CHAPTER Vector-Valued Functions Section. Vector-Valued Functions...................... 9 Section. Differentiation and Integration of Vector-Valued Functions.... Section. Velocit and Acceleration..................... 8 Section. Tangent Vectors and Normal Vectors............... 5 Section.5 Arc Length and Curvature..................... Review Eercises................................. 8 Problem Solving................................. 7
CHAPTER Vector-Valued Functions Section. Solutions to Odd-Numbered Eercises Vector-Valued Functions. rt 5ti tj t k Component functions: ft 5t gt t ht t Domain:,,. rt ln ti e t j tk Component functions: ft ln t gt e t ht t Domain:, 5. 7. rt Ft Gt cos ti sin tj tk cos ti sin tj cos ti tk Domain:, i rt Ft Gt sin t Domain:, j cos t sin t k cos ti sin t cos tj sin tk cos t 9. rt t i t j (a) r i (b) r j (c) rs s i s j s i sj (d) r t r t i tj i j t t i tj i j t t i tj. rt ln ti j tk t (a) r ln i j k (b) r is not defined. ln does not eist. (c) (d) rt lnt i j t k t r t r ln ti j tk i j k t ln ti t j tk 9
Chapter Vector-Valued Functions. rt sin ti cos tj tk 5. rt sin t cos t t t rt ut t t t 8 t t t t t 5t t, a scalar. The dot product is a scalar-valued function. 7. rt ti tj t k, t 9. t, t, t Thus,. Matches (b) rt ti t j e.75t k, t t, t, e.75t Thus,. Matches (d). (a) View from the negative -ais:,, (c) View from the -ais:,, (b) View from above the first octant:,, (d) View from the positive -ais:,,. t 5. t, t 7. cos, sin t 7 5 5 5 9 Ellipse 9. sec, tan. t. cos t, sin t, t 9 9 9 9 Hperbola 9 t t Line passing through the points: (,, ),, 5, 5,, (,, ) (,, 5) 5 t Circular heli 7 5. sin t, cos t, e t 7. t, t, t e t, t (,, ) 5 (,, )
Section. Vector-Valued Functions 9. rt t i tj t k. Parabola Heli 5 rt sin ti cos t t j cos t k. (a) (b) (c) π π 8π π π π π π The heli is translated units back on the -ais. The height of the heli increases at a faster rate. The orientation of the heli is reversed. (d) (e) π π π The ais of the heli is the -ais. The radius of the heli is increased from to. 5. 7. Let t, then t. rt ti tj Let t, then t. rt ti t j 9. 5 Let 5 cos t, then 5 sin t. rt 5 cos ti 5 sin tj 5. Let sec t, tan t. rt sec ti tan tj 5. The parametric equations for the line are t, 5t, 8t. One possible answer is rt ti 5tj 8tk. 8 7 5 (, 8, 8) 5 (,, ) 7 8 55. r t ti, t r, r i r t ti tj, t r i, r j r t tj, t r j, r (Other answers possible)
Chapter Vector-Valued Functions 57. r t ti t j, t 59. r t ti, t r t tj, t (Other answers possible), Let t, then tand t. Therefore, t, t, t. rt ti tj t k ( ) ( ),, 5,,., sin t, cos t sin t t rt sin ti cos tj sin tk., Let sin t, then sin t and. sin t sin t sin t cos t, sin t, sin t ± cos t ± cos t t ± ± ± rt sin ti cos tj sin tk and rt sin ti cos tj sin tk 5., Subtracting, we have or ±. (,, ) Therefore, in the first octant, if we let t, then t, t, t. rt ti tj t k (,, )
Section. Vector-Valued Functions 7. t cos t t sin t t 9. 8 lim t ti t t t j t k i j k since t lim t t t lim t t. t (L Hôpital s Rule) 7 5 8 7. lim t t i tj since cos t k t 7. cos t sin t lim lim. t t t (L Hôpital s Rule) lim t t i cos tj sin tk does not eist since lim t t does not eist. 75. rt ti t j 77. rt ti arcsin tj t k Continuous on, Continuous on,,, 79. 8. rt e t, t, tan t 8. See the definition on page 78. Discontinuous at t n Continuous on n, n rt t i t j tk (a) st rt k t i t j t k (b) st rt i t i t j tk (c) st rt 5j t i t j tk 85. Let rt t tj tk and ut ti tj tk. Then: lim t c rt ut lim t c t t t ti t t t tj t t t tk lim t c t lim t c t lim t c t lim t c t i lim t c t lim t c t lim t c t lim t c t j lim t c lim ti lim tj lim t c t c lim rt lim ut t c t c t lim t c t lim t c t lim t c t k t c tk lim ti lim tj lim t c t c t c tk 87. Let rt ti tj tk. Since r is continuous at t c, then lim rt rc. t c rc ci cj ck c, c, c are defined at c. r t t t lim r c c c rc t c Therefore, r is continuous at c. 89. True
