Calculus I, Fall Solutions to Review Problems II

Transcription

1 Calculus I, Fall Solutions to Review Problems II. Find te following limits. tan a. lim 0. sin 2 b. lim 0 sin 3. tan( + π/4) c. lim 0. cos 2 d. lim 0. a. From tan = sin, we ave cos tan = sin cos = sin From class, as 0, sin. Furter, as 0, cos cos 0 = (cos is a continuous function). Tus, as 0, tan = ( sin ) ( cos Tat is, tan lim =. 0 cos. ) =. b. Note sin 2 sin 3 = ( sin 2 ) ( 3 ) 2 2 sin 3 3. ( ) As 0, certainl 2 0 and 3 0. Tus as 0, sin 2 sin 3 and, and ence also sin 3 =. So te first two terms on te rigt side of ( ) converge to as 0. Terefore sin 2 lim 0 sin 3 = 2 3. c. Let f() = tan. Ten f f(π/4 + ) f(π/4) (π/4) = lim 0 tan(π/4 + ) tan(π/4) = lim 0 tan(π/4 + ) = lim (since tan(π/4) = ). 0

2 2 From class, tan as derivative sec 2. So te limit is sec 2 π/4. Now cos π/4 = / 2 and tus sec π/4 = 2. We conclude tan(π/4 + ) lim = 2. 0 Anoter approac is to use te trig. identit tan(φ + ) = tan(π/4 + ) = A little algebra ten sows tan(π/4) + tan tan(π/4) tan = + tan tan. tan(π/4 + ) = 2 tan ( tan ). tan φ + tan. Tis gives tan φ tan B part a, tan as 0. Furter, tan 0 as 0 (tan is continuous). Terefore, as 0, so te limit is 2. d. From class, cos 2 tan ( tan ) = 2 tan tan 2 0 = 2, cos 2 Tat is, lim 0 cos 2 0 as 0. Tus also = 2 ( cos 2 2 = 0. cos 2 2 ) 2 0 = 0, as Find all points were te tangent line to = cos + sin is orizontal. 0 as 0. Terefore = sin + cos. So te tangent line is orizontal (i.e., = 0) at all points suc tat cos = sin. Tis means tat te point (cos, sin ) on te unit circle corresponds to an angle of π/4 or 5π/4 = π/4 + π. It follows tat = π/4 + nπ for n = 0, ±, ±2, Suppose f() = g() = and f () =, g () = 2. Find te derivative at = of eac function. a. f( 2 ). b. f(g()). c. f() 2 g(f()). d. f(g()). a. Te derivative at is f ( 2 ) 2. So te derivative at is 2f () = 2.

3 3 b. B te cain rule, te derivative at is f (g()) g () = f () g () = 2 = 2. c. Using te product rule and te cain rule, te derivative at is For =, tis is 2f()f () g(f()) + f() 2 g (f())f () = 4. d. Te derivative at is f(g()) 2 f (g()) g (). For =, tis is 2 = True or false. a. Te derivative of f(2) is 2f (). b. If f() = f( ) ten f () = f ( ). c. Te derivative of f( + ) is f ( + ). d. Te derivative of f(/) is /f() 2. a. False. Te derivative of f(2) is 2f (2). b. True. Te derivative of f ( ) is f ( ) d ( ) = f ( ). c. True. Te derivative of f( + ) is f ( + ) d ( + ) = f ( + ). d. False. Te derivative of f(/) is f (/) / 2 = f (/)/ Find d2 2 for = sin. B te cain rule, = cos d ( ) = cos 2 Tus d 2 2 = d Appling te quotient rule, tis is (since d ( ) = 2 ). ( ) d = ( ) cos 2. d (cos ) 2 cos d (2 ). ( ) 4

4 4 Since d (cos ) = sin, te cain rule gives d (cos ) = sin 2. Tus ( ) becomes sin 2 2 cos = 4 sin cos = sin 4 4 cos If + = 3, find. =4 Differentiate bot sides of + = 3 wit respect to. Tis gives or = 0, =. For = 4, = 2 and + = 3 ten implies =. Tus = = Te kinetic energ of an object is K = 2 mv2 were m is te object s mass and v its velocit. If te object is accelerating at a rate of 0 0 m/s 2, at wat rate is te kinetic energ canging wen te object s velocit is 30 m/s? We want to find dk wen v = 30 (meters per second). We re told te object accelerates at te constant rate of 0 (meters per second squared), tat is, dv = 0. B te cain rule, Now dk = dk dv dv = 0 dk dv. dk dv = d ( dv 2 mv2) = mv. Hence dk = 0 mv and te rate of cange of kinetic energ is 300m wen v = 30.

5 5 8. A 0 ft. ladder leaning against a vertical wall starts to slip. If te bottom of te ladder is sliding awa from te wall at a rate of ft./sec., find te rate at wic te top of te ladder is falling along te wall. Write c for te speed of ligt (c = 983, 57, 056 ft./sec.). How ig in terms of c is te top of te ladder from te ground wen it is falling at a rate of 2c ft./sec? Someting is wrong ere! Wat is going on? Write for te distance from te bottom of te ladder to te wall and for te eigt of te top of te ladder along te wall as in te picture. 0 Now and are functions of time t. We re given = and want to find. We ave = 0 2 (Ptagoras Teorem). Differentiating wit respect to t gives Tus = 0. = =. From = 0 2, we see = Hence 00 = 2. ( ) 00 2 So 2c = wen te ladder is falling at twice te speed of ligt. Squaring bot sides, 4c 2 = Tat is, 4c 2 2 = 00 2, or 4c = 00. Solving for, we find = 0 4c2 +.

6 6 Tis is nonsense te ladder does not go faster tan te speed of ligt! Look closel at equation ( ). As 0 +, So for ver small, is essentiall Tus 2 becomes arbitraril large (i.e., goes to infinit) as 0 +. Does tis mean te ladder crases into te ground at infinite speed? No! At some point during its fall, te ladder loses contact wit te wall. From tat point on, our analsis doesn t appl. Comment. Wat if te ladder is constrained to sta in contact wit te wall, sa its top is fied to a smoot groove in te wall, and te bottom moves at constant speed? Ten does te ladder smas into te ground at infinite speed? Tis scenario requires tat forces approacing infinit be applied to te ladder. It is not realistic. 9. An ice cube is melting so tat its volume is srinking at a rate of 0. in. 3 /min. At wat rate is te surface area of te cube canging wen its side is 0.2 in? Write for te sidelengt of te cube, V for its volume and S for its surface area. We ave V = 3 and S = 6 2. We re given dv = (te minus sign is because te volume 0 is srinking) and want to find ds wen = 5. Now Tis means tat Hence dv = dv = (from V = 3 ). = 32, or 30 2 =. ds = ds = 2 (from S = 6 2 ). = 2 ( ) 30 2 = 2 5. Tus wen =, te rate of cange of surface area is 2 (square inces per minute) A person ft. tall walks awa from a street lamp tat is 20 ft. ig. If te lengt of te person s sadow is increasing at a rate of ft./sec., ow fast (in terms of ) is te person walking? Write for te person s distance from te street lamp and s for te lengt of te person s sadow. We ave ds = and want to find. Look closel at te diagram.

7 Note (cot = adj. opp. = ). Tus tan cot = 20 and cot = s 20 = s and = ( 20 ) s. Te person s eigt is a constant. Hence = ( 20 ) ds (in feet per second). = 20 s (since ds = )