Chapter Let. 1) k be a vector-valued function. (a) Evaluate f (0). (b) What is the domain of f () t? (c) Is f () t continuous at t = 1?

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1 Chaper. Le f() = sin i+ ( 3+ ) j ln( + ) k be a vecor-valued funcion. (a) Evaluae f (). (b) Wha is he domain of f ()? (c) Is f () coninuous a =?

2 Chaper. Le f() = sin i+ ( 3+ ) j ln( + ) k be a vecor-valued funcion. (a) Evaluae f (). f () = sin, 3 +, ln( + ) =,, = j + subsiuion (b) Wha is he domain of f ()? The domain of f () is he inersecion of he domains of each componen. The domain of sin is all real numbers; he domain of 3+ is all real numbers; he domain of ln( + ) is >. Hence, he domain of f ( ) is >. (c) Is f () coninuous a =? Since ln( + ) is no defined a =, f () canno be coninuous a = since lim ln( + ) does no exis. + + considers componens + componen domains + ln( + ) no defined a = + raionale

3 Chaper 3. Le 3 f() = e i ( + 4) j+ cos k be a vecor-valued funcion. (a) Evaluae f (). (b) Wha is he domain of f ()? (c) Is f () coninuous a = π?

4 Chaper 4. Le 3 f() = e i ( + 4) j+ cos k be a vecor-valued funcion. (a) Evaluae f (). 3 f() = e, ( + 4), cos =, 4, = i+ 4j+k + subsiuion (b) Wha is he domain of f ( )? The domain of f ( ) is he inersecion of he domains of each componen. The domain of e is all real numbers; he domain 3 of ( + 4) is all real numbers; he domain of cos is all real numbers. Hence, he domain of f () is all real numbers. (c) Is f ( ) coninuous a = π? Since lim ( ) ( ) everywhere. f = f a every real value, f ( ) is coninuous + considers componens + componen domains + () f defined a every real + raionale

5 Chaper 5 3. Le f () = 3 4,,sec be a vecor-valued funcion. (a) Find lim f ( ). (b) Calculae f (). (c) Calculae f () d.

6 Chaper 6 3. Le f () = 3 4,,sec be a vecor-valued funcion. (a) Find lim f ( ). lim f ( ) = lim 3 4,, sec = lim 3 4, lim, lim sec = 3 4,,sec =,, = j+ k (b) Calculae f (). d d d d d f ( ) = 3 4,,sec = 3 4,, sec = d d d d d = 6 4,,sec an. Thus, f () = 6 4,, sec an = 4,, = 4i (c) Calculae f () d. f ( ) d= 3 4,,sec d= 3 4 d, d, secd = ( ) 3 π =, sin +, ln sec+ an =,, ln sec+ an = 4 π = i + j+ ln sec+ an k 4 + componen limis + componen derivaives + f () + subsiuion + componen inegrals + aniderivaives

7 Chaper 7 4. Le () =,an, f be a vecor-valued funcion. (a) Find lim f ( ). (b) Calculae f (). (c) Calculae f () d.

8 Chaper 8 4. Le () =,an, f be a vecor-valued funcion. (a) Find lim f ( ). lim f ( ) = lim, an, + 3 = lim, lim an, lim =, an, =,, = i (b) Calculae f (). d d d d d f (),an, 3, an, 3 d d + d + d d,sec =, + 6. ( + ) Thus, f () =, sec, + 6 =,, = i+ j+ k (+ ) (c) Calculae = + = + = f () d. d = + + d = + d d + d = f (),an, 3, an, 3 ln( ), ln cos, 3 ln, ln cos, ln ln cos = + + = = i j+ k + componen limis + componen derivaives + f () + subsiuion + componen inegrals + aniderivaives

9 Chaper 9 5. The barrel of a cannon makes an angle of 57 above he horizonal. A cannonball is fired f wih iniial speed 8. (Assume ha he cannonball leaves he barrel a a heigh of 3 f s above he ground.) (a) Wrie he posiion funcion r ( ) for he cannonball. (b) How high does he cannonball ge? (c) How far away from he cannon does he cannonball hi he ground?