Chapter Vector-Valued Functions Section. Differentiation and Integration of Vector-Valued Functions. rt t i tj, t. t t, t t r i j rt ti j r i j rt is tangent to the curve. r (, ) r 8 rt cos ti sin tj, t t cos t, t sin t r j rt sin ti cos tj r i r (, ) r rt is tangent to the curve. 5. rt ti t j (a) (b) (c) 8 r r r r i j r i j r r i j r r r This vector approimates r. 8 rt i tj r i j i j i j 7. rt cos ti sin tj tk,, t π (,, ) r r j k r i k rt sin ti cos tj k π r π t 9. rt ti 7t j t k. rt i tj t k rt a cos ti a sin tj k rt a cos t sin ti a sin t cos tj. rt e t i j 5. rt e t i rt t sin t, t cos t, t rt sin t t cos t, cos t t sin t, 7. rt t i t j (a) rt t i tj rt t i j (b) rt rt t t t 8t t
Section. Differentiation and Integration of Vector-Valued Functions 5 9. rt cos ti sin tj. (a) rt sin ti cos tj rt cos t i sin tj (b) rt rt sin t cos t cos t sin t rt t i tj t k (a) rt ti j t k rt i tk (b) rt rt t t t t t. rt cos t t sin t, sin t t cos t, t (a) rt sin t sin t t cos t, cos t cos t t sin t, t cos t, t sin t, rt cos t t sin t, sin t t cos t, (b) rt rt t cos tcos t t sin t t sin tsin t t cos t t 5. rt costi sintj t k, t rt sinti costj tk r r r i j k r r r r r i j k rt costi sintj k r i j k r r r i j k 7. rt t i t j 9. rt ti t j r Smooth on,,, r cos i sin j r cos sin i 9 sin cos j r n Smooth on n n, n an integer.,. r sin i cos j. r cos i sin j r for an value of Smooth on, rt t i t j t k rt i t j tk r is smooth for all t :,,,
Chapter Vector-Valued Functions 5. rt ti tj tan tk rt i j sec tk r is smooth for all t n Smooth on intervals of form n. n, n 7. rt ti tj t k, ut ti t j t k (a) rt i j tk (c) rt ut t t t 5 D t rt ut 8t 9t 5t (e) rt ut t i t t j t t k D t rt ut 8t i t t j t tk (b) (d) rt k rt ut ti 9t t j t t k D t rt ut i 9 tj t t k (f) rt t t t t D t rt t t 9..855.87 rt sin ti cos tj rt cos ti sin tj rt rt 9 sin t cos t cos t sin t 7 sin t cos t cos maimum at minimum at rt rt rt rt 7 sin t cos t 9 sin t cos t9 cos t sin t arccos 7 sin t cos t t.97 5 t.5 9 sin t cos t9 cos t sin t and and for t n n,,,,....57, t.785. t 5.98 7. π 7. rt lim t rt t rt t lim t t t i t t j t i t j t lim t ti tt t j t lim t i t tj i tj. ti j k dt t i tj tk C 5. t i j t k dt ln ti tj 5 t5 k C 7. t i t j tk dt t ti t j t k C 9. sec ti t j dt tan ti arctan tj C
t 5. 8ti tj k dt t i j tk i j k 5. a cos ti a sin tj k dt a sin ti Section. Differentiation and Integration of Vector-Valued Functions 7 a cos tj tk ai aj k 55. rt e t i e t j dt e t i e t j C 57. r i j C i C j rt e t i e t j rt j dt tj C r C i j rt i tj rt i tj dt ti t t j C r C rt ti t t j 59. rt te t i e t j k dt et i e t j tk C r i j C i j k C i j k rt i et et j t k e t i et j t k. See Definition of the Derivative of a Vector-Valued Function and Figure.8 on page 79.. At t t, the graph of ut is increasing in the,, and directions simultaneousl. 5. Let rt ti tj tk. Then crt cti ctj ctk and D t crt cti ctj ctk cti tj tk crt. 7. Let rt ti tj tk, then f trt f tti f ttj fttk. D t f trt f tt ftti f tt fttj ftt fttk f tti tj tk ftti tj tk f trt ftrt 9. Let rt ti tj tk. Then r f t f ti f tj f tk and D t r f t f t fti f t ftj f t ftk (Chain Rule) ft f ti f tj f tk ftr ft.
