x f(x) =x 2 +2x

Transcription

1 . Section P t t 9. t 3 30t +56t = 7. < f() = f() = Te range appears to be y

2 Te range appears to by y (a) 8 feet (b) 3 feet (c) 3 seconds. (a) t r(t) = 6t +96t C = πd =πr 5. A = π D 7. 3 πr = πr 3 9. V = π d , Section P.. y = + 3. y = 5. y = y = ft/sec ft./sec.

3 3. SECTION / 3.(a) 0 3 f() = (b) 8 (c) 5. (a) feet (b) 9.7 feet (c) 3.88 ft/sec. (d) 5.96 ft/sec. 7. (a) 5 people/year (b) 7 people/year (c) It is growing faster at te beginning of te decade since te rate of growt is 5 early in te decade and 7 near te end of te decade. 9. (a) R() = (b) $3,50.00 (c) As canges from 00 to 0 te average rate of cange of revenue is $7.99 per sirt, as canges from 500 to 50 te average rate of cange of revenue is $.99 per sirt. Te revenue is increasing at a greater rate wen 500 sirts are sold. 3. (a).6 inces (b).06 inces/minute (c) 0.88 inces/minute Te water is draining at a slower rate one minute into te eperiment. 33. odd 3. even 35. odd 36. even 37. neiter 38. neiter 39. neiter 0. odd - 3. Section L 0 () =+3; -

4 L 0 () =; Te graps will be omitted in 5-9 but be sure you ceck tem on you graper! 5. L 0 () =+; 7. L 0 () =6 ; 9. L 0 () =+3. L 0 () =+; 3. L 0 () =; 5. L 0 () =; 7. L 0 () =0; 9. L 0 () =0. (a) L 0 () = 3 (b) (c) f() f(0) (a) L 0 () = (b) (c) f() f(0) (a) L 0 () =96 (b) (c) r() r(0) (b) (a) L 0 () =6 (c) r() r(0) L 0 () f() L 0 () =3; rate of cange is 3 dollars per sirt. 33. L 0 () = ; rate of cange is.08 inces per minute (tis is ow fast te water level is dropping wen = L() in part (c) is te linearization of te function. Te table for part (c) is L 0 () f() (a) a +a (b) a 3 +a (c)+ k. Section f() = 3,p=0; Te slope is 0.; L 0 () =0 3. f() = 3 +, p = ;Te slope is 5. ;L () =5 5. f() =( +),p= ; Te slope is 0.; L () = 7. f() = 5,p=; Te slope is 5.; L () =5

5 5. SECTION.3 LIMITS 5 9. f() = 5,p= π; Te slope is 87.05; L () = f() =3,p=; L () passes tru te point (, 3); f(+) =3(+) = o() =3 and te slope of L () is 6; L () =3+6( ) = f() =, p=; L () passes tru te point (, 0). f( + ) =(+) = + ; o() = te slope of L () is ; L () = 0+( ) = + 5. f() = +3,p=0; f(0 + ) =(0+)+3(0+) = +3 ; o() =3 L 0 () = 7. L () = f() =+3 +,p=0; f(0 + ) =+3(0+)+(0+) =+3 + o() = ; L 0 () =+3. f 0 (0) = L 0 () = 3. f 0 () = L () = + 7. f 0 () = 5 L () =5 9. f 0 ( ) = 80 L () = (a) L () = (c) L () f() (a) L () = (c) L () f() π R 0 (500) = y 0 (0) = Section.3 Limits / Form 0 0 ; / 9. Form 0 0 ; Form 0 0 ; d.n.e. 5. 3/3 7. Form 0 0 ; 8 9. Form 0 0 ;. Form 0 0 ; + 3. / 5. 5/ / 3. ( + ) +3=7+ + ; f 0 () = ; o() = 0 = 0 =0 33. ( + ) +(+) =3+ + ; f 0 () = ; o() = 0 = 0 = ( + )+=8+3; f 0 () = 3; o() =0 0 0 = 0 0= ( + ) = ; r 0 () = 3; o() = = 0 6 = ( + ) 6( + ) = ; r 0 () = 6; o() = = 0 6 =0. (a) () = 8 ; (b) 9 (c) 8 (d) 0 in te it te two ladders are approacing two ladders wic touc and ave no area between tem.. (a) () = 3 7 (b) () = 6 7

6 6 [(+) ( )][(+)+( )] 3. = 8 =8 ( )( 7. + ) = 9.,5,9 6. Section.. f() = +3, p = f( + ) f() f(+) f() = f() = , p = f(0 + ) f(0) 0 f(0+) f(0) = f() = , p = f( +) f( ) 0 f( +) f( ) = f() =(3 ) 0, p = f( + ) f() 0 f(+) f() = f() = 6 +7, p = f(3 + ) f(3) 0 f(3+) f(3) = r(t) =0t 6t, p = r(.0+) r(.0) 0 r(.0+) r(.0) =68ft./sec. 7. r(t) =60 6t, p = r(.0+) r(.0) 0 r(.0+) r(.0) = 3 ft./sec.

