cIIT Dept. Applied Mathematics, December 6, 2005 1PRINT Last name: KEYFirst name:Signature: Student ID:Math 152 Exam 2, Fall 2005Instructions. You must show the mathematical steps of the solution in order to receive fullcredit; writing only the answer will receive no credit. Written explanations for each step are notrequired (due to time constraint) but an incorrect solution with a correct written explanationmight receive partial credit.Conditions. No calculators, computers, notes, books, or scratch paper. By writing your nameon the exam you certify that all work is your own, under penalty of all remedies outlined in theIIT student rules. Please do not talk until you are away from the room.Time limit: 50 minutes (strict).NOTE: The topics may not be in order either of increasing difficulty or of th e order they werecovered in the course.POSSIBLY USEFUL FORMULASsec2x = tan2x + 1 Mn= ∆xfx0+x12+ fx1+x22+ · · · + fxn−1+xn2 cos2x =1+cos 2x2Tn=∆x2[f(x0) + 2f(x1) + · · · + 2f(xn−1) + f(xn)]Rdx1+x2= tan−1x + CRtan x dx = − ln | cos x| + CP V = nRT |EM| <K(b−a)324n2(K ≥ f′′(x))F = ρgAd |ET| <K(b−a)312n2(K ≥ f′′(x))|Rn(x)| ≤M(n+1)!|x − a|n+1n · (r − r0) = 0Sn=∆x3[f(x0) + 4f(x1) + 2f(x2) · · · + 2f(xn−2) + 4f(xn−1) + f(xn)]|ES| <K(b−a)5180n4(K ≥ f(4)(x))R∞n+1f(x) dx ≤ s − sn≤R∞nf(x) dxR1√1−x2dx = sin−1x + C11+x2=P∞n=0(−1)nx2nddx(sec−1x) =1x√x2−1sin 2x = 2 sin x cos xVol =Rba2π [(f(x))2− (g(x))2] dxcIIT Dept. Applied Mathematics, December 6, 2005 2SHOW WORK FOR FULL CREDIT NO CALCULATORS1. Find the partial fraction decomposition ofx3+ x2(x2+ 1)2.The denominator is the repeated quadratic irreducible factor (x2+ 1) · (x2+ 1). Thereforethe partial fraction decomposition form isx3+ x2(x2+ 1)2=Ax + Bx2+ 1+Cx + D(x2+ 1)2.Cross-multiply and solve for A, B, C, D.x3+ x2= (Ax + B)(x2+ 1) + Cx + Dx3+ x2= Ax3+ Bx2+ (A + C)x + (B + D).For polynomials to be equal, the coefficients must be equal. ThereforeA = 1 and B = 1 .Plugging these in, we havex3+ x2= x3+ x2+ (1 + C)x + (1 + D)0 = (1 + C)x + (1 + D),from whichC = −1 and D = −1 .2. Given the integralZ∞e1x(ln x)pdx ,(a) Find a value of p so that the integral diverges.(b) Find a value of p so that the integral converges, and compute the integral in this case.(Hint for both parts: Find the indefinite integral only once using the substitution u = ln x.)First the indefinite integral using u = ln x and du = dx/x.Zdxx(ln x)p=Zduup=(u−p+1−p+1+ C, if p 6= 1,ln |u| + C, if p = 1=((ln x)−p+1−p+1+ C, if p 6= 1,ln | ln x| + C, if p = 1 .Soln. to (a). Any value p ≤ 1 will work. Here is the justification for p = 1:limt→∞Ztedxx ln x= limt→∞ln | ln x||te= limt→∞(ln ln t − ln ln e) =∞ .Soln. to (b). Any value p > 1 will work. Here is the justification for p = 2:limt→∞Ztedxx ln x= limt→∞(ln x)−1−1te= limt→∞−1ln t+1ln e=1 .cIIT Dept. Applied Mathematics, December 6, 2005 33. Sketch four particular solutions to the differential equation whose slope field is below:(a) 2 equilibrium solutions, (b) 1 increasing solution, and (c) 1 decreasing solution.0200400600800y(x)100 200 300 400x(a) The two equilibrium so-lutions are the two horizon-tal lines at approximatelyy = 200 and y = 500, re-spectively.