8 Chapter Vector-Valued Functions 7. Let rt ti tj tk, ut ti tj tk, and vt ti tj tk. Then: rt ut vt t t t t t t t t t t t t t t t D t rt ut vt t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t rt ut vt rt ut vt rt ut vt 7. False. Let rt cos ti sin tj k. rt d rt dt rt sin ti cos tj rt Section. Velocit and Acceleration. rt ti t j. rt t i tj vt rt i j at rt t, t, At,, t. v i j, a (, ) v vt rt ti j at rt i t, t, At,, t. v i j a i (, ) a 8 v 5. rt cos ti sin tj 7. vt rt sin ti cos tj at rt cos ti sin tj cos t, sin t, At,, t. v i j a i j a v (, ) rt t sin t, cos t vt rt cos t, sin t at rt sin t, cos t t sin t, cos t (ccloid) At,, t. v, i a, j ( π, ) v a π π
Section. Velocit and Acceleration 9 9. rt ti t 5j tk. vt i j k st vt 9 at rt ti t j t k vt i tj tk st t t 5t at j k. rt ti tj 9 t k 5. vt i j t 9 t k st t 9 t 8 t 9 t 9 at 9 t k rt t, cos t, sin t vt, sin t, cos t i sin tj cos tk st 9 sin t 9 cos t 5 at, cos t, sin t cos tj sin tk 7. (a) rt t, t, t, t t rt, t, r,, (b) r..,.,..,.,.5 t, t, t 9. at i j k, v, r. vt i j k dt ti tj tk C v C, vt ti tj tk, vt ti j k rt ti tj tk dt t i j k C r C, rt t i j k, r i j k i j k at tj tk, v 5j, r vt tj tk dt t j t k C v j k C 5j C 9 j k vt t 9 j t k rt t 9 j t k dt t 9 t j t t k C r j k C C j k rt t 9 t j t t k r 7 j k. The velocit of an object involves both magnitude and direction of motion, whereas speed involves onl magnitude. 5. rt 88 cos ti 88 sin t t j ti t t j 5
5 Chapter Vector-Valued Functions 7. rt v cos ti h v sin t gt j v ti v t t j v t when t v, v, v 9, v 97.98 ftsec The maimum height is reached when the derivative of the vertical component is ero. t tv t t t t t t t t v t t. v v 5 v, v Maimum height: 5 5 5 78 feet 9. t tv cos or t v cos t tv sin t h v v cos sin v cos h tan v sec h. rt ti.t.7t j, or (a)..7 (b) 8.8.7 5.875 (c) 5.875. feet. and (d) From Eercise 9, tan.7 sec v. v sec. cos v 7. ftsec... mph miles feet hr 58 sechour mile ftsec (a) rt cos ti sin t t j (b) Graphing these curves together with shows that. 5 CONTINUED
Section. Velocit and Acceleration 5. CONTINUED (c) We want t cos t and From t, the minimum angle occurs when t cos. Substituting this for t in t ields: sin cos cos, tan sec 7, tan tan 7, tan 8, tan 5,7 t sin t t. tan 8, ± 8,,5,7, tan 8,,,,8 8,8 9.8 5. rt v cos ti v sin t t j (a) We want to find the minimum initial speed v as a function of the angle. Since the bale must be thrown to the position, 8, we have v cos t 8 v sin t t. t v cos from the first equation. Substituting into the second equation and solving for v, we obtain: 5 sin v cos cos 5 v sin cos cos 5 We minimie f. sin cos cos f 5 cos sin sin cos sin cos cos f cos sin Substituting into the equation for v, v 8.78 feet per second. (b) If 5, v cos t v t 8 v sin t t v t t From part (a), v 8 v sin sin 5 cos v cos v sin cos v cos cos 5 tan v cos sin cos cos 5.7 58.8 5 5 v ftsec.
