FINAL EXAM CALCULUS 2 - Department of Mathematics

1 FINAL EXAMCALCULUS 2 MATH 2300 FALL 2018 NamePRACTICE EXAMSOLUTIONSP lease answer all of the questions, and show your must explain your answers to get will be graded on the clarity of your exposition!Date: December 12, the region bounded by the graphs off(x) =x2+1 andg(x) =3 (a).(5 points)Write the integral for the volume of the solid of revolution obtained byrotating this region about thex-axis. Do not evaluate the : We can see the region in question below. 11123xyg(x) =3 x2f(x) =x2+1 Using the washer method, the volume integral is 1 1g(x)2 f(x)2dx= 1 1(3 x2)2 (x2+1) (b).

2 (5 points)Write the integral for the volume of the solid of revolution obtained byrotating this region about the linex=3. Do not evaluate the : Now using the shell method, the integral is equal to 1 12 (3 x)(g(x) f(x))dx=2 1 1(3 x)((3 x2) (x2+1))dx=2 1 1(3 x)(2 2x2)dx226 points2. MULTIPLE CHOICE:Circle the best (a).(1 point)Is the integral 1 11x2dxan improper integral?YesNo2.(b).(5 points)Evaluate the integral: 1 11x2dx=SOLUTION: The function 1/x2is undefined atx=0, so we we must evaluate the im-proper integral as a limit. 1 11x2dx=limc 0 c 11x2dx+limc 0+ 1c 1x2dx=limc 0 1x c 1+limc 0+ 1x 1c=limc 0 (1c 1 1)+limc 0+ (11 1c)=limc 0 (1c+1)+limc 0+ (1 1c).

3 Now, sincelimc 0 (1c+1)=limc 0 1c 1andlimc 0+ (1 1c)=limc 0+1c 1both diverge to , and so the integral does not converge. Thus, the integral the curve parameterized by{x=13t3+3t2+23y=t3 t2for 0 t (a).(6 points)Find an equation for the line tangent to the curve whent= : We first find a general formula for the slope using the chain rule, and thenevaluate att=1, givingdydx t=1=dy/dtdx/dt t=1=3t2 2tt2+6t t=1= (1) =4 andy(1) =0, we need the formula for a line with slope 1/7 that passesthrough(4, 0). This equation isy=17x 473.(b).(3 points)Computed2ydx2att= : Again employing the chain rule,d2ydx2 t=1=ddxdydx t=1=ddtdydxdxdt t=1=(6t 2)(t2+6t) (2t+6)(3t2 2t)(t2+6t)2t2+6t t=1= (c).}

4 (5 points)Write an integral to compute the total arc length of the curve. Do notevaluate the : Arc length is given by 50 (dxdt)2+(dydt)2dt= 50 (t2+6t)2+(3t2 2t) the functionf(x) =x2arctan(x).4.(a).(5 points)Find a power series representation forf(x).SOLUTION: The power series of arctan(x)is n=0( 1)nx2n+12n+1, with interval of conver-gencex [ 1, 1]. Thus,f(x) =x2 n=0( 1)nx2n+12n+1= n=0( 1)nx2n+32n+1forx [ 1, 1].4.(b).(3 points)What isf(83)(0), the 83rd derivative off(x)atx=0?SOLUTION: For a power seriesf(x) = n=0cn(x a)nwith positive radius of conver-gence, we havef(n)(a) =n!

5 Cn. In our power series representationf(x) =x2arctan(x) = n=0( 1)n2n+1x2n+3, which has radius of convergence 1, the coefficient ofx83=x2 40+3is( 1)402 40+1=181, so thatf(83)(x) =83! , using the above power series representation, and formally differentiat-ing, we havef(83)(x) = n=40(2n+3)!(2n+3 83)!( 1)nx2n+3 832n+1= n=40(2n+3)!(2(n 40))!( 1)nx2(n 40)2n+ ,f(83)(0) = (83)!( 1)402 40+1=83 82 (80!).5510 tank contains 200 Lof salt water with a concentration of 4 water with a concentration of 3g/Lis being pumped into the tankat the rate of 8L/min, and the tank is being emptied at the rate of8L/min.

