Solucao Calculo 5ed CAP.13

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(1)Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves. 2. 1. The component functions t , t1 , and 5t are all defined when t1 0 t 1 and 5t 0 t 5 , so the domain of r(t) is 1,5 . t2 2 2 , sin t , and ln (9t ) are all defined when t 2 and 9t >0 t+2 3<t<3 , so the domain of r(t) is ( 3,2 ) ( 2,3) . ln t 1/t =lim =lim t=0 . Thus 3. lim cos t=cos 0=1 , lim sin t=sin 0=0 , lim tln t=lim 2 + + + + 1/t + + t 0 t 0 t 0 t 0 t 0 1/t t 0 2. The component functions. lim cos t,sin t,tln t = +. lim cos t,lim sin t,lim tln t +. t 0. +. t 0 t. = 1,0,0 .. +. t 0. t 0. t. e 1 e =lim =1 [ using l’Hospital’s Rule], 4. lim t 0 t t 0 1 1+t 1 =lim t t 0. lim t 0. 1+t 1. t. 1+t +1 1+t +1. 1 3 1 = , lim =3 . 1+t +1 2 t 0 1+t. =lim t 0. 1 ,3 . 2 t1 1 1 = , lim 5. lim t+3 =2 , lim 2 =lim t 1 t 1 t 1 t 1 t+1 2 t 1 1 Thus the given limit equals 2, ,tan 1 . 2 Thus the given limit equals. 6. lim arctant= t . Thus lim. t . 1,. tan t t. =tan 1 .. ln t 1/t 2t =lim =0 [ by l’Hospital’s Rule]. , lim e =0 , lim 2 t t t t 1 2t. arctant,e ,. ln t t. =. ,0,0 2. . 4. 7. The corresponding parametric equations for this curve are x=t +1 , y=t . We can make a table of 4. values, or we can eliminate the parameter: t=y x=y +1 , with y

(2) R . By comparing different values of t , we find the direction in which t increases as indicated in the graph.. 3. 2. 8. The corresponding parametric equations for this curve are x=t ,y=t . We can make a table of 3. 3. 2. values, or we can eliminate the parameter: x=t t= x y=t =. ( 3 x ) 2=x2/3 , with t

(3) R x

(4) R . By 1.

(5) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves. comparing different values of t , we find the direction in which t increases as indicated in the graph.. 9. The corresponding parametric equations are x=t , y=cos 2t , z=sin 2t . Note that 2. 2. 2. 2. 2. 2. y +z =cos 2t+sin 2t=1 , so the curve lies on the circular cylinder y +z =1 . Since x=t , the curve is a helix.. 10. The corresponding parametric equations are x=1+t , y=3t , z=t , which are parametric equations of a line through the point ( 1,0,0 ) and with direction vector 1,3,1 .. 2. 2. 2. 2. 11. The parametric equations give x +z =sin t+cos t=1 , y=3 , which is a circle of radius 1 , center ( 0,3,0 ) in the plane y=3 .. 12. The parametric equations are x=t , y=t , z=cos t . Thus x=y , so the curve must lie in the plane x=y . Combine this with z=cos t to determine that the curve traces out the cosine curve in the vertical plane x=y .. 2. 4. 6. 13. The parametric equations are x=t , y=t , z=t . These are positive for t 0 and 0 when t=0 . So 2. the curve lies entirely in the first quadrant. The projection of the graph onto the xy plane is y=x , 3. 3. 2. y>0 , a half parabola. On the xz plane z=x , z>0 , a half cubic, and the yz plane, y =z . 2. 2. 2. 2. 2. 14. The parametric equations give x +y +z =2sin t+2cos t=2 , so the curve lies on the sphere with radius 2 and center ( 0,0,0 ) . Furthermore x=y=sin t , so the curve is the intersection of this sphere with the plane x=y , that is, the curve is the circle of radius 2 , center ( 0,0,0 ) in the plane x=y .. 15. Taking r = 0,0,0 and r = 1,2,3 , we have from Equation 13.5.4 0. 1. r(t)=(1t) r +t r =(1t) 0,0,0 +t 1,2,3 , 0 t 1 or r(t)= t,2t,3t , 0 t 1 . 0. 1. Parametric equations are x=t , y=2t , z=3t , 0 t 1 . 2.

(6) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves. 16. Taking r = 1,0,1 and r = 2,3,1 , we have from Equation 13.5.4 0. 1. r(t)=(1t) r +t r =(1t) 1,0,1 +t 2,3,1 , 0 t 1 or r(t)= 1+t,3t,1 , 0 t 1 . 0. 1. Parametric equations are x=1+t , y=3t , z=1 , 0 t 1 . 17. Taking r = 1,1,2 and r = 4,1,7 , we have r(t)=(1t) r +t r =(1t) 1,1,2 +t 4,1,7 , 0 t 1 0. 1. 0. 1. or r(t)= 1+3t,1+2t,2+5t , 0 t 1 . Parametric equations are x=1+3t , y=1+2t , z=2+5t , 0 t 1 . 18. Taking r = 2,4,0 and r = 6,1,2 , we have r(t)=(1t) r +t r = ( 1t ) 2,4,0 +t 6,1,2 , 0. 1. 0. 1. 0 t 1 or r(t)= 2+8t,45t,2t , 0 t 1 . Parametric equations are x=2+8t , y=45t , z=2t , 0 t 1 . 2. 2. 2. 2. 19. x=cos 4t , y=t , z=sin 4t . At any point ( x,y,z ) on the curve, x +z =cos 4t+sin 4t=1 . So the curve lies on a circular cylinder with axis the y axis. Since y=t , this is a helix. So the graph is VI. t. 2. 2. 20. x=t , y=t , z=e . At any point on the curve, y=x . So the curve lies on the parabolic cylinder 2. y=x . Note that y and z are positive for all t , and the point ( 0,0,1 ) is on the curve (when t=0 ). As t , ( x,y,z ) ( , ,0 ) , while as t , ( x,y,z ) ( , , ) , so the graph must be II. 2. 2. 21. x=t , y=1/(1+t ) , z=t . Note that y and z are positive for all t . The curve passes through ( 0,1,0 ) when t=0 . As t , ( x,y,z ) ( ,0, ) , and as t , ( x,y,z ) ( ,0, ) . So the graph is IV. t. t. t. 22. x=e cos 10t , y=e sin 10t , z=e . 2. 2. 2t. 2. 2t. 2t. 2. 2. 2t. 2. 2. 2. 2. 2. x +y =e cos 10t+e sin 10t=e (cos 10t+sin 10t)=e =z , so the curve lies on the cone x +y =z . Also, z is always positive; the graph must be I. 2. 2. 2. 2. 23. x=cos t , y=sin t , z=sin 5t . x +y =cos t+sin t=1 , so the curve lies on a circular cylinder with axis the z axis. Each of x , y and z is periodic, and at t=0 and t=2 the curve passes through the same point, so the curve repeats itself and the graph is V. 2. 2. 2. 2. 24. x=cos t , y=sin t , z=ln t. x +y =cos t+sin t=1 , so the curve lies on a circular cylinder with axis the z axis. As t 0 , z , so the graph is III. 2. 2. 2. 2. 2. 2. 2. 2. 25. If x=tcos t , y=tsin t , and z=t , then x +y =t cos t+t sin t=t =z , so the curve lies on the cone 2. 2. 2. z =x +y . Since z=t , the curve is a spiral on this cone.. 3.

