Math 122 Fall Solutions to Homework #5. ( ) 2 " ln x. y = x 2

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1 Math 1 Fall 8 Solutions to Homework #5 Problems from Pages (Section 7.) 6. The curve in this problem is defined b the equation: = ( ) " ln and we are interested in the part of the curve between = and =. The derivative is given b: d = " 1, and the integral for the arc length will be: * # Arc length = 1+ % " 1, & ) ( d =, $ ', +, ( ) 16( 1+ ) + ln( ) / / /. / = 6 + ln( ). 18. The curve in this problem is the ellipse defined b the equation: a + b =1. # Rearranging to make the subject of this equation gives: = ± b 1" & % (. The $ a ' derivative of with respect to is given b: = ± b a " $ $ b 1# '' & & )) % % a (( #1 = ± b a a #. The integral for the length of the portion of the ellipse that lies above the -ais is: a # b & a a ) " a + b 1+ % ( d = ) d. $ a a " ' a " a "a "a

2 The total arc length of the entire ellipse (including the half above the -ais and the half below the -ais) will be twice this integral.. There are man different was to attack this problem. The solution given here has the virtue of being relativel short but ma not relate well in our mind to what we studied when we learned about arc length in lecture. If so, come to office hours to discuss this (or alternative) methods of solution. Here we will interchange the roles of and in the usual arc length integration formula so that: Arc length = 1+ d " % ( $ '. # & = b = a The integral that we will approimate using Simpson s rule will then be: = " Arc length = % ( $ '. =1 # & The approimate value of this integral given b Simpson s rule with n = 1 is: Arc length (This figure was obtained b calculating a midpoint and a trapezoid sum for the above integral using n = 5 and then combining one third of the trapezoid total with two thirds of the midpoint total.) Calculator integration on a TI-83 Plus (see below) gave the result shown below, which is almost eactl the same value furnished b Simpson s rule. Problems from Pages (Section 7.5) 1. In writing this solution I assume that at the end of the chain lifting operation, there are still four meters of chain ling on the ground in a horizontal position and the top of the suspended part of the chain is si meters off the ground (see diagrams

3 below). If our assumptions about the final configuration of the chain are different then ou ma have set up the problem in a different wa. Before lifting After lifting d 6 - Let s begin b focusing on the work done to raise the small piece of chain (shaded in the above diagram) that lies between and + d. The vertical distance that this piece of chain moves is the distance 6 meters. The force eerted b gravit on this small piece of chain is g times the mass of the small piece of chain. This force is given b ( 9.8) ( 8 1)d. Multipling these two together gives the work to raise the small piece of chain: Work to raise small piece of chain = ( 6 " ) ( 9.8) 8 ( )d. The work done to raise the first si meters of chain can be epressed as the Riemann sum: N$1 Total work = Lim N "# 6 $ k % 6$ 8 &( N ) % ( 9.8) % ( 1) % k % 6$ k= As we take the limit as N, the total work is given b the definite integral: 1 N. Total work = 6 # ( 6 " ) ( 9.8) ( 8 1)d = " 1 [ ] 6 =111. Nm. 16. The swimming pool is illustrated hereafter. To calculate the amount of work needed to empt the pool, we will slice the water up into pieces. The defining characteristic of these pieces is that each molecule of water contained in the piece move approimatel the same vertical distance to leave the pool. This means that the pieces will be thin, horizontal slices of water as shown in the diagram (dark gra).

4 The distance moved b all the water molecules in the dark gra horizontal slice will be 5 feet. To get the force eerted b gravit on the dark gra slice we need to multipl the densit of 6.5 pounds per cubic foot b the volume of the dark gra slice (epressed in cubic feet). Note that we do not have to multipl b 9.8 m/s (or its imperial equivalent of 3 ft/s ) because pounds are alrea units of force. The volume of the dark gra slice (a disk with radius 1 feet and thickness feet) is " #1 # cubic feet so the force eerted b gravit is 6.5" # "1 " pounds. Multipling distance moved b force gives work, so the work (in units of foot pounds) required to remove the dark gra slice of water from the swimming pool will be: Work to remove dark gra slice = ( 5 " ) # 6.5# $ #1 #. The total amount of work to remove all of the water from the pool will be the integral of this from the value of where the water begins ( = ) to the value of where the water ends ( = ). This integral is: Total work = % ( 5 " ) # 6.5 # $ #1 # = 9# $ # 5 " 1 [ ] = ft-lbs. 18. The method of slicing the water in the hemispherical tank is eactl the same as the clindrical swimming pool featured in the previous problem. The main difference between this problem and the previous one is that the radius of this tank varies with (rather than staing constant).

