NAME                                    MATH 106 PRACTICE MIDTERM 2   Please work out each of the given problems.  Credit will be based on the steps that you show towards the final answer.  Show your work.   PROBLEM 1  Set up the integrals that solve the following problems.  Sketch the appropriate diagram for each.  Then use a calculator to finish the problem. A)   (10 Points) Find the volume of the solid that is formed by revolving the region bounded by y  =  x3 + x and y  =  x2 + x  around the y-axis. Solution We choose a vertical cross-section since we do not want to solve for y.  The picture shows that this produces a cylinder.  The Surface area of a cylinder is         A  =  2prh         r  =  x         h  =  (x2 + x) - (x3 + x)  =  x2 - x3 To find the limits, we set the equations equal to each other:         x2 + x  =  x3 + x         x2  =  x2                x  =  0    or     x  =  1 We get the integral           =  .3142  B)    (10 Points). Find the volume of the solid that is formed by revolving the region bounded by  y  =  x4  and y  =  x between x  =  0 and  x  =  1/2  about the line y  =  -10. Solution   We choose a vertical cross-section again since we do not want to solve for y.  The picture shows that this produces a washer.  The area of a washer is         A  =  p(R2 - r2)         R  =  x + 10         r  =  x4 + 10 We get the integral           =   7.59 C)   (10 Points) Find the volume of the solid that is formed by revolving the region bounded by y  =  x3 - x  and y  =  3x around the line x  =  5. Solution             From the picture, we can see that since the top curve changes we will need to perform two integrations.  We can also see that we will need to use the method of cylindrical shells.  The Surface area of the bottom cylinder is         A  =  2prh         r  =  5 - x         h  =  (x3 - x) - (3x)  =  x3 - 4x The Surface area of the top cylinder is         A  =  2prh         r  =  5 - x         h  =  3x - (x3 - x)  =  4x - x3  To find the limits, we set the equations equal to each other:         x3 - x  =  3x         x3 - 4x   =  0          x(x - 2)(x + 2)               x  =  -2    or     x  =  0    or    x  =  2 We get the integral                    =   80p D)   (10 Points) Find the area of the region bounded by the curves         y  =  x3 - 3x2 - 9x + 12 and y  =  x + 12 Solution   Notice that there are two regions.  We set         x3 - 3x2 - 9x + 12  =  x + 12         x3 - 3x2 - 10x  =  0         x(x - 5)(x + 2)  =  0         x  =  -2    or    x  =  0    or    x  =  5 We integrate    =  101.75          E)    (10 Points) Find the length of the curve y  =  sin x for  0 <   x  < 2p. We use the arc length formula:         1 + y'2  =  1 + cos2 x So that the length is         F)    (10 Points) Find the volume of the sphere of radius 2. Solution     We can find the volume of the sphere by revolving the curve         about the x-axis.  The picture shows that we can use the disc method with         A  =  pr2          r2  =  4 - x2 We get the integral                  PROBLEM 2 You have been called as an expert witness in the case involving the recent murder that occurred in room E106.  It is clear that at the time of death the victim was healthy with a temperature of 98.6 degrees.  It is also know that a human body in this situation will cool down to 90 degrees in one hour.  When the body was discovered at 10:00 PM the corpse had a body temperature of 85 degrees.  During the entire day, the temperature of the room was a constant 65 degrees.  A) (10 points) Use the Newton’s Law of Cooling (the rate of change of the temperature of the body is proportional to the difference between the body’s temperature and the ambient temperature) to write down a differential equation for this situation.   Solution         dT                     =  k(T - 65),        T(0)  =  98.6,    T(1)  =  90         dt   B) (10 points) Solve the differential equation from part A. Solution Separating variables gives            dT                            =  kdt         T - 65 Now integrate         ln|T - 65|  =  kt + C Now use T(0)  =  98.6         ln(98.6 - 65)  =  C Now use T(1)  =  90         ln|90 - 65|  =  k + ln(65) Solving gives         k  =  ln(25/33.6)  Exponentiating the solution produces         T - 65  =  33.6 eln(25/33.6)t    C) (10 points) What was the time of death? The body temperature is 85 degrees, hence           ln|85 - 65|  =  ln(25/33.6)t + ln(33.6) Now solve for t                    ln(20) - ln33.6                 t  =                                  =  1.75476                       ln(25/33.6) Now convert 0.75476 to minutes by multiplying by 60 to get that the murder occurred one hour and 45 minutes ago or at 8:15 PM.     PROBLEM 3 Solve the following differential equations A.  (15 Points) y(1 + x2)y'  -  x(1 + y2)  =  0,    y(0)  =  Solving for dy/dx gives         dy               x(1 + y2)                   =                                   dx                y(1 + x2) Separating variables gives          y dy                  x dx                        =                               1 + y2               1 + x2  Integrating both sides yields         ln(1 + y2)  =  ln(1 + x2) + C The initial value   y(0)  =    gives         ln 4  =  C Using the sum to product property of ln gives         ln(1 + y2)  =  ln(4 + 4x2) Exponentiating both sides produces         1 + y2  =  4 + 4x2 or                B.  (15 Points)                 x3 + y3                           y'  =                                                              xy2   This is a homogeneous differential equation.  We can write                         1 + (y/x)3            y'  =                                                     (y/x)2   Letting          v  =  y/x,        y  =  xv,        y'  =  v + xv' gives                                 1 + v3            v + xv'  =                                                       v2   or               x dv                 1                              =                             dx                 v2   Separating gives                                             dx                         v2 dv    =                                                           x  Integrating both sides gives            1/3 v3  =  ln|x| + C1 Multiplying by 3, letting C = 3C1, and taking the 1/3 power produces             v  =  (3ln|x| + C)13 Substituting v = y/x and multiplying by x gives             y  =  x(3ln|x| + C)1/3   Problem 4   (20 Points) When completed, the International Space Station orbiting at 238 miles above the surface of the earth will weigh one million pounds (at the surface of the earth 4000 miles from the center).  How much total work will it take to send the entire station in orbit?   Solution We use the fact that          F  =  k / x2            1,000,000  =  k / 40002          k  =  1.6 x 1013      So that            It will take 220 million mile pounds to send the entire station in orbit. Problem 5  (20 Points) A 30 foot chain that weighs 3 pounds per foot is used to lift a 200 pound piece of sheet metal from the ground to the top of a 30 foot tall building.  How much work is required? Solution We have         Work  =  Work for the sheet metal + Work for the chainWork for the sheet metal  =  (200)(30)  =  6,000To find the work for the chain, we have        DW  =  (3Dy)(30 - y)The total work is the integral        The total work is         6000 + 1350  =  7350 foot pounds   Questions, Comments and Suggestions:  Email:  greenl@ltcc.edu