(7) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves. 2. 2. 2. 2. 2. 2. 26. Here x =sin t=z and x +y =sin t+cos t=1 , so the curve is the intersection of the parabolic 2. 2. 2. cylinder z=x with the circular cylinder x +y =1 .. 2. 27. r(t)= sin t,cos t,t. 4 2. 2. 28. r(t)= t t +1,t,t. 2. 29. r(t)= t , t1 , 5t. 30. We have the computer plot the parametric equations x=sin t , y=sin 2t , z=sin 3t , 0 t 2 . The 4.

(8) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves. shape of the curve is not clear from just one viewpoint, so we include a second plot drawn from a different angle.. 31.. x=(1+cos 16t)cos t , y=(1+cos 16t)sin t , z=1+cos 16t . At any point on the graph, 2. 2. 2. 2. 2. 2. x +y =(1+cos 16t) cos t+(1+cos 16t) sin t 2. 2. =(1+cos 16t) =z , so the graph lies on the cone 2. 2. 2. x +y =z . From the graph at left, we see that this curve looks like the projection of a leaved two dimensional curve onto a cone. 32.. 2. 2. x= 10.25cos 10t cos t , y= 10.25cos 10t sin t , z=0.5cos 10t . At any point on the graph, 5.

(9) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves. 2. 2. 2. 2. 2. 2. 2. x +y +z =(10.25cos 10t)cos t 2. +(10.25cos 10t)sin t+0.25cos t 2. 2. =10.25cos 10t+0.25cos 10t=1 , 2. 2. 2. so the graph lies on the sphere x +y +z =1 , and since z=0.5cos 10t the graph resembles a trigonometric curve with ten peaks projected onto the sphere. The graph is generated by t

(10) 0,2 . 33. If t=1 , then x=1,y=4,z=0 , so the curve passes through the point ( 1,4,0 ) . If t=3 , then x=9,y=8,z=28 , so the curve passes through the point ( 9,8,28 ) . For the point ( 4,7,6 ) to be on the 3. curve, we require y=13t=7 t=2. But then z=1+(2) =7 6 , so ( 4,7,6 ) is not on the curve. 2. 2. 34. The projection of the curve C of intersection onto the xy plane is the circle x +y =4,z=0 . Then we can write x=2cos t,y=2sin t,0 t 2 . Since C also lies on the surface z=xy , we have z=xy=(2cos t)(2sin t)=4cos tsin t , or 2sin (2t) . Then parametric equations for C are x=2cos t,y=2sin t,z=2sin (2t),0 t 2 , and the corresponding vector function is r(t)=2cos t i+2sin t j+2sin (2t) k , 0 t 2 . 2. 2. 35. Both equations are solved for z , so we can substitute to eliminate z : x +y =1+y 1 2 2 2 2 2 x +y =1+2y+y x =1+2y y= (x 1) . We can form parametric equations for the curve C of 2 1 2 1 2 1 2 intersection by choosing a parameter x=t , then y= (t 1) and z=1+y=1+ (t 1)= (t +1) . Thus a 2 2 2 1 2 1 2 vector function representing C is r(t)=t i+ (t 1) j+ (t +1) k . 2 2 2. 36. The projection of the curve C of intersection onto the xy plane is the parabola y=x ,z=0 . Then 2. 2. 2. we can choose the parameter x=t y=t . Since C also lies on the surface z=4x +y , we have 2. 2. 2. 22. 2. 2. 4. z=4x +y =4t +(t ) . Then parametric equations for C are x=t , y=t , z=4t +t , and the corresponding 2. 2. 4. vector function is r(t)=t i+t j+(4t +t ) k . 37.. 6.

(11) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves. 2. 2. The projection of the curve C of intersection onto the xy plane is the circle x +y =4,z=0 . Then we 2. can write x=2cos t , y=2sin t , 0 t 2 . Since C also lies on the surface z=x , we have 2. 2. 2. 2. z=x =(2cos t) =4cos t . Then parametric equations for C are x=2cos t , y=2sin t , z=4cos t,0 t 2 . 38.. 2. 2. 2. 2. 2. 4. x=t y=t 4z =16x 4y =16t 4t z=. 1 t 2. 4. 2 4. t. . Note that z is positive because the 2. intersection is with the top half of the ellipsoid. Hence the curve is given by x=t , y=t , 1 2 4 z= 4 t t . 4 2. 2. 39. For the particles to collide, we require r (t)=r (t) t ,7t12,t 2. 1 2. 2. 2. 2. = 4t3,t ,5t6 . Equating 2. components gives t =4t3 , 7t12=t , and t =5t6 . From the first equation, t 4t+3=0 (t3)(t1)=0 so t=1 or t=3 . t=1 does not satisfy the other two equations, but t=3 does. The particles collide when t=3 , at the point ( 9,9,9 ) . 2 3. 40. The particles collide provided r (t)=r (t) t,t ,t 1. 2. = 1+2t,1+6t,1+14t . Equating components 7.

