and 8 y, respectively. It follows that the force exerted across each of these vertical surfaces is given by F ρ(8 y) 4y 5 y

12 Hence the force exerted on one end of the trough is F F = ρ(8 y) 4y 5 y Thus F = ˆ 8 0 ρ(8 y) 4y 5 dy =. = ρ ft-lb Does the length of the trough affect the result?

APPLICATIONS OF INTEGRATION

6 APPLICATIONS OF INTEGRATION APPLICATIONS OF INTEGRATION 6.4 Work In this section, we will learn about: Applying integration to calculate the amount of work done in performing a certain physical task.