5 Chapter Vector-Valued Functions 7. rt v cos ti v sin t t j v sin t t when The range is v cos t v cos v sin Hence, sin t and v t v sin. sin. sin 5.9. 9. (a),(b) rt cos ti sin t t j rt 5ti.t t j Maimum height:.5 feet Range:.557 feet 5 v ftsec, rt cos ti sin t t j rt.78ti 5.5t t j Maimum height:. feet Range: 7.88 feet 5 v ftsec 5 (c) 5, rt cos 5ti sin 5t t j rt.7ti.7t t j Maimum height:. feet Range:.5 feet v ftsec (d) 5, rt cos 5ti sin 5t t j rt.ti.t t j Maimum height:.5 feet Range:.5 feet v ftsec 8 (e), rt cos ti sin t t j rt ti 57.t t j Maimum height: 5.7 feet Range: 7.888 feet v ftsec (f), rt cos ti sin t t j rt 7ti.t t j Maimum height: 9.797 feet Range: 57.88 feet v ftsec
Section. Velocit and Acceleration 5. rt v cos ti h v sin t.9t j cos ti.5 sin t.9t j The projectile hits the ground when.9t t.5 t. seconds. The range is therefore cos. 88. meters. The maimum height occurs when ddt. sin 9.8t t 5. sec The maimum height is.5 sin 5..95. 9. meters.. rt bt sinti b cos tj vt b cos ti b sin t j b cos ti b sin tj at b sin ti b cos tj b sinti costj vt b cost at b (a) vt when t,,,.... (b) vt is maimum when then vt b. t,,..., 5. vt b sinti b costj rt vt b sint cost b sint cost Therefore, rt and vt are orthogonal. 7. at b costi b sintj b costi sintj rt at is a negative multiple of a unit vector from, to cos t, sin t and thus at is directed toward the origin. 9. at b m F m b radsec vt b 8 ftsec 5. To find the range, set t h v then gt sin t gt v sin t h. B the Quadratic Formula, (discount the negative value) t v sin v sin gh g At this time, v sin v sin gh. g t v cos v sin v sin gh g v cos g v sin v sin gh v v cos g sin sin gh v.
5 Chapter Vector-Valued Functions 5. rt ti tj tk Position vector vt ti tj tk Velocit vector at ti tj tk Acceleration vector Speed vt t t t Orthogonal C, C is a constant. d dt t t t tt tt tt tt tt tt vt at 55. rt cos ti sin tj (a) vt rt sin ti cos tj vt sin t 9 cos t sin t cos t sin t at vt cos ti sin tj (c) 9 9 (b) t Speed (d) The speed is increasing when the angle between v and a is in the interval, The speed is decreasing when the angle is in the interval.,. Section. Tangent Vectors and Normal Vectors. rt t i tj. rt ti j, rt t t Tt rt ti j rt t ti j t T i j i j rt cos ti sin tj rt sin ti cos tj rt sin t cos t Tt rt sin ti cos tj rt T i j 5. rt ti t j tk 7. rt i tj k When t, r i k, t at,,. T r r i k Direction numbers: a, b, c Parametric equations: t,, t rt cos ti sin tj tk rt sin ti cos tj k When t, r j k, t at,,. T r r 5 j k 5 Direction numbers: a, b, c Parametric equations:, t, t
Section. Tangent Vectors and Normal Vectors 55 9. rt cos t, sin t, rt sin t, cos t, When t, r,,, t r T r,, Direction numbers: a, b, c Parametric equations: t, t, at,,.. rt t, t, t rt, t, t When t, r,, 8, t at, 9, 8. T r r,, 8 9 8 5 9 9 5 8 Direction numbers: a, b, c 8 Parametric equations: t, t 9, 8t 8. rt ti ln tj tk, t 5. rt i t j t k r i j k T rt i j k rt i j k r,, u8,, Hence the curves intersect. rt, t, r,, 8, Tangent line: t, t, t rt. r..i.j.5k.,.,.5 us,, s, u8,, cos r u8.97 r u8.95. 7. rt ti t j, t 9. rt i tj Tt rt i tj rt t Tt t t i t j T i 5 5 j N T T 5 i j i 5 5 5 5 j rt sin ti cos tj Tt rt sin ti cos tj rt Tt cos ti sin tj, Tt N rt cos ti sin tj k, t i j