(12) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves. 2. 3. gives t=1+2t , t =1+6t , and t =1+14t . The first equation gives t=1 , but this does not satisfy the other equations, so the particles do not collide. For the paths to intersect, we need to find a value for t 2 3. and a value for s where r (t)=r (s) t,t ,t 1. 2. 2. = 1+2s,1+6s,1+14s . Equating components, t=1+2s ,. 3. 2. t =1+6s , and t =1+14s . Substituting the first equation into the second gives ( 1+2s ) =1+6s 1 1 2 4s 2s=0 2s(2s1)=0 s=0 or s= . From the first equation, s=0 t=1 and s= t=2 . 2 2 Checking, we see that both pairs of values satisfy the third equation. Thus the paths intersect twice, at 1 the point ( 1,1,1 ) when s=0 and t=1 , and at ( 2,4,8 ) when s= and t=2 . 2 41. Let u(t)= u (t),u (t),u (t) and v(t)= v (t),v (t),v (t) . In each part of this problem the basic 1. 2. 3. 1. 2. 3. procedure is to use Equation 1 and then analyze the individual component functions using the limit properties we have already developed for realvalued functions. (a) lim u(t)+lim v(t)= lim u (t),lim u (t),lim u (t) + lim v (t),lim v (t),lim v (t) t a. t a. 1. t a. t a. 2. t a. 3. 1. t a. 2. t a. t a. 3. and the. limits of these component functions must each exist since the vector functions both possess limits as t a . Then adding the two vectors and using the addition property of limits for realvalued functions, we have that lim u(t)+lim v(t) = t a t a. lim u (t)+lim v (t),lim u (t)+lim v (t),lim u (t)+lim v (t) t a. 1. 1. 1. t a. 2. t a. u (t)+v (t) ,lim. lim. =. t a. 1. t a. 2. t a. u (t)+v (t) ,lim 2. t a. 2. 3. 3. t a. u (t)+v (t). t a. 3. 3. u (t)+v (t),u (t)+v (t),u (t)+v (t) [using(1)backward. = lim. 1. t a. 1. 2. 2. 3. 3. u(t)+v(t) = lim t a (b) lim cu(t) =lim t a. cu (t),cu (t),cu (t) = lim cu (t),lim cu (t),lim cu (t) 1. t a. 2. 3. t a. 1. = clim u (t),clim u (t),clim u (t) =c t a. =clim t a. 1. 2. t a. t a. 3. t a. 2. 3. t a. lim u (t),lim u (t),lim u (t) t a. 1. t a. 2. t a. 3. u (t),u (t),u (t) =clim u(t) 1. 2. 3. t a 8.

(13) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves. (c) lim u(t) lim v(t) = lim u (t),lim u (t),lim u (t). t a. t a. 1. t a. = lim u (t) 1. t a. 2. t a. t a. lim v (t),lim v (t),lim v (t). 3. t a. 1. t a. 2. t a. 3. lim v (t) + lim u (t). lim v (t) + lim u (t). lim v (t). t a. t a. t a. 1. 2. t a. 2. t a. 3. 3. =lim u (t)v (t)+lim u (t)v (t)+lim u (t)v (t) 1. t a. =lim. 1. t a. 2. 2. 3. t a. 3. u (t)v (t)+u (t)v (t)+u (t)v (t) =lim u(t) v(t) 1. t a. 1. 2. 2. 3. 3. t a. (d) . lim u(t)lim v(t) = lim u (t),lim u (t),lim u (t) t a. t a. 1. t a. =. 2. t a. t a. 3. lim v (t),lim v (t),lim v (t) 1. t a. t a. lim u (t). lim v (t) lim u (t). lim v (t) ,. t a. t a. t a. 2. 3. t a. 3. lim v (t) lim u (t). lim v (t) ,. t a. t a. t a. 3. 1. t a. 1. lim v (t) lim u (t). lim v (t). t a. t a. t a. = lim. 2. t a. u (t)v (t)u (t)v (t) ,lim 2. t a. lim. 3. 3. 2. 1. u (t)v (t)u (t)v (t) ,. t a. 3. 1. 1. 3. u (t)v (t)u (t)v (t) 1. t a. t a. 2. 3. 3. lim u (t) 1. t a. 2. lim u (t). =lim. 2. 2. 2. 1. u (t)v (t)u (t)v (t),u ( t ) v (t)u (t)v (t), 2 3 3 2 3 1 1 3. u (t)v (t)u (t)v (t) 1. 2. 2. 1. =lim u(t) v(t) t a. 42. The projection of the curve onto the xy plane is given by the parametric equations x=(2+cos 1.5t)cos t , y=(2+cos 1.5t)sin t . If we convert to polar coordinates, we have 2 2 2 2 r = x +y = (2+cos 1.5t)cos t + (2+cos 1.5t)sin t. 2. = (2+cos 1.5t)2(cos 2t+sin 2t) = (2+cos 1.5t)2 9.

(14) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves. y (2+cos 1.5t)sin t = =tan t =t . x (2+cos 1.5t)cos t Thus the polar equation of the curve is r=2+cos 1.5 . At =0 , we have r=3 , and r decreases to 1 as 2 2 4 increases to . For , r increases to 3 ; r decreases to 1 again at =2 , increases 3 3 3 8 10 to 3 at = , decreases to 1 at = , and completes the closed curve by increasing to 3 at 3 3 =4 . We sketch an approximate graph as shown in the figure. r=2+cos 1.5t . Also, tan =. We can determine how the curve passes over itself by investigating the maximum and minimum 5 values of z for t=

(15) 0,4 . Since z=sin 1.5t , z is maximized where sin 1.5t=1 1.5t= , , or 2 2 9 2 5 7 11 3 7 , , or 3 . z is minimized where sin 1.5t=1 1.5t= , , or , or t= t= , 3 3 2 2 2 3 11 . Note that these are precisely the values for which cos 1.5t=0 r=2 , and on the graph of the 3 projection, these six points appear to be at the three selfintersections we see. Comparing the maximum and minimum values of z at these intersections, we can determine where the curve passes over itself, as indicated in the figure.. We show a computerdrawn graph of the curve from above, as well as views from the front and from the right side.. 10.

(16) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves. Top view. Front view. Side view The top view graph shows a more accurate representation of the projection of the trefoil knot on the xy plane (the axes are rotated 90 ). Notice the indentations the graph exhibits at the points corresponding to r=1 . Finally, we graph several additional viewpoints of the trefoil knot, along with two plots showing a tube of radius 0.2 around the curve.. 11.

(17) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves. 43. Let r(t)= f ( t ) ,g ( t ) ,h ( t ) and b= b ,b ,b. . If lim r(t)=b , then lim r(t) exists, so by (1),. 1 2 3. b=lim r(t)= lim f (t),lim g(t),lim h(t) t a. t a. t a. t a. t a. . By the definition of equal vectors we have lim f (t)=b ,. t a. 1. t a. lim g(t)=b and lim h(t)=b . But these are limits of realvalued functions, so by the definition of 2. t a. 3. t a. limits, for every >0 there exists >0 , >0 , >0 so f (t)b < /3 whenever 0< ta < , g(t)b < /3 1. 2. 3. 1. 1. 2. whenever 0< ta < , and h(t)b < /3 whenever 0< ta < . Letting = minimum of 2. { , , } 1 2 3. r(t)b = =. 3. 3. , we have f (t)b + g(t)b + h(t)b < /3+ /3+ /3= whenever 0< ta < . But 1. 2. f (t)b ,g(t)b ,h(t)b 1. 2. 2. 3. 2. 2. ( f (t)b ) +(g(t)b ) +(h(t)b ) 1. 2. 3. = f (t)b + g(t)b + h(t)b 1. 2. 3. 2. [ f (t)b ] + 1. 2. 2. [g(t)b ] +. [h(t)b ]. 2. 3. . Thus for every >0 there exists >0 such that. 3. r(t)b f (t)b + g(t)b + h(t)b < whenever 0< ta < . Conversely, if for every >0 , 1. 2. 3. there exists >0 such that r(t)b < whenever 0< ta < , then f (t)b ,g(t)b ,h(t)b 1. 2. 2. 3. 2. < . 2. 2. 2. 2. [ f (t)b ] +[g(t)b ] +[h(t)b ] < 1. 2. 3. 2. [ f (t)b ] +[g(t)b ] +[h(t)b ] < whenever 0< ta < . But each term on the left side of this 1. 2. 3. 2. 2. 2. 2. 2. 2. inequality is positive so [ f (t)b ] < , [g(t)b ] < and [h(t)b ] < whenever 0< ta < , or 1. 2. 3. taking the square root of both sides in each of the above we have f (t)b < , g(t)b < and 1. 2. h(t)b < whenever 0< ta < . And by definition of limits of realvalued functions we have 3. 12.