7 6. SECTION r(t) =.5t.9t, p = r(.0+) r(.0) r(.0+) r(.0) =.7 ft./sec.. r(t) = 0 + 6t 6t, p = r(.0+) r(.0) 0 r(.0+) r(.0) =0ft./sec. 3. r(t) = 0 + 8t 6t, p = r( + ) f() r(+) f() = 6 ft./sec i) ft. ii) as te time after te object was trown approaces sec te position above te ground approaces ft. iii) 3 ft./sec. 7. (a) v 0 =0(it as zero vertical velocity initially as it leaves te table orizontallly) and r 0 =ft. (b). ft./sec. (c) 5.8 ft./sec (d) v ave (e) 6 ft./sec. 9. (a) t =3.6 sec (b) t =. sec (c) v ave v.6 ft./sec. (d) v ave v 7.68 ft./sec. (e) 7.5 ft./sec. 3. (a) slope is 75 (b) (c). f(5 + ) f(5) f(5+) f(5) (d) 0 =75 (e) Yes, te slope of te tangent line at an input value is equal to te instantaneous rate of cange of te function at tat input value. 33. (a) $/tree (b) $3.50 Approimate cost to grow te 0 st tree. (c) $.60 (d) $3.00 (e) $ (a) R() = and C() = so tat P () = R() C() = (b) $,00.00 (c) $/tree (d) $.50 /tree 37.(a) (t) = V (t) πr = t 0.0t 6π (b).98 feet. (c) V 0 (0) = 0 ft. 3 /sec., rate at wic te water is being pumped into te tank. After 0 minutes, te water

8 8 ceases to be pumped into te tank. (d) 0 (0) = 0 ft./sec.,rateatwictewater level is rising. After 0 minutes te water level is no longer rising. 7. Self-Test.-. ST. (a)true (b)false (c)true (d)true (e)false (f)false (g)false () True (i)true (j)true (k)false (l)true (m)false ST. c ST3. a ST. b ST5. d ST6. (a) 9 (b) 9 (c) 0 ST7.(a) 9 (b) (c) appro ST8. (a) L () =9 (b) L () =3 36 (c) L () = ST9.(a) 5 feet (b) 7.6 ft/sec (c) 6.6 ft./sec (d) 6 ft./sec., falling ST. f( + ) f() (b) 8 ST. (a) $.70 (b) $6.9 ST.(a) 3 (b) 6 people/year (c) people/year. Te population is increasing at te rate of persons per year after Þve years. ST3.(a) 8 (b) 3 (c) 0 ST.(a) L = + (b) L = ST5.(a) 3 πr ST6.(a) v ave feet/sec. (b) feet 8. Section.5. f 0 () =5, f 00 () = f 0 () = +, f 00 () = 5. f 0 () = 7 +, f 00 () = f 0 () =6 +5 8, f 00 () = 8 9. f 0 () =5 + 3, f 00 () = f 0 () =8 +, f 00 () =8 3. f 0 () =, f 00 () =0 5. f 0 () =0.3, f 00 () =0 7. f 0 () = , f 00 () = f 0 () = , f 00 () = f 0 () = /, f 00 () =/ 3 3. v(t) = 3t, a(t) = 3 5. v(t) =50 9.6t, a(t) = v(t) =35 0t 3, a(t) = 0t 9. v(t) =56 6t +3t, a(t) = 6 + 6t 3. (a) 0 (b) 0 (c) 0 (d) 0 (e) 0 (f) 0

9 . SECTION (a) r(t) =30 6t (b) r(t) =5.5+0t 6t (c) r(t) = 6t 38 6t 0 <t<.7 (d) r(t) =.7 <t<3.7 6t 3.7 <t 35. Mistake on problem in tet: r(t) =7.5+73t 6t likewise cange s(t). (a) ft. (b) sec. 33.L () =5, f 00 () =5 35.L () =/ +/ 37.L () = (b) d d ( +)( ) + ( +) d d ( ) ( ) + ( +) = 3 alternatively d d ( ) = 3 d. d (e )= d d e + d d e = e ( + ) d d 5. d ( ln()) = d ln()+ d d ln() =ln()+ d 6. d ( sin() = d d sin()+ d d sin() = sin()+ cos() 50. (a) F 0 (r) = 0.07r r (b) F (0.5) = 6.9, F 0 (0.5) = (c) It will increase since F 0 (0.5) > 0. (d) r = Section.6. f 0 () = =5 3. f 0 () = 0+ 3 = 3 5. f 0 () = + ( +)= + 7. f 0 () =(+)3+(3+) = + 9. f 0 () =(+)+(+) = 8 +. f 0 () =(3 3 +)9 +9 (3 3 +)= / 5. 3 / (7 ) 3. 5 (5+) 3. 5( +.) (5+) ( +. ) ( +. ) (+) 37. g()+ g 0 () 39. g 0 () + g(). g() g/() g() 3/ g 0 () 3. 3 / g()+ ) /. Section.7. ( +) 3. 8( +) ( +) ( +9) 9. (. ( 3 ) /3 3. 3³ 3 ( +3+) 7. ( + k ) 9. k (k +) / ( ) 3/ 8s µ +q (+ ) q (+ )