(b) Any solution with initialcondition y(0) between 200and 500 will yield an increas-ing solution.(c) Any solution with ini-tial condition y(0) between 0and 200 or between 500 and800 will yield a decreasingsolution.4. Plutonium-238, a satellite power source, has a half-life of 88 years. Compute the time atwhich 10% of a sample of plutonium-238 remains. You may start from the general solutionfor exp on ential decay: y(t) = A · ekt.Starting from the general solution, solve for k by plugging in y(88) = A/2.y(88) =A2= Aek·88, or12= ek·88ln12= 88k, ork =ln1288.Now the model is y(t) = A · exp(tln1288) . We use this to solve y(t) = .1A for t:y(t) = .1A = A · exp(tln1288).1 = exp(tln1288)ln .1 = tln1288t =88ln .1ln .5.cIIT Dept. Applied Mathematics, December 6, 2005 45. For this problem, use the differential equation for Newton’s L aw of Cooling:dTdt= k(T − Ts) ,where T is the temperature, t is time, Tsis the ambient temperature, and k is a constant.Question: Water at temperature 60oF is placed in a freezer which is at 30oF. After 1hour, the water is at temperature 40oF. How long does it take the water to freeze (i.e., at32oF)?First solve the differential equation byseparation of variables.dTdt= k(T − 30)dTT − 30= k dtln |T − 30| = kt + C|T − 30| = ekt+CT − 30 = ±eCekt= AektT (t) =30 + Aekt.Now find k u sing the data T (0) = 60 and T (1) =40:T (0) = 60 = 30 + Aek·030 = 30 + A, soA = 30 .T (1) = 40 = 30 + 30ek·113= ekk =ln13.Finally, using the above solve T (t) = 32 for t.T (t) = 32 = 30 + 30 exp(t ln13)230= exp(t ln13)ln115= t ln13, sot =ln(1/15)ln(1/3).cIIT Dept. Applied Mathematics, December 6, 2005 56. Find the general solution to the differential equationx2y′+ 3xy = ex,where x > 0 (this means the general solution is defined only for x > 0).First put the diff. eq. into linear form: y′+3xy =exx2. Then the integrating factor isI(x) = exp(Z3xdx) = exp(3 ln |x|),which since we are given x > 0 yieldsI(x) = exp(3 ln x) = x3. Multiplying through bythe integrating factor, we havex3y′+ x23y= xex(yx3)′= xexyx3=Zxexdx = xex− ex+ Cy =xex− ex+ Cx3.The integral of xexis computed by parts.u = x dv = exdxdu = dx v = exUsing the formulaRu dv = uv −Rv du,Zxexdx = xex−Zexdx = xex− ex+ C .cIIT Dept. Applied Mathematics, December 6, 2005 67. The plot below is of the parametric equationsx = t · cos(3t) −13sin(3t)y =t2· sin(2t) +14cos(2t)0 < t < π .(a) Find all values of t in the interval (0, π) which correspond to horizontal or verticallines. (Do not check t = 0 or t = π.)(b) Label the corresponding horizontal and vertical tangent lines on the plot above withthe appropriate values of t.Soln. to (a). We solve x′(t) = 0 and y′(t) = 0 and consider the results. Using the ChainRule,dxdt= cos(3t) − 3t sin(3t) − cos(3t) = −3t sin(3t) .dxdt= 0 when either t = 0 or sin(3t) = 0, which for 0 < t < π is whent = π/3, 2π/3 .dydt=t22 cos(2t) +12sin(2t) −142 sin(2t) = t cos(2t) .dxdt= 0 when either t = 0 or cos(2t) = 0, which for 0 < t < π is whent = π/4, 3π/4 .Because these zeros of dx/dt and dy/dt do not overlap, these zeros for dx/dt correspondto vertical tangent lines,