(18) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves. lim f (t)=b , lim g(t)=b and lim h(t)=b . But by (1), lim r(t)= lim f (t),lim g(t),lim h(t) t a. 1. lim r(t)= b ,b ,b t a. 2. t a 1 2 3. t a. 3. t a. t a. t a. , so. t a. =b .. 13.

(19) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.2 Derivatives and Integrals of Vector Functions. 1. (a). r(4.5)r(4) =2[r(4.5)r(4)] , so we draw a vector in 0.5 the same direction but with twice the length of the r(4.2)r(4) (b) =5[r(4.2)r(4)] , so vector r(4.5)r(4) . 0.2 we draw a vector in the same direction but with 5 times the length of the vector r(4.2)r(4) (c) (d) r(4+h)r(4) / By Definition 1, r (4)=lim . h h 0 T(4) is a unit vector in the same. /. T(4)=. r (4) /. /. .. direction as r (4) , that is, parallel to the tangent line to the curve at r(4) with length 1 .. r (4). 2. 2. (a) The curve can be represented by the parametric equations x=t , y=t , 0 t 2 . Eliminating the 2. parameter, we have x=y , 0 y 2 , a portion of which we graph here, along with the vectors r(1) , r(1.1) , and r(1.1)r(1) .. 1.

(20) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.2 Derivatives and Integrals of Vector Functions. 2. /. /. (b) Since r(t)= t ,t , we differentiate components, giving r (t)= 2t,1 , so r (1)= 2,1 . 1.21,1.1 1,1 r(1.1)r(1) = =10 0.21,0.1 = 2.1,1 . 0.1 0.1. /. As we can see from the graph, these vectors are very close in length and direction. r (1) is defined to r(1+h)r(1) r(1.1)r(1) be lim , and we recognize as the expression after the limit sign with h=0.1. h 0.1 h 0 r(1.1)r(1) / Since h is close to 0 , we would expect to be a vector close to r (1) . 0.1 3. (a), (c). 2.

(21) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.2 Derivatives and Integrals of Vector Functions. (b). /. r (t)= sin t,cos t. 4. (a), (c). (b). /. r (t)=. 1,. 2. 1 2 t. 2. 5. Since (x1) =t =y , the curve is a parabola. (a), (c). (b). /. r (t)=i+2t j. 6. (a), (c). (b). /. t. t. r (t)=e ie j. 7.. 3.

(22) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.2 Derivatives and Integrals of Vector Functions. (a), (c). (b). t. /. 3t. r (t)=e i+3e j 2. 2. 2. 2. 8. x=2sin t , y=3cos t , so ( x/2 ) + ( y/3) =sin t+cos t=1 and the curve is an ellipse . (a), (c). (b). /. r (t)=2cos t i3sin t j d 2 d d t , 1t , dt dt dt. /. 9. r (t)=. t. =. 2t,1,. 1 2 t. /. 10. r(t)= cos 3t,t,sin 3t r (t)= 3sin 3t,1,3cos 3t 4t. /. 4t. 4t. 11. r(t)=i j+e k r (t)=0 i+0 j+4e k=4e k 1. 1. /. 2. 12. r(t)=sin t i+ 1t j+k r (t)=. 1t t. 2. t. /. 2. 13. r(t)=e i j+ln (1+3t) k r (t)=2te i+. t. i 2. 1t. j 2. 3 k 1+3t. 14. /. 2. 2. r (t) =[at(3sin 3t)+acos 3t] i+b 3sin tcos t j+c 3cos t(sin t) k 2. 2. =(acos 3t3atsin 3t) i+3bsin tcos t j3ccos tsin t k /. 15. r (t)=0+b+2t c=b+2t c by Formulas 1 and 3 of Theorem 3. 4.

(23) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.2 Derivatives and Integrals of Vector Functions. /. 2. /. 16. To find r (t) , we first expand r(t)=t a (b+t c)=t(a b)+t (a c) , so r (t)=a b+2t(a c) . /. 4. 2. /. /. 2. 2. 2. 17. r (t)= 30t ,12t ,2 r (1)= 30,12,2 . So r (1) = 30 +12 +2 = 1048 =2 262 and /. T(1)=. r (1). =. /. r (1). 1 2 262. .. 2 / i+2t j+k r (1)=2 i+2 j+k . Thus t. /. 18. r (t)= /. T(1)=. 15 6 1 , , 262 262 262. 30,12,2 =. r (1). 1. =. /. 2. r (1). (2 i+2 j+k)= 2. 2. 2 +2 +1. /. 1 2 2 1 (2 i+2 j+k)= i+ j+ k . 3 3 3 3. /. 19. r (t)=sin t i+3 j+4cos 2t k r (0)=3 j+4 k . Thus /. T(0)=. r (0). 1. =. /. 2. r (0). 2. 2. 0 +3 +4. /. 2. 20. r (t)=2cos t i2sin t j+sec t k r. Thus T. r. 4. = r 2 3. /. r (t) r. / /. =. i j k 1 2t 3t2 0 2 6t. (t) =. 2. 4. /. 1 2 2. 2. (. = 2 i 2 j+2 k and 2 i 2 j+2 k ) =. r. 4. /. = 2+2+4 =2 2 .. 1 1 1 i j+ k. 2 2 2. /. /. 2. 2. 2. . Then r (1)= 1,2,3 and r (1) = 1 +2 +3 = 14 , so. 1 1,2,3 = 14. =. r (1) /. /. /. /. T(1)=. 4 4. /. r (t)= 1,2t,3t. 21. r(t)= t,t ,t r (1). 1 3 4 (3 j+4 k)= j+ k . 5 5 5. (3 j+4 k)=. 1 2 3 , , 14 14 14 2t 2. =. 2. 2. 3t 6t. i. .r. 1 0. / /. (t)= 0,2,6t , so 2. 3t 6t. 1 0. j+. 2t 2. k. 2. =(12t 6t ) i(6t0) j+(20) k= 6t ,6t,2 . 2t 2t. 2t. 22. r(t)= e ,e ,te. /. 2t. 2t. 2t. r (t)= 2e ,2e ,(2t+1)e. /. 0. 0. 0. r (0)= 2e ,2e ,(0+1)e. = 2,2,1 and 5.