5 Chapter Vector-Valued Functions. rt ti vt i at O Tt vt i vt i Tt O Nt Tt is undefined. Tt The path is a line and the speed is constant.. rt t i vt 8ti at 8i Tt vt 8ti vt 8t i Tt O Nt Tt is undefined. Tt The path is a line and the speed is variable. 5. rt ti j, vt i v i j, t t j, 7. at a j t j, Tt vt vt t t i t j t t i j T Nt Tt Tt N i j i j i j i j a T a T a N a N t t i t t j t t t i t j rt e t cos ti e t sin tj vt e t cos t sin ti e t cos t sin tj at e t sin ti e t cos tj At t T v v i j i j., Motion along r is counterclockwise. Therefore, N i j i j. a T a T e a N a N e 9. rt cos t t sin t i sin t t cos t j vt t cos t i t sin t j at cos t t sin t i t cos t sin t j Tt v v cos t i sin t j Motion along r is counterclockwise. Therefore Nt sin t i cos t j. a T a T a N a N t t
Section. Tangent Vectors and Normal Vectors 57. rt a costi a sintj. Speed: vt a vt a sinti a costj The speed is constant since a T. at a costi a sintj Tt vt vt sinti costj Nt Tt Tt costi sintj a T a T a N a N a 5. rt ti t j, t 7. rt ti tj tk t, t rt i t j Tt t i j t Nt i t j t r i j T 7 i j 7 N 7 i j 7, N T vt i j k at Tt v v i j k i j k Nt T is undefined. T are not defined. a T, a N 9. rt ti t j t k. rt ti cos tj sin tk vt i tj tk v i j k at j k Tt v v 5ti tj tk T i j k 5ti j k Nt T T 5t 5ti j k 5 5 5t 5t N 5i j k a T a T 5 a N a N vt i sin tj cos tk v i j at cos tj sin tk a k Tt v v i sin tj cos tk 5 T i j 5 Nt T cos tj sin tk T N k a T a T a N a N T N π π
58 Chapter Vector-Valued Functions. Tt rt rt Nt Tt Tt If at a T Tt a N Nt, then a T is the tangential component of acceleration and is the normal component of acceleration. a N 5. If a N, then the motion is in a straight line. 7. rt t sin t, cos t The graph is a ccloid. (a) rt t sin t, cos t vt cos t, sin t at sin t, cos t t = t = t = Tt vt vt cos t cos t, sin t Nt Tt Tt cos t sin t, cos t a T a T a N a N cos t sin t cos t cos t sin t sin t cos t sin cos t t cos t cos t cos t cos t cos t When t : a T, a N When t : a T, a N When t : a T, a N (b) Speed: s vt cos t ds dt sin t cos t a T When t : a T > the speed in increasing. When t : a T the height is maimum. When t : a T < the speed is decreasing.
Section. Tangent Vectors and Normal Vectors 59 9. rt cos ti sin tj t k, t rt sin ti cos tj k Tt 7 7 sin ti cos tj k Nt cos ti sin tj r j 7 T 7 i k N j B k T N 7 i k 7 i 7 7 j k 7 7 7 7 i 7 7 7 k i k 7 N B T (,, π ) 5. From Theorem. we have: rt v t cos i h v t sin t j vt v cos i v sin tj at j Tt v cos i v sin tj v cos v sin t Nt v sin ti v cos j v cos v sin t a T a T v sin t v cos v sin t v a N a N cos v cos v sin t (Motion is clockwise.) Maimum height when v sin t ; (vertical component of velocit) At maimum height, a T and a N. 5. rt cos t, sin t, t, t (a) rt sint, cost, rt sin t cos t 5 mihr (b) a T and a N a T because the speed is constant. 55. rt a cos ti a sin tj From Eercise, we know a T and a N a. (a) Let. Then a N a a a or the centripetal acceleration is increased b a factor of when the velocit is doubled. (b) Let a a. Then a N a a a or the centripetal acceleration is halved when the radius is halved.