10 Chaper 5. The barrel of a cannon makes an angle of 57 above he horizonal. A cannonball is fired f wih iniial speed 8. (Assume ha he cannonball leaves he barrel a a heigh of 3 f s above he ground.) (a) Wrie he posiion funcion r ( ) for he cannonball. r() = [ vcosθ ] i+ h+ vsinθ g j = [ 8cos 57 ] 3 8sin 57 = i j = = i j (b) How high does he cannonball ge? This will occur when he derivaive of he second componen is zero. d Thus, d + = gives us ha =, or = = 3.9 seconds. So, h max = (3.9) 6(3.9) = 6 = 56.7 fee. (c) How far away from he cannon does he cannonball hi he ground? We need o find when he cannonball his he ground. This will occur when he second componen is equal o zero. Hence, = gives us ha =.3 seconds or = 6.5 seconds. So, x max = 64.67(6.5) = fee. + formula + subsiuions + locaion idea + calculaion + locaion idea + calculaion

11 Chaper 6. A quarerback hrows a fooball wih iniial speed m s a an angle of 43 above he horizonal. (Assume ha he quarerback releases he fooball a a heigh of m above he ground.) (a) Wrie he posiion funcion r ( ) for he fooball. (b) How high does he fooball ge? (c) How far away from he quarerback does he fooball hi he ground?

12 Chaper 6. A quarerback hrows a fooball wih iniial speed m s a an angle of 43 above he horizonal. (Assume ha he quarerback releases he fooball a a heigh of m above he ground.) (a) Wrie he posiion funcion r ( ) for he fooball. r() = [ vcosθ ] i+ h+ vsinθ g j = [ cos 43 ] sin 43 = i j = = 6.9 i j (b) How high does he fooball ge? This will occur when he derivaive of he second componen is zero. Thus, d d + = gives us ha =, or = = seconds. So, h max = + 5.4(.59) 4.95(.59) = 9.8 = meers. (c) How far away from he quarerback does he fooball hi he ground? We need o find when he fooball his he ground. This will occur when he second componen is equal o zero. Hence, = gives us ha =.8 seconds or = 3.87 seconds. So, x max = 6.9(3.87) = 5.79 meers. + formula + subsiuions + locaion idea + calculaion + locaion idea + calculaion

13 Chaper 3 7. Le r( ) = ( + ) i e j represen a posiion vecor. (a) Find he velociy vecor v () and he acceleraion vecor a () associaed wih r (). (b) Find he uni angen vecor T () and he principal uni normal vecor N () a =. (c) Find he angenial and normal componens of acceleraion a T and a N a =.

14 Chaper 4 7. Le r( ) = ( + ) i e j represen a posiion vecor. (a) Find he velociy vecor v () and he acceleraion vecor a () associaed wih r (). d d d v() = r () = ( +, ) ( e ) =, e d d d d d d a() = v () =, ( e ) =, e d d d (b) Find he uni angen vecor T () and he principal uni normal vecor N () a =. + v () + a () v(), e T() = = =, e v() + e + e T () =, =,. +, so ha + T () + T () + T () + N () + N () Also, e e, 3/ 3/ T () ( + e ) ( + e ) N() = = = T () 4 e e + ( + e ) ( + e ) 3 3 e e =, = 3/ 3/ e e e + ( + e ) 3 ( ) ( + ) ( + e ) + e e e e =, =, e + e + e ( + e ) ( + e ) 3/ 3/ e N () =, =,. + e + e, so ha

15 Chaper 5 (c) Find he angenial and normal componens of acceleraion a T and a N a =. e e at = at =, e, =, so ha + e + e + e a e T = = =. + e e e an = an =, e, =, so ha + e + e + e a e N = = =. + e + a T + a N

16 Chaper 6 8. Le r( ) = ln( ) i+ ( 3) j represen a posiion vecor. (a) Find he velociy vecor v () and he acceleraion vecor a () associaed wih r (). (b) Find he uni angen vecor T () and he principal uni normal vecor N () a =. (c) Find he angenial and normal componens of acceleraion a T and a N a =.