(24) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.2 Derivatives and Integrals of Vector Functions. /. 2. 2. /. r ( 0). 2. r (0) = 2 +(2) +1 =3 . Then T(0)=. 1 2,2,1 = 3. =. /. r ( 0) r. / /. 2t. 2t. 2t. (t)= 4e ,4e ,(4t+4)e. /. r (t) r. / /. 2t. r. / /. 2t. 0. 0. 0. (0)= 4e ,4e ,(0+4)e. 2t. 2t. 2t. 2 2 1 , , 3 3 3. .. = 4,4,4 . 2t. ( t ) = 2e ,2e ,(2t+1)e 4e ,4e ,(4t+4)e 2t. 2t. 2t. 2t. 2t. 2t. =(2e )(4e )+(2e )(4e )+((2t+1)e )((4t+4)e ) 4t. 4t. 2. 4t. 2. 4t. 4t. =8e 8e +(8t +12t+4)e =(8t +12t+12)e 8e 5 4 3. 23. The vector equation for the curve is r(t)= t ,t ,t. /. 4. 3. , so r (t)= 5t ,4t ,3t. 2. . The point ( 1,1,1 ). /. corresponds to t=1 , so the tangent vector there is r (1)= 5,4,3 . Thus, the tangent line goes through the point ( 1,1,1 ) and is parallel to the vector 5,4,3 . Parametric equations are x=1+5t , y=1+4t , z=1+3t . 2. 2. /. 24. The vector equation for the curve is r(t)= t 1,t +1,t+1 , so r (t)= 2t,2t,1 . The point ( 1,1,1 ) /. corresponds to t=0 , so the tangent vector there is r (0)= 0,0,1 . Thus, the tangent line is parallel to the vector 0,0,1 and parametric equations are x=1+0 t=1 , y=1+0 t=1 , z=1+1 t=1+t . t. t. t. 25. The vector equation for the curve is r(t)= e cos t,e sin t,e t. /. t. t. t. , so. t. r (t) = e (sin t)+(cos t)(e ),e cos t+(sin t)(e ),(e ) t. t. t. = e (cos t+sin t),e (cos tsin t),e. .. The point ( 1,0,1 ) corresponds to t=0 , so the tangent vector there is /. 0. 0. 0. r (0)= e (cos 0+sin 0),e (cos 0sin 0),e = 1,1,1 . Thus, the tangent line is parallel to the vector 1,1,1 and parametric equations are x=1+(1)t=1t , y=0+1 t=t , z=1+(1)t=1t . 2. /. /. 26. r(t)= ln t,2 t ,t , r (t)= 1/t,1/ t ,2t . At ( 0,2,1 ) , t=1 and r (1)= 1,1,2 . Thus, parametric equations of the tangent line are x=t , y=2+t , z=1+2t . /. 27. r(t)= t, 2 cos t, 2 sin t r (t)= 1, 2 sin t, 2 cos t . At r. /. 4. ,1,1 4. = 1,1,1 . Thus, parametric equations of the tangent line are x=. , t=. and 4. +t , y=1t , z=1+t . 4 6.

(25) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.2 Derivatives and Integrals of Vector Functions. 2t. 2t. /. 2t. 2t. /. 28. r(t)= cos t,3e ,3e , r (t)= sin t,6e ,6e . At ( 1,3,3) , t=0 and r (0)= 0,6,6 . Thus, parametric equations of the tangent line are x=1 , y=3+6t , z=36t .. 3 4 5. r (t)= 3t ,4t ,5t. /. 4 5. r (t)= 3t +1,4t ,5t. 29. (a) r(t)= t ,t ,t 3. 2. /. (b) r(t)= t +t,t ,t. 2. 3. 3. 4. /. , and since r (0)= 0,0,0 =0 , the curve is not smooth. 4. /. . r (t) is continuous since its component functions are. /. /. continuous. Also, r (t) 0 , as the y and z components are 0 only for t=0 , but r (0)= 1,0,0 0 . Thus, the curve is smooth. 3. 3. /. 2. 2. (c) r(t)= cos t,sin t r (t)= 3cos tsin t,3sin tcos t . Since /. 2. 2. r (0)= 3cos 0sin 0,3sin 0cos 0 = 0,0 =0 , the curve is not smooth. 30. (a) p250pt The tangent line at t=0 is the line through the point with position vector r(0)= sin 0,2sin 0,cos 0 = 0,0,1 , /. and in the direction of the tangent vector, r (0)= cos 0,2 cos 0, sin 0 = ,2 ,0 . So an equation of the line is /. x,y,z =r(0)+u r (0)= 0+ u,0+2 u,1 = u,2 u,1 . (b). 7.

(26) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.2 Derivatives and Integrals of Vector Functions. 1 2. r r. /. = 1 2. sin. ,2sin ,cos 2 2 2. = 1,2,0 ,. ,2 cos , sin 2 2 2. = cos. .. = 0,0,. So the equation of the second line is x,y,z = 1,2,0 +v 0,0, = 1,2, v . The lines intersect where u,2 u,1 = 1,2, v , so the point of intersection is ( 1,2,1 ) . 31. The angle of intersection of the two curves is the angle between the two tangent vectors to the /. curves at the point of intersection. Since r (t)= 1,2t,3t. 2. /. and t=0 at ( 0,0,0 ) , r (0)= 1,0,0 is a 1. 1. /. tangent vector to r at ( 0,0,0 ) . Similarly, r (t)= cos t,2cos 2t,1 and since r (0)= 0,0,0 , 2 1 2 /. r ( 0 ) = 1,2,1 is a tangent vector to r at ( 0,0,0 ) . If is the angle between these two tangent 2 2 vectors, then cos =. 1 1 1 1,0,0 1,2,1 = and =cos 6 1 6. 1 6. 66 .. 32. To find the point of intersection, we must find the values of t and s which satisfy the following 2. 2. three equations simultaneously: t=3s , 1t=s2 , 3+t =s . Solving the last two equations gives t=1 , s=2 (check these in the first equation). Thus the point of intersection is ( 1,0,4 ) . To find the angle of intersection, we proceed as in Exercise 31 . The tangent vectors to the respective curves at ( 1,0,4 ) 1 6 1 / / are r (1)= 1,1,2 and r (2)= 1,1,4 . So cos = and 11+8 ) = = ( 1 2 6 18 6 3 3 1 1 =cos. 55 . 3 Note: In Exercise 31 , the curves intersect when the value of both parameters is zero. However, as seen in this exercise, it is not necessary for the parameters to be of equal value at the point of intersection. 33..