Chapter Vector-Valued Functions 57. v 9.5.8 misec 59. v 9.5 85.7 misec. Let Tt cos i sin j be the unit tangent vector. Then M sin i cos j cos i sin j and is rotated counterclockwise through an angle of from T. If ddt >, then the curve bends to the left and M has the same direction as. Thus, M has the same direction as which is toward the concave side of the curve. Tt dt dt N T T, dt dt sin i cos j d dt T d d M d dt. If ddt <, then the curve bends to the right and M has the opposite direction as T. Thus, again points to the concave side of the curve. N T T M T φ M T φ N. Using a a T T a N N, T T O, and T N, we have: v a va N T N Thus, a N v a vt a T T a N N va T T T va N T N va N T N va N v a. v Section.5 Arc Length and Curvature. rt ti tj. d, dt d, dt s 9 dt dt t d dt rt a cos ti a sin tj d dt a cos t sin t, a sin t cos t dt a d dt a sin t cos t s a cos tsin t a sin t cos t dt sin t dt a cos t a (, ) a 8 a a (, ) 8 a
Section.5 Arc Length and Curvature 5. (a) rt v cos ti h v sin t gt j cos 5ti sin 5t t j 5ti 5t t j (b) vt 5i 5 tj 5 t t 5 Maimum height: 5 5 5 8.5 ft (c) 5t t t. Range: 5. 5.5 feet. (d) s 5 5 t dt.9 feet 7. rt ti tj tk d dt s dt d dt, d dt dt t (,, ) (,, ) 9. rt a cos ti a sin tj btk. d a sin t, dt s ( a,, πb) πb a sin t a cos t b dt a b dt a b t d a cos t, dt d dt b a b rt t i tj ln tk d t, dt s t t dt t t dt t d, dt d dt t t t t dt 8.7 πb ( a,, ). rt ti t j t k, t (a) r,,, r,, 8 distance 8 8 9.5 CONTINUED
Chapter Vector-Valued Functions. CONTINUED (b) r,, r.5.5,.75,.5 r,, r.5.5,.75,.75 r,, 8 distance.5.5.5.5.75.875.5.5.75.5.75.5.578.5.7.97 9.59 (c) Increase the number of line segments. (d) Using a graphing utilit, ou obtain 9.5757. 5. rt cos t, sin t, t t (a) s u u u du t sin u cos u du t t 5 du 5u 5t (c) When s 5: When s : cos.8 sin.8.8,.8,. cos. 5 sin.95 5 (b) s 5 t cos s 5, sin s 5, rs cos s 5 i sin s j s 5 s 5 5 k 7..789 5.,.95,.789 (d) rs 5 5 sin s 5 5 cos s 5 5 5 rs s i s j 9. rs i j Ts rs rs rs and Ts K Ts rs (The curve is a line.) rs cos s 5 i sin s j s 5 Ts rs 5 sin s i 5 Ts 5 cos s 5 i 5 sin s 5 j K Ts 5 5 k 5 cos s j 5 5 k
Section.5 Arc Length and Curvature. rt ti tj. rt ti t j vt i j Tt Tt i j 5 K Tt rt (The curve is a line.) vt i t j v i j at t j a j Tt t i j t Nt t i t j N i j K a N v 5. rt costi sintj7. rt 8 sinti 8 costj Tt sinti costj Tt costi sintj K Tt rt 8 rt a costi a sintj rt a sinti a costj Tt sinti costj Tt costi sintj K Tt rt a a 9. rt e t cos ti e t sin tj. rt e t sin t e t cos ti e t cos t e t sin tj Tt sin t cos ti cos t sin tj Tt cos t sin ti sin t cos tj rt cos t t sin t, sin t t cos t From Eercise, Section., we have: a N t K at Nt v t t t K Tt rt et et. rt ti t j t rt i tj tk Tt Tt K Tt rt i tj tk 5t 5ti j k 5t k 5. 5 5t 5t 5 5t rt ti cos tj sin tk rt i sin tj cos tk Tt i sin tj cos tk 5 Tt cos tj sin tk 5 K Tt 5 rt 5 5
Chapter Vector-Valued Functions 7., Since is undefined. K, and the radius of curvature 9. K.57 7 7 7.5 K (radius of curvature). a a a At : a a K K a a a (radius of curvature). (a) Point on circle: Center:, Equation:, (b) The circles have different radii since the curvature is different and r K. 5.,, 7. K Radius of curvature. Since the tangent line is horiontal at,, the normal line is vertical. The center of the circle is unit above the point, at, 5. Circle: 5 (, ) e, e, e, K r, K The slope of the tangent line at, is. The slope of the normal line is. Equation of normal line: or The center of the circle is on the normal line units awa from the point,. Since the circle is above the curve, and. Center of circle:, Equation of circle: 8 8 ± (, )