6 Assume the contents of the tank are being mixed thoroughlyand continuously. LetS(t)be the amount of salt (measured in grams) in the tank at timet(measured in minutes).5.(a).(1 points)What is the amount of salt in the tank at timet=0?SOLUTION:S(0)g=200L 4g/L= (b).(2 points)What is the rate at which salt enters the tank?SOLUTION: 8L/min 3g/L=24g/min5.(c).(2 points)What is the rate at which salt leaves the tank at timet?SOLUTION: As the volume of water is a constant 200L, this isS(t)g200L8 Lmin=S(t) (d).(1 points)What isdSdt, the net rate of change of salt in the tank at timet?SOLUTION: Net change is given by gain minus loss, so using parts (b) and (c),dSdtgmin=24 S(t) (e).

7 (4 points)Write an initial value problem relatingS(t)anddSdt. Solve the initial : The initial value problem isdSdt=24 S(t)25, withS(0) =800. Since thisdifferential equation is separable, we can solve by separating and then integrating: 124 125 SdS= dt 25 ln 24 125S =t+C,Note that 24 125S 0, so we can write this as 25 ln(125S 24)=t+C, so that125S 24=Ae 125t. From this we getS=Ae 125t+600. Settingt=0, and using (a), we find theanswer isS=200e 125t+600668 the following (a).(4 points) sin3(x)cos2(x)dxSOLUTION: First, using the pythagorean identity, sin3(x)cos2(x)dx= sin(x)(1 cos2(x))cos2(x)dx= sin(x)cos2(x)dx sin(x)cos4(x) , letu=cos(x), so thatdu= sin(x)dx.

8 Then the above equation is equal to u2du+ u4du= u33+u55+ , reversing our substitution, we find that sin3(x)cos2(x)dx= cos3(x)3+cos5(x)5+ (b).(4 points) x+1x2(x 1)dxSOLUTION: We start by using partial fractions:x+1x2(x 1)=Ax+Bx2+Cx 1,which givesx+1=Ax(x 1) +B(x 1) +Cx2= (A+C)x2+ (B A)x B,from which we deduceA+C=0,B A=1, and B=1. Therefore,B= 1,A= 2,andC=2. Thus, x+1x2(x 1)dx= 2x+ 1x2+2x 1dx= 2 ln|x|+1x+2 ln|x 1|+ slope field for the differential equationy =2y(1 y3)is shown below. 10123456 2 101234xy(0, 1)(a)(0, 1)(b)y =2y(1 y3)7.(a).(2 points)Sketch the graph of the solution that satisfies following initial the solution as (a).

9 Y(0) =17.(b).(2 points)Sketch the graph of the solution that satisfies following initial the solution as (b).y(0) = 17.(c).(2 points)Show that fory(0) =c 0, we have limx y(x)is : That this should be true is evident from the picture above. To see that it is infact true, we argue as follows. First, ifP0=0, thenP(t) =P0for allt, and limt P(t) =0. IfP06=0, consider the general solution to the logistics equation:dPdt=kP(1 PM)P(t) =M1+ (MP0 1)e ktThe functionP(t)is well-defined, so long as the denominator is non-zero. We focus hereon the casek,M>0. If 0<P0 M, so that(MP0 1) 0, then the denominator is clearlynever zero, and we have limt P(t) =M.

10 IfP0>M, then it is also easy to see that thedenominator is never zero fort 0, and so again, one easily computes limt P(t) = however, that ifP0<0, then we have 1+(MP0 1)e kt=0 t=1klog(1 MP0)In fact, it is not hard to check that forP0<0, we have limt 1klog(1 MP0)P(t) = 886 the series n= (a).(3 points)Use the Remainder Estimate for the Integral Test to find an upper boundfor the error in usingS10(the 10th partial sum) to approximate the sum of this : IfR10denotes the error described above, the Remainder Estimate for theIntegral Test tells us thatR10 101x4dx=limc 1 31x3 c10=limc 1 31c3 1 31103= (b).