(27) 0 (16t 1. 3. 2. 4. i9t j+25t k)dt =.

(28) 0 16t 1. = 4t 34.

(29). 1 0. 4 1+t. 2. j+. 2t 1+t. 2. k. 4 1 0. 3.

(30) 0 9t 1. dt i. i 3t 1. 3 1 0. 2. j+ 5t. dt j+ 5 1 0 2.

(31) 0 25t 1. 4. dt k. k=4 i3 j+5 k. dt= 4tan t j+ln (1+t ) k. 1 0. 8.

(32) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.2 Derivatives and Integrals of Vector Functions. 1. 4. 1. = 4tan 1 j+ln 2 k 4tan 0 j+ln 1 k =4 /2. 35.

(33) =. 2. j+ln 2 k0 j0 k= j+ln 2 k. 2. (3sin tcos t i+3sin tcos t j+2sin tcos t k)dt. 0. /2.

(34) 0. 2. 3sin tcos t dt i+. 3 /2. = sin t. 0. 3 /2. i+ cos t. 0. /2.

(35) 0. 2. 3sin tcos t dt j+ 2. j+ sin t. /2 0. /2.

(36) 0. 2sin tcos t dt k. k. =(10) i+(0+1) j+(10) k=i+ j+k 36.

(37). ( 1 4. t. 16 2 3 3. =. 2. 2 3/2 1 t it k 3. ). t i+te j+t k dt=. 4. +. 1 4 1. t 4. 4 t. +

(38) e dt j. te. 1. 1. 1 14 3 4 1 1 3 1 k+(4e +e e +e ) j= i+e (25e ) j+ k 4 3 4. i. ) (

(39) et dt ) i+ (

(40) 2t dt ) j+ (

(41) ln t dt ) k. (t. 37.

(42) e i+2t j+ln t k dt= t. 2. =e i+t j+ ( tln tt ) k+C , where C is a vector constant of integration. 38..

(43) (cos t i+sin t j+t k)dt = (

(44) cos t dt ) i+ (

(45) sin t dt ) j+ (

(46) t dt ) k =. 1 1 2 1 sin t i cos t j+ t k+C 2 . 1 3 4 1 3 t i+t j t k+C , where C is a constant vector. 3 3 1 3 4 1 3 1 3 1 3 4 But j=r(0)=(0) i+(0) j(0) k+C . Thus C= j and r(t)= t i+t j t k+ j= t i+(t +1) j t k . 3 3 3 3 /. 2. 3. 2. 39. r (t)=t i+4t jt k r(t)=. /. 2. 40. r (t)=sin t icos t j+2t k r(t)=(cos t) i(sin t) j+t k+C . But i+ j+2k=r(0)=i+(0) j+(0) k+C . 2. Thus C=2 i+ j+2 k and r(t)=(2cos t) i+(1sin t) j+(2+t ) k . 41. d d u(t)+v(t) = dt dt =. u (t)+v (t),u (t)+v (t),u (t)+v (t) 1. 1. 2. 2. 3. 3. d d d u (t)+v ( t ) , u (t)+v (t) , u (t)+v (t) 1 2 3 dt 1 dt 2 dt 3 9.

(47) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.2 Derivatives and Integrals of Vector Functions. /. /. /. /. /. /. 1. 1. 2. 2. 3. 3. /. /. /. = u (t)+v (t),u (t)+v (t),u (t)+v (t) /. /. /. /. /. = u (t),u ( t ) ,u (t) + v (t),v (t),v (t) =u (t)+v (t) . 1 2 3 1 2 3 42. d d f (t) u(t) = dt dt. f (t)u (t),f(t)u (t),f(t)u (t) 1. 2. 3. d dt. f (t)u (t) ,. =. d dt. =. f (t)u (t)+ f (t)u (t),f (t)u (t)+ f (t)u (t),f (t)u (t)+ f (t)u (t). f (t)u (t) , 1. /. 2. /. 1. .. f (t)u (t) 3. /. /. 1. 2. /. /. 2. /. 3. /. /. /. 1. 2. 3. 3. = f (t) u (t),u (t),u (t) + f (t) u (t),u (t),u (t) 1. 2. 3. /. /. = f (t) u(t)+ f (t) u (t) d u(t) v(t) dt. 43. =. d dt. d dt. u (t)v (t)u (t)v (t),u (t)v (t)u (t)v (t),u (t)v (t)u (t)v (t) 2. 3. 3. /. 2. 3. /. /. 3. 3. /. /. 1. 1. 3. 1. 2. 2. 1. /. = u v (t)+u (t)v (t)u (t)v (t)u (t)v (t), 2 3. 2. /. 2. 3. 2. /. u (t)v (t)+u (t)v ( t ) u (t)v (t)u (t)v (t), 3 1 3 1 1 3 1 3 /. /. /. 2. 2. /. u (t)v (t)+u (t)v (t)u (t)v (t)u (t)v (t) 1. 2. 1. /. /. 1. 1. 2. /. /. /. /. = u (t)v (t)u (t)v ( t ) ,u (t)v (t)u (t)v (t),u (t)v (t)u (t)v (t) 2 3 3 2 3 1 1 3 1 2 2 1 /. /. /. /. /. /. + u (t)v (t)u (t)v (t),u (t)v ( t ) u (t)v (t),u (t)v (t)u (t)v (t) 2 3 3 1 1 2 3 2 1 3 2 1 /. /. =u (t) v(t)+u(t) v (t) Alternate solution: Let r(t)=u(t) v(t) . Then r(t+h)r(t) = u(t+h) v(t+h) u(t) v(t) = u(t+h) v(t+h) u(t) v(t) + u(t+h) v(t) u(t+h) v(t) 10.

(48) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.2 Derivatives and Integrals of Vector Functions. =u(t+h) v(t+h)v(t) + u(t+h)u(t) v(t) (Be careful of the order of the cross product.) Dividing through by h and taking the limit as h 0 we have u(t+h) v(t+h)v(t) u(t+h)u(t) v(t) / +lim r (t) = lim h h h 0 h 0 = u(t) v /(t)+u /(t) v(t) by Exercise 14.1.41(a) and Definition 1. 44. d d u( f (t)) = dt dt =. d d d u ( f (t)) , u ( f (t)) , u ( f (t)) dt 1 dt 2 dt 3. u ( f (t)),u ( f (t)),u ( f (t)) = 1. /. 2. /. 3. /. /. /. /. /. /. f (t)u ( f (t)),f (t)u ( f (t)),f (t)u ( f (t)) = f (t) u (t) 1. 2. 3. 45. d u(t) v(t) =u /(t) v(t)+u(t) v /(t) dt 2. 2. 3. 2. 3. =(4t j+9t k) ( t i+cos t j+sin t k ) +(i2t j+3t k) (isin t j+cos t k) 2. 2. 3. =4tcos t+9t sin t+1+2t sin t+3t cos t 2. 3. =14tcos t+11t sin t+3t cos t 46. d u(t) v(t) =u /(t) v(t)+u(t) v /(t) dt 2. =(4t j+9t k) (t i+cos t j+sin t k)+(i2t j+3t k) (isin t j+cos t k) 2. 3. 2. =(4tsin t9t cos t) i+(9t 0) j+(0+4t ) k 2. 3. 3. 2. +(2t cos t+3t sin t) i+(3t cos t) j+(sin t+2t ) k 3. 2. 3. 2. =[(sin t)(3t 4t)11t cos t] i+(12t cos t) j+(6t sin t) k d / / / r(t) r (t) =r (t) r (t)+r(t) r dt (see Example 13.4.2 ). Thus, 47.. / /. /. /. (t) by Formula 5 of Theorem 3. But r (t) r (t)=0. 11.