Section.5 Arc Length and Curvature 5 9. π π B A 5., K, (a) K is maimum when or at the verte,. π (b) lim K 5. 57., 9, 55. K 9 9 9 (a) K as. No maimum (b) lim K Curvature is at,. b 59. s K The curvature is ero when. a K 9 rt dt. The curve is a line. at.. Endpoints of the major ais: Endpoints of the minor ais: 8 ±,, ± K Therefore, since, K is largest when ± and smallest when. 5. f (a) K (b) For, K. f. At,, the circle of curvature has radius. Using the smmetr of the graph of f, ou obtain. For, K 5 5. f. At,, the circle of curvature has radius 5 K. Using the graph of f, ou see that the center of curvature is,. Thus, f 5. To graph these circles, use ± and ± 5. CONTINUED
Chapter Vector-Valued Functions 5. CONTINUED (c) The curvature tends to be greatest near the etrema of f, and K decreases as However, f and K do not have the same critical numbers. ±. 5 Critical numbers of f:, ± ±.77 Critical numbers of K:, ±.77, ±.8 7. (a) Imagine dropping the circle k into the parabola. The circle will drop to the point where the tangents to the circle and parabola are equal. and k k 5. Taking derivatives, k and Hence, Thus, k k. 5 5. k k k k Thus, k 5.75. Finall, k.5, and the center of the circle is.5 units from the verte of the parabola. Since the radius of the circle is, the circle is.5 units from the verte. (b) In -space, the parabola or has a curvature of K at,. The radius of the largest sphere that will touch the verte has radius K. 9. Given f : K R K The center of the circle is on the normal line at a distance of R from,. Equation of normal line: Thus,,,. For e, When : Center of curvature:, (See Eercise 7) e, e, e e e e.
Section.5 Arc Length and Curvature 7 7. 75. r sin r cos r sin K r rr r r r cos sin sin sin cos sin sin 8 sin sin r e a, a > r ae a r a e a K r rr r a e a a e a e a r r a e a ea e aa (a) As, K. (b) As a, K. 7. 77. r a sin r a cos r a sin K r rr r r r a cos a sin a sin a cos a sin a a a, a > r sin r 8 cos At the pole: K r 8 79. f t 8. gt d d d dt gt ft d dt ftgt gtft ft ftgt gtft ft K d dt gt d ft dt ftgt gtft ft ft gt ft ftgt gtft ft ft gt ft ftgt gtft ft gt a sin a cos a sin K a cos cos a sin a cos a sin cos cos a cos a sin a cos a cos cos a cos a cos a csc Minimum: a Maimum: none K as 8. a N mk ds dt 55 lb 58 ft ftsec ft sec 7.5 lb
8 Chapter Vector-Valued Functions 85. Let r ti tj tk. Then r r t t t and Then, r dr dt t t t t t t tt tt tt r r. r ti tj tk. tt tt tt 87. Let r i j k where,, and are functions of t, and r r. d dt r r rr rdrdt rr rr rr r r r rr r r r (using Eercise 77) i j k i j k r r i j k i j k r r r r r 89. From Eercise 8, we have concluded that planetar motion is planar. Assume that the planet moves in the -plane with the sun at the origin. From Eercise 88, we have Since and r are both perpendicular to L, so is e. Thus, e lies in the -plane. Situate the coordinate sstem so that