VANDERBILT UNIVERSITY MAT 55B, FALL SOLUTIONS TO THE PRACTICE FINAL. Important: These solutions should be used as a guide on how to solve the problems and they do not represent the format in which answers should be given in the test. In many questions below only a brief eplanation containing the main ideas is given, while in the eam you are epected to fully justify your answers. Question. Find the derivative of f(). (a). f() arctan(ln ) (b). f() e+sin cos f() ln(arcsin(arctan )) ( cos ) (d). f() tan e arctan Solutions. Using direct application of the chain and quotient rules, we find (b). (d). (a). ( + (ln ) ) ( + cos )e +sin e +sin cos + (cos ) arcsin(arctan )( + ) (arctan ) ( sec cos ) ( + ) sin + cos e arctan e arctan ( + ) Question. If f() tan(cos (ln )), find the values of for which f() is invertible, and compute (f ) (). Computing, Then where y f (). We want cos (ln ) ( π, π ) [, π] [, π ), hence ln (, ], so (, e]. f () sec (cos (ln )) (ln ). (f ) () f (f ()) y (ln y) sec (cos (ln y)),
MAT 55B SOLUTIONS PRACTICE FINAL Question 3. Compute the following its. ln( + 7) (a). (b). ( 4) csc( 6) 4 ln( + 7) L H + 7. ( 4) 4 4 csc( 6) 4 sin( 6) L H 4 cos( 6) 8. e ln ( (d). + 9 ) + 9 e ln + e ln L H e. ( )( + 9 + 9 + ) + 9 + ( + 9 ) 9 + 9 + 9. 9 + 9 + sin (e). + sin (f). ( ) + y sin + sin sin + sin. ( ) + ln y ( ) ln + So + ln y ( ) ln ln + L H ( ) + y e 6. 3 ( ) ( + )( ) 6.
MAT 55B SOLUTIONS PRACTICE FINAL 3 Question 4. Evaluate the following integrals. (a). d ( + ) 3 Using u +, d 3 ( + ) 3 ( + ) 3 + C. (b). e sin cos d Using u sin, e sin cos d e sin + C. cos sin d Using half-angle identities, + cos() cos sin d 8 8 3 + (d). + d Use long division to get so sin() d 4 cos(4) d 8 sin(4) 3 3 + + 3 + + d + + +, + d + arctan ln( + ) + C. + C. ( cos ()) d + d (e). tan d Integrate by parts with u tan and dv d, to find tan d tan ln( + ) + C. (f). e d Integrate by parts with u and dv e d, e d e e + C.
4 MAT 55B SOLUTIONS PRACTICE FINAL (g). cos(3)e d Integrate by parts with u cos 3 and dv e d, so cos(3)e d e cos(3) 3 sin(3)e d. Integrate by parts again with u sin 3 and dv e d to get cos(3)e d e cos(3) + 3e sin(3) 9 cos(3)e d, hence (h). + d so (i). cos d cos(3)e d e cos(3) + 3e sin(3) + d + +, d + C. d arctan + C. + Integrate by parts with u and dv cos d, cos d sin sin d. Integrate by parts again with u, dv sin d, cos d sin + cos sin + C. (j). cos d Put z, so that d z dz and cos d z cos z dz. Now integrate by parts with u z, dv cos z dz to get cos d sin + cos + C. (k). (ln ) 3 d
MAT 55B SOLUTIONS PRACTICE FINAL 5 Put u ln, so (ln ) 3 d u 3 du 4 (ln )4 + C. (l). ln d Use integration by parts with u ln and dv d, ln d ln + d ln + C. (m). sin d Use integration by parts with u and dv sin d, sin d cot + cot d cot + ln sin + C. (n). + 3 + d Use partial fractions to find so (o). cos 3 sin d + 3 + + + 3 +, () d ln + + 3 ln + + C. () + 3 + Write cos 3 sin d ( sin ) sin cos d. Making the substitution u sin we find, cos 3 sin d sin3 sin5 + C. 3 5 (p). 3 3 3 d Notice that Write 3 3 3 + 4 + ( + + )( ) ( + ) ( ). 3 3 3 A + + B ( + ) + C.