10 3. f() =( +),f 0 () = , f 00 () = = f() =( +9) 5, f 0 () =0 +9, f 00 () = = f() =, f 0 () =, f 00 () = 3 + = ( ) ( ) ³ (( )(+)) 3 ³ ( ) 9. g() = 3p ( 3 ), g 0 () = ³ 3, g 00 () = ³ (3 ) 3 + (3 ) 5 ³ 3 = ³ ( 3 ) 3 (( )( ++)) f 0 () = (g()+0) / g 0 () 35. f 0 () =6 g 0 (3 5) 37. f 0 () =g 0 (g( d )) d g( )=g 0 (g( )) g 0 ( ) dy 39. =3.5 =.5 dt =( 5) = d dt dy. dt = dp dt =6+ dw dt ft./sec π cm 3 d / sec 7. dt = 0.05 in/min ds 9. dt =. ft./sec. 5. (a) + y = (b) p/ p p (c) p p /p (d) negative reciprocals, te tangent to a point on te circle is perpendicular to te radius of te circle. 53. (a) y =5 (b) above (c) yes, tey are practically te same as one zooms in at te point (0,5) (d) f 00 (0) = Q 00 o(0) = (a) c 0 (t) =., g 0 (c) =0.06, c 0 (t) g 0 (c) =0.030 (b) g(c(t)) = 0.030t (c) R(g(c(t))) = t = t, R0 using te cain rule. (t) = dr dg dg dt = 3 g0 (t) = dg dc 3 dc dt. Section.8. y 0 = 3. y 0 = y 7. y 0 = y y 0 = y +y 5. y 0 = 7. y 0 = 3 +6y+3y 8y 3 +6y+3y y 3 y 0 =(y +)(y + y 0 ),Solvingfory 0 yields y 0 = y y y 3 y. 3(y +) (y + y 0 )=y + y 0,Solvingfory 0 yields : y 0 = 3 y 3 +6y +y 3 3 y +6 y+ 3. 3y y 0 +6yy 0 =3 +6, Solvingfory 0 yields y 0 = 3 6 3y +6y 5. L () = L.90 () = L () = y 0 = y Left: y 0 Rigt: p y + d d ( +3 +) ( +3 +)+ +3 p + +9 ( +3) y 0 = y. 3 3y 3. Left: y 0 y Rigt: d d ( + ) ( + ) ( +) +

11 . SELF-TEST Left: y 0 y Rigt: 3 d d (3 +) ( 3 +) 3 (3 +) Left: y 0 + y Rigt: 3 d d (3 +/)+( 3 +/) 3 (3 / )+ 3 +/ 3 3 / + 3 +/ 3. Self-Test.5-.8 Note: Problems 8 and ave typograpical errors so no correct answer appears. ST. (a) False, derivative rules are consistent (b) True (c) False, te product rule as been misapplied. (d) True (e) False, o( )=3 + 3 (f) False, te solution to a d.e. is a function. (g) False, te coefficient of 3 sould be 5. However f 0 is correct. () True (i) False, + 3 =5not 3. (j) True (k) True (l) False, no rate of cange does not mean zero rate of cange in matematics. ST. a. f( + ) =+5 + +( +5) + ;sof 0 () = +5 b. f 0 () = +3 ST3. (d) ST. (b) ST5. (d) ST6. (a) ST7. (c) ST π =. 639 π (no correct answer sown) ST9. (a) ST. R() = ; R () = +;sor 0 (0) =.37 (no correct answer sown) ST.. 5 feet,. v(t) = 3 3t ; a(t) = ft/sec.. t =sec.; feet. ST.(a) y = (3 +5 +) y 0 = (6 +5)=9 + + (b) y = + y0 = 0 ( +) (c) y = +3 5 y0 = = 7 (5 ) (5 ) (d) y =( + )3 y 0 =3 + + (e) y = k+ y 0 = k ³ (k+) 3 ST3. (a) +y 0 =0 (b) y ( ) + y ( ) y0 =0 y 0 = y (c) y +yy 0 + y + y 0 =0 (y + )y 0 = y y y 0 = y y y+

12 ST (a) n(n ) n,notetatn is a constant. (b) y = + y 0 = (+ ) y 00 = ³ 3 + = ³ 3 (+ ) (+ ) (+ ) (c) y = + y 0 = + = (+) (+) y 00 = + = (+) (+) 3 (+) 3 ST5 (a) f() = 5+ f 0 5 = So f 0 () = 5 5 Hence y = ³ (5+) 3 ( ) = (b) y =5 y +yy 0 =0 y 0 = y So y 0 = at (5, ). Hence y = + ( 5) = 3 + (c) f() = +k f 0 k () = (k+) So f 0 (0) = k Hence y =+ k