(49) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.2 Derivatives and Integrals of Vector Functions. d / r(t) r (t) =r(t) r dt. / /. (t) .. 48. d d / v(t) w ( t ) ( u(t) v ( t ) w(t) ) =u (t) v(t) w(t) +u(t) dt dt /. /. /. /. /. =u (t) v(t) w(t) +u(t) v (t) w(t)+v(t) w (t) /. =u (t) v(t) w(t) +u(t) v (t) w(t) +u(t) v(t) w (t) /. /. /. =u (t) v(t) w(t) v (t) u(t) w(t) +w (t) u(t) v(t) 49.. d d r(t) = r(t) r(t) dt dt. 1/2. =. 1 r(t) r(t) 2. 1/2. /. 2r(t) r (t) =. 1 / r(t) r (t) r(t). d d 2 r(t) r(t) = r(t) . Thus r(t) dt dt , is a constant, and hence the curve lies on a sphere with center the origin. /. /. 50. Since r(t) r (t)=0 , we have 0=2r(t) r (t)=. /. 51. Since u(t)=r(t) r (t) r =0+r(t) r. / /. (t) r. / /. /. / /. /. /. /. (t) , u (t)=r (t) r (t) r. (t)+r (t) r. / / /. /. (t) =r(t) r (t) r. / /. / / /. (t) +r(t). d / r (t) r dt. 2. / /. , and so r(t). (t). (t). 12.

(50) .. Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.3 Arc Length and Curvature. /. .. /. 2. 2. 2. 1. r (t)= 2cos t,5,2sin t r (t) = (2cos t) +5 +(2sin t) = 29 . Then using Formula 3, we have 10. L= . 10. 10. r (t) dt= /. 29 dt=. 10. 29 t. 10. =20 29 .. 10. /. 2. r (t)= 2t,cos t+tsin tcos t,sin t+tcos t+sin t = 2t,tsin t,tcos t /. 2. 2. 2. 2. 2. 2. 2. r (t) = (2t) +(tsin t) +(tcos t) = 4t +t (sin t+cos t) = 5 t = 5 t for 0 t . Then using . . Formula 3, we have L= r (t) dt= 5 t dt= /. 0 t. t. /. 0. 3. r (t)= 2 i+e je k /. (. r (t) = 1. ). 2. 2. t 2. t2. 1. t. t. t. 0. 0. .. t. 0. t 2. e. /. e. 2. 1. =ee. 0. . 2. 2. 2. /. 2. 4. 2. for 1 t e .. 2. 5. r (t)=2t j+3t k r (t) = 4t +9t =t 1. t. t. 1 2 e 2 +2t dt= ln t+t =e 1 t. 1+2t dt= L= t 1 1 /. 2. t. t. 1+2t 1+2t 4. r (t)= 2t,2,1/t , r (t) = 4t +4+(1/t) = = t t /. 5 2 . 2. =. = (e +e ) =e +e (since e +e >0 ).. t 1. Then L= r (t) dt= (e +e )dt= e e /. 2t. 2t. +(e ) +(e ) = 2+e +e. . 2. t 5 2. 1. Then L= r (t) dt= t /. 0. 2. 4+9t dt=. 0. /. /. 4+9t (since t 0 ).. 1 2 2 3/2 (4+9t ) 18 3. 1. =. 0. 1 1 3/2 3/2 3/2 (13 4 )= (13 8) . 27 27. 2. 2. 6. r (t)=12 i+12 t j+6t k r (t) = 144+144t+36t = 36(t+2) =6 t+2 =6(t+2) for 0 t 1 . 1. 1. 1. Then L= r (t) dt= 6(t+2)dt= 3t +12t =15 . /. 0. 0. 2. 0. 2. 7. The point ( 2,4,8 ) corresponds to t=2 , so by Equation 2, L=. 2. 2. 22. (1) +(2t) +(3t ) dt . If. 0 2. 4. f (t)= 1+4t +9t , then Simpson’s Rule gives 1.

(51) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.3 Arc Length and Curvature. L. 20 f (0)+4 f (0.2)+2 f (0.4)+ +4 f (1.8)+ f (2) 9.5706 . 10 3. 8. Here are two views of the curve with parametric equations x=cos t , y=sin 3t , z=sin t :. The complete curve is given by the parameter interval 0,2 , so 2. 2. L= . (sin t) +(3cos 3t) +(cos t) dt= . 0. 0. 2. 2. 2. 2. 1+9cos 3t dt 13.9744 . t. t. ds / / = r (t) = 4+9+16 = 29 . Then s=s(t)= r (u) du= 29 du= 29 t . 9. r (t)=2 i3 j+4 k and dt 0 0 1 s , and substituting for t in the original equation, we have Therefore, t= 29 3 4 2 si+ 1 s j+ 5+ s k. r(t(s))= 29 29 29 /. /. 2t. 2t. .. 10. r (t)=2e (cos 2tsin 2t) i+2e (cos 2t+sin 2t) k , ds / 2t 2 2 2t = r (t) =2e (cos 2tsin 2t) +(cos 2t+sin 2t) =2e dt t. t. s=s(t)= r (u) du= 2 2 e du= /. 0. 2u. 0. s 1 2t +1=e t= ln 2 2 r(t(s)). 2. =e. 1 ln 2. s 2. s +1 2 +1. cos 2. 2u t. 2e. 2. 2. 2t. 2cos 2t+2sin 2t =2 2 e .. 2t. = 2 (e 1). 0. . Substituting, we have 1 ln 2. s +1 2. 2. i+2 j+e. 1 ln 2. s 2. +1. sin 2. 1 ln 2. s +1 2 2. k.

