ࡱ> 79 6} $'bjbj55 .Z__$)((((((($*,r(((''' ('('''0#nC~'((0)'"-?"-'"-'>,'$((A#~)"- : 172 Calculus 2 Review 3 1. The curve EMBED Equation.DSMT4 is revolved about the x-axis. Is the volume of the resulting solid finite or infinite? Solution. We apply the disks method and see that the question is about the convergence or divergence of the following improper integral.  EMBED Equation.DSMT4  . We will prove that the integral converges and find its value using integration by parts. Let EMBED Equation.DSMT4 . Then EMBED Equation.DSMT4 , and  EMBED Equation.DSMT4 . The first term in this expression is equal to 0. Indeed, at the lower limit we have  EMBED Equation.DSMT4 and at the upper limit we have to look at  EMBED Equation.DSMT4 . The last limit is an indeterminate form  EMBED Equation.DSMT4 and we can apply the LHospitals rule.  EMBED Equation.DSMT4 . Therefore  EMBED Equation.DSMT4 . We apply integration by parts once again; this time we take  EMBED Equation.DSMT4 . Then EMBED Equation.DSMT4 , and EMBED Equation.DSMT4 . Like in the previous expression we see that the first term is 0 and it remains to evaluate  EMBED Equation.DSMT4 . So, the volume is finite and its value is EMBED Equation.DSMT4 . 2. The same question about the area of the surface of revolution. Solution. The area of the surface of revolution can be expressed as the following integral.  EMBED Equation.DSMT4 . We will prove, using the comparison test that the integral diverges to  EMBED Equation.DSMT4 . To this end notice the simple inequality  EMBED Equation.DSMT4  whence  EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . Let  EMBED Equation.DSMT4 then  EMBED Equation.DSMT4 and  EMBED Equation.DSMT4 . Therefore  EMBED Equation.DSMT4 and the area of the surface of revolution is infinite. 3. Prove that the integral  EMBED Equation.DSMT4  converges and find its value. Solution.  EMBED Equation.DSMT4  (Recall that EMBED Equation.DSMT4 ) 4. Use the limit comparison test to show that the integral  EMBED Equation.DSMT4  diverges. Solution. We can write EMBED Equation.DSMT4 . The first term is a proper integral and its value is finite. To the second term we apply the limit comparison test. The leading term in the numerator is  EMBED Equation.DSMT4 and in the denominator -  EMBED Equation.DSMT4 . Therefore the integral  EMBED Equation.DSMT4 converges or diverges at the same time as the integral EMBED Equation.DSMT4  , where  EMBED Equation.DSMT4 It proves that the integral  EMBED Equation.DSMT4 diverges. 5. Use an appropriate test to find out whether the series  EMBED Equation.DSMT4  converges or diverges. Solution. By the limit comparison test the series  EMBED Equation.DSMT4 behaves in the same way as the series EMBED Equation.DSMT4 . The terms of the last series are positive and decreasing and therefore the series converges or diverges at the same time as the improper integral EMBED Equation.DSMT4  . Therefore the series  EMBED Equation.DSMT4 converges. 6. Use the Ratio test to find out whether the series  EMBED Equation.DSMT4  converges or diverges. Solution. Let  EMBED Equation.DSMT4 be the nth term of our series. Then the ratio  EMBED Equation.DSMT4 can be simplified as  EMBED Equation.DSMT4  By the ratio test the series converges. 7. Find the radius of convergence of the power series EMBED Equation.DSMT4 . What happens at the ends of the interval of convergence? Solution. We will apply the ratio test to find the radius of convergence. Let  EMBED Equation.DSMT4 . Then EMBED Equation.DSMT4 . Next notice that EMBED Equation.DSMT4 . Indeed,  EMBED Equation.DSMT4 . To find the limit  EMBED Equation.DSMT4 we will apply the LHospitals rule, EMBED Equation.DSMT4 . Therefore EMBED Equation.DSMT4 . By the ratio test the series converges absolutely if  EMBED Equation.DSMT4 and diverges if EMBED Equation.DSMT4 . We have just proved