6 MAT 55B SOLUTIONS PRACTICE FINAL Writing a system as usual, we find A 3, B, C 3. Then 3 3 3 d ln + 3 + + ln + C. 3 Question 4. Determine whether the integrals below are improper of type I, improper of type II, or neither. When they are improper, determine whether they are convergent or divergent, evaluating them when they are convergent. (a). d Improper of type II and convergent because d. (b). e d Improper of type I and convergent because e d e d e None because e for [, ], so (d). 3 d. e d ln e ln e. Improper of type II and divergent because (e). e d 3 3 + t. t + t 3 t + Improper of type I and convergent because letting u we find and e d e e d.,
MAT 55B SOLUTIONS PRACTICE FINAL 7 So e d. (f). 4 d Improper of type II and divergent because and 4 d 4 4 d + 4 4 d 4 4 d 4 ln 4t. t 4 (g). 5 5 π d Improper of type II and divergent because 5 5 and both these integrals diverge. π π d 5 5 π d + π π d Question 5. Let y f(), a b, be a curve on the plane, written in Cartesian coordinates. Its length is given by b ( ) dy L + d. d a (a). Suppose that the same curve is given in parametric form by f(t) and y g(t), t t t. Show that its length can then be computed by t (d ) ( ) dy L +. What is the relation between a, b and t, t? Using the chain rule, t dy dy d d dy d dy d.
8 MAT 55B SOLUTIONS PRACTICE FINAL So + ( ) dy d d ( ) dy + d d (d ) + ( d ( ) dy ) + ( dy ( d ( d ) d ) d (d ) ( ) dy ( + d (d ) ( ) dy + d d (d ) ( ) dy +. ) ) d From this the result follows after noticing that (t ) a and (t ) b. (b). Assume now that the same curve is given in polar form by r r(θ), θ θ θ. Show that in this case θ ( ) dr L r + dθ. dθ What is the relation between a, b and θ, θ? θ Since r cos θ r(θ) cos θ and y r sin θ r(θ) sin θ, the curve is parametrized by θ and we can use the previous formula with θ replacing t, θ (d ) ( ) dy L + dθ. dθ dθ Computing we find and θ d dθ dr cos θ r sin θ, dθ dy dθ dr sin θ + r cos θ, dθ and using these epressions into the above formula for L we obtain the result. Question 6. A tank with gallons of water contains % of alcohol (per volume). Water with 5% of alcohol enters the tank at a rate 3 gal/min. The miture is kept homogeneous and is pumped out at the same rate. Find an epression that describes the percentage of alcohol in the tank after t minutes.
MAT 55B SOLUTIONS PRACTICE FINAL 9 Denote the amount of alcohol (in gallons) at time t by q(t). To find an equation for dq, notice that dq (rate in) (rate out) The volume remains constant equal to gallons. We have The rate out is given by Hence or Computing the integrals we find (rate in) 5% of 3 gal/min.5 gal/min. (rate out ) q(t) 3 3q(t) gal/min dq.5 3q 5 3q dq 5 q 3 dq 5 q 3 (5 q), 3. ln 5 q.3t + C q Ae.3t + 5. To find A we need to use the initial condition. At time zero there is % of alcohol, which corresponds to % of gal gal Hence q() and Therefore Ae + 5 A 3. q 3e.3t + 5. This is an epression for the amount of alcohol in gallons; the percentage is q/. Question 7. Determine whether the series below are convergent or divergent. (a). ln(n + ) n Since ln(n + ) n + for large n, ln(n + ) n +, and the series diverges by comparing with the harmonic series. (b). n 6n + n Converges by the it comparison test with n. n (n + ) n ln(n (ln n + ln(n + 4))n n + 3)
MAT 55B SOLUTIONS PRACTICE FINAL Notice that But Hence n (n + ) n ln(n + 3) (ln n + ln(n + 4))n n n ln(n + 4n) n ln(n + 3) L H n n and the series diverges by the divergence test. (d). sin n n n ( ) ln n + ln(n + 4) n + n ln(n + 3) n ( + n) n ln(n + 4n) n ln(n + 3) ln(n + 4n) e n ln(n + 3). n+4 n +4n n n +3 n n + 4 n (n + ) n ln(n + 3) (ln n + ln(n + 4))n n e, Diverges by using the it comparison test with n. (e). n ( ) n ln(n + 7) Converges by the alternating series test. (f). n n Diverges by the ratio test. (g). n n 3 n 9 + n 3 Converges by the it comparison test with (h). n 3 n Converges by the it comparison test with 3 n. n 3. n + 3 n + 4n. (i). n tan ( )n n
Since tan is an odd function, we have MAT 55B SOLUTIONS PRACTICE FINAL tan ( )n n ( ) n tan n. Because tan is an increasing function on [, ], and n is decreasing for all n, we have that tan n is decreasing, and since tan n as n, the series converges by the alternating series test. (j). n ( ) n+ n 4 n Converges by the ratio test. (k). n4 (n + ) (n ln n) Using the it comparison test with n(ln n) : But n (n+) (n ln n) n(ln n) converges by the integral test, so the series converges. (n + )n(ln n) n(n + ) n (n ln n) n n. n4 n4 n(ln n) (n + ) (n ln n) (l). n e en Use the ratio test a n+ a n (n+)! e en+ e en Since e.7..., we see that (n + )!een e en+ (n + )e en e n+ (n + )e en ( e). e en ( e) e en. But (n + n + n )e en L H n e en n e en e n, so a n+ n a n by the squeeze theorem, and the series converges by the ratio test.