17 Chaper 7 8. Le r( ) = ln( ) i+ ( 3) j represen a posiion vecor. (a) Find he velociy vecor v () and he acceleraion vecor a () associaed wih r (). v() = d r () = d ln(), d ( 3 ) =, d d d d d d a() = v () =, =, d d d (b) Find he uni angen vecor T () and he principal uni normal vecor N () a =. + v () + a (), v() T() = = =, =, v() T () =, =, , so ha + T () + T () + T () + N () + N () Also, 4, 3/ 3/ T () ( + 4 ) ( + 4 ) N() = = = T () ( + 4 ) ( + 4 ) 4 =, = 3/ 3/ ( + 4 ) 3 ( 4 ) ( + ) ( + ) =, =, ( + 4 ) ( + 4 ) 3/ 3/ N () =, =, , so ha

18 Chaper 8 (c) Find he angenial and normal componens of acceleraion a T and a N a =. at = at =,, =, so ha a T = = = an = an =,, =, so ha a N = = = a T + a N

19 Chaper 9 9. Le f( ) = 3cos( ) i+ 8j + 3sin( ) k define a curve. (a) Find he arc lengh from π = o = π. (b) Calculae he arc lengh funcion s () for in he inerval [, ). (c) Find he curvaure Κ of f () a any value of.

20 Chaper 9. Le f( ) = 3cos( ) i+ 8j + 3sin( ) k define a curve. (a) Find he arc lengh from π = o = π. b π [ ()] [ ()] [ ()] π [ 6sin() ] 8 [ 6cos() ] a L= x + y + z d = + + d = π = + + = + = = π π π 36sin ( ) 64 36cos ( ) d π d π 5 π. (b) Calculae he arc lengh funcion s ( ) for in he inerval [, ). [ ] [ ] [ ] s( ) = x ( u) + y ( u) + z ( u) du = 36sin ( u) cos ( u) du = a = du =. (c) Find he curvaure Κ of f ( ) a any value of. s Since s =, we have =. Then, f 4 () 3cos s 8 s 3sin s s 3cos s s 3sin s = i j k= 5 i 5 j 5 k Hence, f s s () s = sin + + cos 5 5 i 5 j 5 5 k. Then, f () s 3 s 4 3 s T() s = = sin i+ j+ cos k since f () s =. Thus, f () s formula + derivaives + formula + solve for + f ( s) + T ( s) + κ 3 s 3 s 3 s 3 s Κ = T () s = cos sin = cos + sin = 5 5 i 5 5 k s s = cos + sin = =

21 Chaper. Le f( ) = 4 i+ sin(3 ) j cos(3 ) k define a curve. (a) Find he arc lengh from = o = 3. (b) Calculae he arc lengh funcion s () for in he inerval [, ). (c) Find he curvaure Κ of f () a any value of.

22 Chaper. Le f( ) = 4 i+ sin(3 ) j cos(3 ) k define a curve. (a) Find he arc lengh from = o = 3. b 3 [ ()] [ ()] [ ()] 4 [ 3cos(3) ] [ 3sin(3) ] a cos (3 ) 9sin (3 ) d 6 9 d 5( 3 ). L= x + y + z d = + + d = = + + = + = = + formula + derivaives (b) Calculae he arc lengh funcion s ( ) for in he inerval [, ). [ ] [ ] [ ] s( ) = x ( u) + y ( u) + z ( u) du = 4 + 9sin (3 u) + 9cos (3 u) du = a = 5 du = 5. (c) Find he curvaure Κ of f ( ) a any value of. s Since s = 5, we have =. Then, 5 f s s s ( s) = 4 + sin 3 cos 3 s + sin s cos s 5 i 5 j 5 k= 5 i 5 j 5 k Hence, f () s = cos s sin s 5 i j 5 5 k. Then, f () s T() s = = i+ cos s j+ sin s k since f () s =. Thus, f () s Κ = T () s = sin s + cos s = 5 5 j 5 5 k = sin s + cos s = = cos s + sin s = = formula + solve for + f ( s) + T ( s) + κ