that the radius of convergence of the series  EMBED Equation.DSMT4  is 1. To see what happens at the ends of the interval  EMBED Equation.DSMT4 let us first put EMBED Equation.DSMT4 . Then our series becomes EMBED Equation.DSMT4  . The terms of the series are positive and decreasing, hence we can apply the integral test. The series converges or diverges at the same time as the improper integral  EMBED Equation.DSMT4 and we already have seen when solving Problem 4 that the last integral diverges; so our power series diverges to  EMBED Equation.DSMT4 at the right end of the interval of convergence. Now let us put EMBED Equation.DSMT4 . The series becomes EMBED Equation.DSMT4  . This is an alternating series. Indeed, the sign of its terms alternates, the absolute values of its terms are decreasing, and its terms tend to 0 when EMBED Equation.DSMT4 . Therefore the series converges. We have just seen that the series of its absolute values diverges, hence the series converges conditionally. 8. Find the value of  EMBED Equation.DSMT4  with accuracy  EMBED Equation.DSMT4  using an appropriate Taylor series. Solution. A brute force solution would be to use the Maclaurin series for  EMBED Equation.DSMT4 at the point 0.499. The solution suggested below is much more effective. As often in such problems we will try to state the problem in such a way that we can use Maclaurin series of known pattern. To this end notice that EMBED Equation.DSMT4 . Next we will use the formula EMBED Equation.DSMT4 . We can easily prove this formula if we notice that  EMBED Equation.DSMT4  and recall that EMBED Equation.DSMT4 . In our case we get  EMBED Equation.DSMT4 . Now we will use the Maclaurin series EMBED Equation.DSMT4 . If we compute the sum of the first two terms of the series above for  EMBED Equation.DSMT4 we will get EMBED Equation.DSMT4 . All the digits are correct. Remark. You will not be required to do it on the test but it is not difficult to estimate the error of our approximation. Indeed,  EMBED Equation.DSMT4   EMBED Equation.DSMT4  9. Write the Maclaurin polynomial of degree four for the function EMBED Equation.DSMT4 . Solution. We will look at two ways to solve the problem. The first one is straightforward. We use the formula  EMBED Equation.DSMT4 . Successive differentiations provide  EMBED Equation.DSMT4  Plugging in  EMBED Equation.DSMT4  we get  EMBED Equation.DSMT4 and  EMBED Equation.DSMT4 . The second way makes use of standard Maclaurin series for  EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . We know that  EMBED Equation.DSMT4  Therefore  EMBED Equation.DSMT4 Next notice that  EMBED Equation.DSMT4 whence  EMBED Equation.DSMT4 and  EMBED Equation.DSMT4 . Finally notice that the function  EMBED Equation.DSMT4 is even ( EMBED Equation.DSMT4 ) and therefore its Maclaurin series contains only even powers of x. Combining all the terms with exponent of x not greater than 4 we get  EMBED Equation.DSMT4 , where the ellipsis indicates the terms with powers of x greater or equal than five. 10. Estimate the value of  EMBED Equation.DSMT4  with accuracy 0.000001 using an appropriate Taylor series. Solution.  EMBED Equation.DSMT4 . Therefore  EMBED Equation.DSMT4 and  EMBED Equation.DSMT4 . Next we have to decide how many terms of the last series we want to use for approximation. The series is an alternating one so when we stop the error we make is not greater than the absolute value of the next term. We can compute that for  EMBED Equation.DSMT4 the absolute value of the fifth term is  EMBED Equation.DSMT4 which is enough for the accuracy we need. 