MAT 55B SOLUTIONS PRACTICE FINAL Question 8. Evaluate the sum. (). ( ) n 5 n n n ( ) n 5 n n ln( + 5 ). n (b). n n + 3n + n n + 3n + ( n + ) n + n + terms that cancel out (telescoping). n ( ) n (4n + 6n + ) (n + )! n ( ) n (4n + 6n + ) (n + )! n ( ) n (4n + 6n + ) (n + )(n + )(n)! n ( ) n (n)! cos. Question 9. Find the Maclaurin series for the functions below, along with their radius of convergence. (a). ln( + 7) ( ) (b). 3 tan 4 ln( + 7) ( ) n+ 7n n n, R 7. n ( ) 3 tan 3 4 n n ( ) n 4 n+ (n + ) 4n+ ( ) n 4 n+ (n + ) 4n+5, R. e
MAT 55B SOLUTIONS PRACTICE FINAL 3 e ( + n n ( ) n n ( ) n n ), R. n ( ) n n (d). arcsin Recall that arcsin d. But ( )( )( ) ( n + )( ) n n with R, from which we get arcsin n n (n)! n () n, ( ) n (n)! n () (n + ) n+, R. (e). ( 3) 5 ( 3) 5 4 8 4 8 4 8 d 4 d 4 3 d 4 4 8 d 4 3 n n n 3 n n(n )(n )(n 3) n 4 n 3 n n(n )(n )(n 3) n 4, R. n4 (f). cos( ) + cos( ) + n ( ) n (n)! ( ) n n ( ) n (n)! n, R (g). f() { sin if, 3 6 if.
4 MAT 55B SOLUTIONS PRACTICE FINAL sin 3 3 3 ( ) n n+ (n + )! ( ) n n+3 (n + 3)! 3 n n 6 + ( ) n n (n + 3)!. Since this epression reduces to 6 when, we have f() ( ) n n (n + 3)!, R. n n (h). e + e e + e n n n + ( ) n n ( ) n + n, R. n Question. Find the Taylor series for f() centered at the given value of a. You do not have to find its radius or interval of convergence. (a). f() 5, a Compute so and then f () 5 3, f () 5 3, f () 5 3, f () 5 ( 3 ) 3, f () 5 ( 3 ) ( 3 ) 5, f () 5 ( 3 ) ( 3 ) ( 5 ) 7, f (n) () 5( )n+ 8 f (n) () 5( )n+ 8 3 (n 7) n 3 n 5, n 4. 3 (n 7) n 3, n 4.
MAT 55B SOLUTIONS PRACTICE FINAL 5 Hence 5 5 + ( ) + 5 3! ( ) + 5 3 3! ( 5( ) n+ )3 + 8 (b). f() 5, a 3 Compute so f () 5 6, n4 f () ( 5)( 6) 7, f () ( 5)( 6)( 7) 8 f (n) () ( ) n 5 6 (n + 4) (n+5) n (n + 4)! ( ) 4 f (n) n (n + 4)! (3) ( ) 4 3 n+5, 5 n (n + 4)! ( ) 4 3 n+5 ( 3)n n n (n + 4)(n + 3)(n + )(n + ) ( ) 4 3 n+5 ( 3) n. n 3 (n 7) n 3 ( ) n. (n+5) f() cos, a π 4 Write cos cos( π 4 + π 4 ) cos( π 4 ) cos π 4 sin( π 4 ) sin π 4 Hence cos n cos( π 4 ) ( ) n (n)! ( π 4 )n sin( π 4 ). n ( ) n (n + )! ( π 4 )n+.