e lies along the positive -ais and is the angle between e and r. Let e e. Then r e r e cos re cos. Also, Thus, r L GM r r L L L L r r L r r L r GM e r r L GM r e cos r e. GM r e r r r GMre cos r and the planetar motion is a conic section. Since the planet returns to its initial position periodicall, the conic is an ellipse. Sun θ e r Planet 9. A r d Thus, d d da da dt dt r d dt L and r sweeps out area at a constant rate. Review Eercises for Chapter. rt ti csc tk. (a) Domain: t n, n an integer (b) Continuous ecept at t n, n an integer rt ln ti tj tk (a) Domain:, (b) Continuous for all t >
Review Eercises for Chapter 9 5. (a) r i (b) r i j 8 k (c) rc c i c j c k c i c j c k (d) r t r t i t j t k i j k ti tt j t t tk 7. rt cos ti sin tj 9. rt i tj t k. rt i sin tj k t cos t, t sin t, sin t, t t t. rt ti ln tj t k 5. One possible answer is: r t ti tj, t r t i tj, t r t ti, t 7. The vector joining the points is 7,,. One path is 9. rt 7t, t, 8 t.,, t t, t, t rt ti tj t k 5. lim t t i t j k i k
7 Chapter Vector-Valued Functions. rt ti t j, ut ti t j t k (a) rt i j (b) rt (c) rt ut t t t t t (d) ut rt 5ti t t j t k D t rt ut t t D t ut rt 5i t j t k (e) rt t t (f ) rt ut t t i t j t t tk D t rt t t t D t rt ut 8 t t i 8t j 9t t k 5. t and t are increasing functions at t t, and t is a decreasing function at t t. 7. cos ti t cos tj dt sin ti t sin t cos tj C 9. cos ti sin tj tk dt t dt t t ln t t C. rt ti e t j e t k dt t i e t j e t k C r j k C i j 5k C i j k rt t i e t j e t k. ti t j t k dt t t i j t k j e t i t j k dt e t i t j tk 5. e i 8j k 7. rt cos t, sin t, t vt rt cos t sin t, sin t cos t, vt 9 cos t sin t 9 sin t cos t 9 cos t sin tcos t sin t cos t sin t at vt cos tsin t cos t cos t, sin t cos t sin tsin t, cos t sin t cos t, sin t cos t sin t, 9. rt lnt, t, t rt t, t, r, 8,, t. Range v feet sin 75 sin 5 direction numbers Since r,,, the parametric equations are t, 8t, t. rt. r..,.8,.5. v 89.8 Range v.9 msec 9.8 sin 8 sin
Review Eercises for Chapter 7 5. rt 5ti 7. vt 5i vt 5 at Tt i Nt does not eist a T a N does not eist (The curve is a line.) rt ti tj vt i t j t vt t at tt j Tt Nt i tj ti j t t t i tj t a T ttt a N tt 9. rt e t i e t j 5. vt e t i e t j vt e t e t at e t i e t j Tt et i e t j e t e t Nt et i e t j e t e t a T et e t e t e t a N e t e t rt ti t j t k vt i tj tk v 5t at j k Tt Nt 5t a T 5t a N i tj tk 5t 5ti j k 5 5t 5 5 5t 5 5t 5. rt cos ti sin tj tk, cos t, sin t, t When t,,, rt sin ti cos tj k. Direction numbers when t a, b, c, t, t, t 55. v 9.5.5 misec 57. rt ti tj, t 5 rt i j s b rt dt 5 9 dt a 5 t 5 8 (, ) 8 (, 5)
7 Chapter Vector-Valued Functions 59. rt cos ti sin tj rt cos t sin ti sin t cos tj rt cos t sin t sin t cos t cos t sin t s cos t sin t dt sin t. rt ti tj tk, t rt i j k b s a rt dt 9 dt 9 dt 9 8 ( 9,, ) (,, ) 8. rt 8 cos t, 8 sin t, t, t 5. rt ti sin tj cos tk, t rt < 8 sin t, 8 cos t,, rt 5 rt dt 5 5 dt b s a 8 (8,, ) π (, 8, ) 8 π rt i cos tj sin tk s rt dt cos t sin t dt 5 dt 5 t 5 7. rt ti tj 9. Line k rt ti t j t k rt i tj tk, r 5t rt j k r r i j t K r r r k t j k, r r 5 5t 5t 7. 7. ln K At, K 7 and r 7 77., K ] At, K and r. 75. The curvature changes abruptl from ero to a nonero constant at the points B and C.