Name                                       MATH 106 PRACTICE MIDTERM 1   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  Please answer the following true or false.  If false, explain why or provide a counter example.  If true, explain why or state the proper theorem. A)   (10 points)  The differential equation  is a homogeneous differential equation   True, dividing numerator by y2 gives         which is a function of x/y. B) (10 points)  If f is a differentiable function such that both f and f ' are positive for all x, then g(x)  =  ln(f(x)) is increasing for all values of x. True,  since                             f '(x)         g '(x)  =                                                   f(x) is positive if both f ' and f are positive   PROBLEM 2  Calculate the derivatives of the following functions. A)   (10 points)        d                                 (2x)1-x                         dx Solution         (2x)1-x  =  e(1 - x)ln(2x) Now use the chain and product rules.  The derivative of          (1 - x)ln(2x)  is          (1 - x)(1/x) - ln(2x)  =  1/x - 1 - ln(2x) Hence the derivative of           e(1 - x)ln(2x) is          [1/x - 1 - ln(2x)](2x)1-x   B)    (10 points)       d                                  eln(sin x)                         dx First use the inverse property of e and ln to get         eln(sin x)  =  sin x Now the derivative is simply          cos x   C)   (10 points)        d       23t                                                                          dt        t Solution We use the quotient rule to get           t(23t)' -23t                                          t2 Now use the chain rule and the fact that          bx '  =  bx ln b to get           3t ln2 (23t) - 23t                                                              t2 PROBLEM 3 Find the following integrals A)   (10 points)    Let          u  =  1 - x    du  =  -x dx        x  = 1 - u The substitution produces           B)    (10 points)    Let          u  =  2 - 3x    du  =  -3 dx The substitution produces           PROBLEM 4   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 5    Let  f(x) = 2x3 + 4x + 5 (10 Points)  Prove that f(x) has an inverse. Solution We calculate          f '(x)  =  6x2 + 4 which is always positive, hence f(x) has an inverse. (10 Points)  Find       d                                          f -1(11)                                  dx   We use the inverse formula              d                                     1                       f -1(11)   =                                                dx                              f '(f -1(11)) Since          f(1)  =  11 We have         f -1(11)  =  1 and          f '(1)  =  6(1)2 + 4  =  10 Hence              d                                     1                         1                       f -1(11)   =                                  =                             dx                              f '(f -1(11))                10     PROBLEM 6   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