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FMathType 5.0 Equation MathTyOle >CompObjHJ?iObjInfoKAEquation Native B!pe EFEquation.DSMT49q,\DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  arcsin(0.499)_1235815959[~NF#nC#nCOle GCompObjMOHiObjInfoPJ FMathType 5.0 Equation MathType EFEquation.DSMT49q,\DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  10 "-7Equation Native K_1235811642:LSF#nC#nCOle PCompObjRTQi FMathType 5.0 Equation MathType EFEquation.DSMT49q,,\DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  arcsinObjInfoUSEquation Native T_1235811719`XF#nC#nCOle X FMathType 5.0 Equation MathType EFEquation.DSMT49q,\DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  arcsin0CompObjWYYiObjInfoZ[Equation Native \*_1235811990Ve]F#nC#nC.5== p6 FMathType 5.0 Equation MathType EFEquation.DSMT49q,±,\DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_EOle aCompObj\^biObjInfo_dEquation Native e_A  arcsina"-arcsinb==arcsina 1"-b 2 "-b 1"-a 2 () FMathType 5.0 Equation MathType EFEquation.DSMT49q_1235811899bF#nC#nCOle mCompObjacniObjInfodp,%,\DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  sin(arcsina"-arcsinb)==sin(arcsina)cos(arcsinb)0Equation Native qA_1235812050ogF#nC#nCOle {CompObjfh|i"-cos(arcsina)sin(arcsinb) FMathType 5.0 Equation MathType EFEquation.DSMT49q,’\DSMT5WinAllBasicCodePagesObjInfoi~Equation Native _1259686533lF#nC#nCOle Times New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  sin(arcsinx)==x andcos(arcsinx)== 1"-x 2 FMathType 6.0 Equation MathTyCompObjkmiObjInfonEquation Native _1235812636qFw#nCw#nCpe EFEquation.DSMT49qPRtDSMT6WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A   p6"-arcsin(0.499)==arcsin 12 1"-0.499 2 "-0.499 1"- 12() 2 ()H"arcsin0.0011543159 FMathType 5.0 Equation MathType EFEquation.DSMT49qOle CompObjpriObjInfosEquation Native ,,\DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  arcsinx==x++p 1"3"5"& K"(2n"-1)2 n n!(2n++1) n==1" " x 2n++1 FMathType 6.0 Equation MathType EFEquation.DSMT49q lPRtDSMT6WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_1259686532jvFw#nCw#nCOle CompObjuwiObjInfoxEquation Native '_1235817283{Fw#nCw#nCOle CompObjz|i_A  x==0.0011543159 FMathType 5.0 Equation MathType EFEquation.DSMT49q,DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APObjInfo}Equation Native 2_1235816310yFw#nCw#nCOle G_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  arcsin(.499)H" p6"-0.0011543159"- 12"3(.0011543159) 3 H".5224444594CompObjiObjInfoEquation Native _1235816211Fw#nCw#nC FMathType 5.0 Equation MathType EFEquation.DSMT49q,\DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  |Error|d"p 1"3"& K"(2n"-1)2 n n!(2n++1) n==2" " (0.0011543159) 2n++1 d"p(0.0011543159) 2n++1n==2" " ====(0.0011543159) 5 p[(0.0011543159) 2 ] nn==0" " == (0.0011543159) 5 1"-(0.0011543159) 2 H".204938695710 -14Ole CompObjiObjInfoEquation Native  FMathType 5.0 Equation MathType EFEquation.DSMT49q,<DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  _1235742221Fw#nCw#nCOle PIC LMETA HL X <  .1   & MathType@@XCG Times-!f@!x@J !x@!CG Times-!(@!)@g!cos@~!(@!sin@' !)@Symbol-!="Systemn- FMathType 5.0 Equation MathType EFEquation.DSMT49qCompObjiObjInfoOle10NativedEquation Native '` f(x)=cos(sinx), $DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  f(x)==cos(sinx)_1235817591QFw#nCw#nCOle CompObjiObjInfo FMathType 5.0 Equation MathType EFEquation.DSMT49q,\DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  M 4 (x)==f(0)++f(0)xEquation Native ;_1235819818Fw#nCw#nCOle CompObji++ f(0)2!x 2 ++ f(0)3!x 3 ++ f (4) (0)4!x 4 FMathType 5.0 Equation MathType EFEquation.DSMT49qObjInfoEquation Native )_1235818262Fw#nCw#nCOle       #&'()*+.123458;<=@CDEHKLMNORUVWXYZ]`abcdehklmnopqtwxyz{|}, \DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  f(x)==-sin(sin(x))cos(x),f(x)==-cos(sin(x))cos 2 (x)++sin(sin(x))sin(x),f(x)==sin(sin(x))cos 3 (x)++3cos(sin(x))cos(x)sin(x)++sin(sin(x))cos(x),f (4) (x)==cos(sin(x))cos 4 (x)-6sin(sin(x))cos 2 (x)sin(x)-3cos(sin(x))sin 2 (x)++4cos(sin(x))cos 2 (x)-sin(sin(x))sin(x). 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