16. 17. r t te 2t i t 1. 18 19 Find the derivative of the vector function. 19. r t e t cos t i e t sin t j ln t k. 31 33 Evaluate the integral.



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SECTION.7 VECTOR FUNCTIONS AND SPACE CURVES.7 VECTOR FUNCTIONS AND SPACE CURVES A Click here for answers. S Click here for soluions. Copyrigh Cengage Learning. All righs reserved.. Find he domain of he vecor funcion r ln i j e k 5 Find he limi.. lim cos l. lim l cos. 5. 8 Skech he curve wih he given vecor equaion. Indicae wih an arrow he direcion in which increases.. r 7. 8. ; 9 Use a compuer o graph he curve wih he given vecor equaion. Make sure you choose a parameer domain and viewpoins ha reveal he rue naure of he curve. 9. r. lim i l s j an lim l e i j an k r r sin cos r s (a) Skech he plane curve wih he given vecor equaion. (b) Find r. (c) Skech he posiion vecor r and he angen vecor r for he given value of.. r. r e i e j e. r sec i an j 7 Find he domain and derivaive of he vecor funcion.. r 5. r s s k. 7. 8 9 Find he derivaive of he vecor funcion. 8. r ln i s j e k 9. Find he uni angen vecor T a he poin wih he given value of he parameer.. r s an. r i sin j cos k. r e cos i e sin j e k. r. r e e e 5 Find parameric equaions for he angen line o he curve wih he given parameric equaions a he specified poin. 5. x y z ;. x y z ; 7. x cos y sin z ; 8. x sin y s z cos ; 9. x y s cos z s sin ;. x cos y e z e ; Evaluae he inegral.... r i an j sec k r e i j an k r e cos i e sin j ln k y i j k d y i j k d y cos i sin j sin k d ( )

SECTION.7 VECTOR FUNCTIONS AND SPACE CURVES.7 ANSWERS E Click here for exercises. S Click here for soluions.. ( ) ( )... an 5.... 7.. (b) 8.. (b) e i e j 9.. (b) sec an sec j R r () = 5. { } r () =. { (n +) nan ineger} r () = ( sec ) j + (sec an ) k Copyrigh Cengage Learning. All righs reserved.

SECTION.7 VECTOR FUNCTIONS AND SPACE CURVES 7. 8. { } r () =(+) e ( +) j + + k r () = + j e k 9. r () = e (cos + sin ) e (cos sin ) j + k.. j k 5 5 5. j + k.. 5. x =+ y =+ z =+. x =+ y =+ z = 7. x = y = + z =+ 8. x = y =+ z = 9. x = + y = z =+. x = y =+ z =... j + k i 5 j k j + k Copyrigh Cengage Learning. All righs reserved.

SECTION.7 VECTOR FUNCTIONS AND SPACE CURVES.7 SOLUTIONS E Click here for exercises.. The componen funcions ln ande are all defined when > and so he domain of r () is ( ) ( ). 8. The parameric equaions give x + z =sin +cos = y = so he curve lies on he cylinder x + z =.Sincey = he curve is a helix.. lim cos = lim lim cos lim = cos. lim lim e / = cos lim lim e / =. lim += lim = lim + = ( ) an lim =an. 9. r () = Thus he given limi equals an. 5. lim e = lim = lim + an = sohe given limi equals.. The parameric equaions are x = y = z =and he curve is hus given by x = y z = which is a parabola in he plane z =wih verex ( ) and axis z =y =.. r () = 7. The corresponding parameric equaions are x = y = z = which are parameric equaions of a line hrough he origin and wih direcion vecor. Copyrigh Cengage Learning. All righs reserved.

SECTION.7 VECTOR FUNCTIONS AND SPACE CURVES 5.. (b) r () = x = e = ysoy =/x x>. r () =. + r () =.Thus ( r () = ) +( ) + ( ) = and T () = r () r () = / = = r () =cos j sin k r ( ) = j. k. Thus T ( ) ( = +( j ) k) +( /) = 5/ ( j k) = 5 5 j 5 k. (b) r () =e i e j x y = sec an = so he curve is a hyperbola.. r () =e (cos sin j + k)+e ( sin cos j) = e [( cos sin ) ( sin +cos) j +k] r ( ) = e ( j +k) Thus T ( ) e = e 9 ( j +k) = j + k. Copyrigh Cengage Learning. All righs reserved. (b) r () = sec an sec j The domain of r is R and r () =.. 5. The domain of r is { and } or { } and r () = ( ) / ( ) / ( ) =. Since an and sec are no defined for odd muliples of he domain of r is { (n +) nan ineger}. r () = ( sec ) j + (sec an ) k. 7. Since + is no defined for = (and an is defined for all real ) he domain is { }. r () =(+) e ( +) j + + k. r () = 8. + j e k 9. r () = e (cos + sin ) e (cos sin ) j + k r () =. r () = r () =. Thus T () = r () r () = 88 =. r () =. e e ( + ) e r () =. Thus T () = 9 =. 5. Thevecorequaionofhecurveisr () = j + k so r () = j + k. A he poin ( ) =so he angen vecor here is j +k. The angen line goes hrough he poin ( ) and has direcion vecor j +k. Thus parameric equaions are x =+ y =+ z =+. r () =. + + + r () = +. A ( ) =and r () =. Thus he angen lines goes hrough he poin ( ) and has direcion vecor. The parameric equaions are x =+ y =+ z =. 7. r () = cos sin r () =cos sin sin + cos. A ( ) = and r ( ) = + =. Thus parameric equaions of he angen line are x = y = + z =+.

SECTION.7 VECTOR FUNCTIONS AND SPACE CURVES r () = sin cos 8. r () = cos / ( ) sin.a( ) =and r () =. Thus parameric equaions of he angen line are x = y =+ z =. r () = cos sin 9. r () = sin cos.a ( ) = and r ( ) =. Thus parameric equaions of he angen line are x = + y = z =+. r () =. cos e e. r () = sin e e.a( ) =and r () =. Thus parameric equaions of he angen line are x = y =+ z =. ( j + k ) d ( ) ( ) = d [ ] [ = = j + k d ] [ j + ( j + ] k d ) k. [( + ) i j ( ) k ] d = [( + ) i 5 5 j ( ) k ] = [( ) + 8 i 8 j ( 8 ) k ] 5 [( ) + i j ( ) k ] 5 = i 5 j k /. (cos sin j + sin k) d = [ sin i cos j] / [ + [ cos ] / + ] / cos d k = j + [ cos + sin ] k = j + ( ) k = j + k Copyrigh Cengage Learning. All righs reserved.