5 A sidewas view of the hemispherical tank (with a horizontal slice of water shown in gra) is given in the following diagram. From the appearance of the diagram, the vertical distance moved b the slice as it is pumped out of the tank is. + = 5-5 As we are again dealing with imperial units, the force eerted on the slice b gravit will be given b 6.5 pounds per cubic foot times the volume of the slice. The slice is shaped like a disk with radius and thickness so the volume of the slice will be: Volume of the slice = " # # = " # ( 5 $ ) #. The force eerted b gravit on the slice will be 6.5 " # " ( 5 $ ) " and the work done (in units of foot pounds) to remove this slice from the tank will be " # 6.5# $ # ( 5 " ) #. The total work to remove all of the water from the tank will be the integral of this from the -value where the water begins ( = 5) to the -value where the water ends ( = ). This integral is as follows: "5 ( ) # [ ] "5 Total work = % " # 6.5 # $ # 5 " "5 = # + 1 = ft-lbs.. To compute the hdrostatic force eerted on one side of the triangular metal plate, we need to slice the triangular plate into pieces so that each point of each piece eperiences approimatel constant pressure. As pressure depends on depth, these pieces will be horizontal rectangles as shown in the following diagram.

6 1 Surface of the water = 3 " 3 The force eerted on the horizontal rectangle is the product of pressure and area. The pressure is the product of the densit of water (6.5 pounds per cubic foot) and the depth below the surface. As shown in the previous diagram, the depth below the surface is given b 1, so the pressure is 6.5 (1 ). The area of the rectangle is the width of the rectangle multiplied b. The width of the rectangle can be written as so that the area of the rectangle is. However, this will not be the most convenient epression for the area when we set up the integral to give the total force. To create a more convenient epression, we need to substitute for the in this area formula. The tool that enables us to make this substitution is an equation for the straight line that forms the diagonal side of the rectangle in the previous diagram. This is the equation for the line that joints the points (, 3) and (, ) which is: = 3 " 3 or = 3 ( + 3). ( ) " and the ( ) " " ( ) ". With this epression, the area of the horizontal rectangle is force eerted on the horizontal rectangle is 6.5 " 1# The total force eerted on one side of the triangular plate will be the integral of this epression from the -value where the plate begins ( = 3) to the -value where the plate ends ( = ). This integral is: $ 6.5 " ( 1# ) " " ( ) " = 75 lbs. #3

7 3. As with the previous problem, we will slice the area of the gate into horizontal rectangles as each point on a thin horizontal rectangle eperiences almost the same pressure. This is because the points are all at about the same depth and pressure depends on the depth of the water. A diagram showing the situation with a thin, horizontal rectangle drawn on the circular gate is shown below = As in the previous eample, we will calculate the force eerted on a single horizontal rectangle and then integrate this to find the total hdrostatic force. The pressure eerted on the rectangle shown in the previous diagram is the product of gravitational acceleration, the densit of water and the depth. In smbols this is 9.8"1" 1 # ( ) N/m. The area of the horizontal rectangle is given b " = " # ". Multipling pressure and area gives the following epression for the force eerted on the horizontal rectangle: Force eerted on horizontal rectangle = 9.8 "1 " ( 1 # ) " " # ". The total force eerted on the semi-circular gate is the integral of this epression from the -value where the gate begins ( = ) and the -value where the gate ends ( = ). This integral is evaluated using trigonometric substitution ( = sin(θ)) and a u-substitution (u = ) to obtain:

8 Total force = $ 9.8"1 " ( 1 # ) " " # " = 98" % # 16 3 ( ) N. Problems from Pages -7 (Section 7.6). If ou draw the slope field corresponding to the differential equation: d = + "1, ou will obtain the picture shown below. From the alternatives shown at the bottom of page 5, Diagram IV is the closest match. 3. The slope field corresponding to the differential equation: d = " #, is shown in the diagram below. The curve shown on the slope field is the solution curve that passes through the point (1, ).

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