(52) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.3 Arc Length and Curvature. s +1 cos 2. =. s +1 2. ln. s +1 sin 2. i+2 j+. t. s +1 2. ln. k.. t. 11. r (t) = (3cos t) +16+(3sin t) = 9+16 =5 and s ( t ) = r (u) du= 5du=5t t(s)= /. 2. 2. /. 0. 0. 1 s. 5. Therefore, 1 4 s i+ s j+3cos 5 5. r(t(s))=3sin. /. 12. r (t)=. 2. 4t 2. 1 s k. 5. i+. 2. (t +1). 2t +2 2. 2. j,. (t +1). 4t. ds / = r (t) = dt. 2. 2. +. 2. (t +1) 4. =. 2. 2. (t +1). =. 2. 2. 2t +2 2. 4. =. 2. (t +1). 2. 4t +8t +4 2. 4. 2. =. (t +1). 2. 4(t +1) 2. 4. (t +1). 2 2. t +1. Since the initial point ( 1,0 ) corresponds to t=0 , the arc length function t. t. . s(t)= r (u) du= 0. /. 0. 2 2. du=2arctan t . Then arctan t=. u +1 2tan. 1 s 2. 1 1 s t=tan s . Substituting, we have 2 2 1tan. 2. 1 s 2 1 s 2. 2tan. 1 s 2 1 s 2. 2 1 i+ j= i+ j 1 1 2 2 2 2 s +1 tan s +1 1+tan sec tan 2 2 1 2 1tan s 1 2 1 2 = i+2tan s cos s j 1 2 2 2 sec s 2 1 1 1 1 2 2 = cos s sin s i+2sin s cos s j=cos s i+sin s j 2 2 2 2 With this parametrization, we recognize the function as representing the unit circle. Note here that the curve approaches, but does not include, the point ( 1,0 ) , since cos s=1 for s= +2k ( k an integer) 1 but then t=tan s is undefined. 2 r(t(s)) =. 3.

(53) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.3 Arc Length and Curvature. /. /. 2. 2. 13. (a) r (t)= 2cos t,5,2sin t r (t) = 4cos t+25+4sin t = 29 . Then /. T(t)=. r (t) /. 2 5 2 cos t, , sin t 29 29 29. 1 2cos t,5,2sin t or 29. =. r (t) 1 1 / / T (t)= 2sin t,0,2cos t T (t) = 29 29 /. N(t)=. T (t). 1/ 29. =. /. /. T (t). (b) (t)=. 2/ 29. =. /. 2. 2 . Thus 29. 2sin t,0,2cos t = sin t,0,cos t .. 2/ 29. T (t). 2. 4sin t+0+4cos t =. .. =. 29. r (t). 2 . 29. /. /. 2. 2. 2. 2. 2. 2. 14. (a) r (t)= 2t,tsin t,tcos t r (t) = 4t +t sin t+t cos t = 5t = 5 t (since t>0 ). /. r (t). Then T(t)=. 1 1 1 / 2t,tsin t,tcos t = 2,sin t,cos t . T (t)= 0,cos t,sin t 5t 5 5. =. /. r (t) /. T (t ) =. 1 5. 2. /. T (t). Thus N(t)=. =. /. T (t) /. T (t). (b) (t)=. =. /. 1/ 5. 1/ 5 5t. /. r (t) /. T (t). =. 1 t. 1 . 5t /. t. t. 2 ,e ,e. t. =. 1. t 2. t. 1. t. 2 e ,e ,1. after multiplying by. 1. 2t. 2t. 2. 2e. t. and. t 2t. 2 e ,e ,1. ( e2t+1) 2 t. 2t. 2t. 2t. 2 e ,2e ,0 2e. (e +1). e e. 2t. 2 e ,2e ,0 . 2t. t. t. t 2t. 2t. e +1 =. t. = (e +e ) =e +e . Then. e +1. e +e =. 2t. 2t. r (t) = 2+e +e. 2 ,e ,e. /. T(t)=. =. t. t. /. 15. (a) r (t)=. 0,cos t,sin t = 0,cos t,sin t .. 1/ 5. r (t). r (t). 1 . 5. 2. 0+cos t+sin t =. t 2t. 2 e ,e ,1. (e +1) =. 1 2t. t. 2. (. 2t. 2 e 1e. ) ,2e2t,2e2t. (e +1). 4.

(54) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.3 Arc Length and Curvature. /. 1. =. Then T (t). 2t. 2t. 2t. 4t. 4t. 1. 4t. 2e (12e +e )+4e +4e =. 2. ( e2t+1) 2. (e +1) 1. =. 2t. 2t. 2e. 2. t. 2t 2. ( 1+e ). =. 2t. /. 2t. T (t). e +1. =. /. 1. t. T (t). 2e 2t. e +1. t. 2t. 4t. 2t. 2t. 2t. 2 e (1e ),2e ,2e. 2. 2 e (e +1). 1. =. =. (e +1). (e +1) Therefore N(t) =. 2. 2t. t. 2t. 2 e (1+e ). 2t. 2e (1+2e +e ). t. 2t. 2t. 2t. 2 e (1e ),2e ,2e. t 2t. 2 e (e +1) 1 2t t t 1e , 2 e , 2 e = 2t e +1 t. /. T (t). (b) (t)=. 2e. =. /. r (t). 2t. . t. t. /. 16. (a) T(t)=. =. e +1 e +e. r (t) /. 2e. t. t. 3t. 4t. t. 2t. 2. .. (e +1). 2t,2,1/t . But since the k component is. 2. 2. 2e. =. 2t. e +2e +1. 2t,2,1/t =. 2. 2t. 2e. =. e +2e +e. 1. =. 2t. t. 1. 2t +1 r (t) 4t +4+(1/t) ln t , t is positive, t =t and 1 2 2t ,2t,1 . Then T(t)= 2 2t +1 2 1 / 2 2 2 1 T (t)= 2 4t,2,0 2t +1 (4t) 2t ,2t,1 = 4t,24t ,4t , so 2 2 2t +1 (2t +1). (. 2. /. N(t)=. T (t). 4t,24t ,4t. =. /. 2. T (t) T (t) /. r (t) /. 22. 17. r (t)=2t i+k , r . Then. 2. =. 2. 2. 2t +1 / /. t 2. 2t +1 /. 1. =. (4t) +(24t ) +(4t) /. (b) (t)=. ). =. 2. 2t,12t ,2t .. 2. 2t +1 2t 2. 2. (2t +1) 2. 2. 2. 2. /. (t)=2 i , r (t) = (2t) +0 +1 = 4t +1 , r (t) r. / /. /. (t)=2 j , r (t) r. / /. (t) =2. 5.

(55) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.3 Arc Length and Curvature. /. / /. r (t) r. (t)=. (t). /. r (t) /. /. / /. 2. /. / /. 4t +1. r (t). 2. 2. 2. 2. /. / /. (t)=2 i2 j ,. 2. (t). 2 2. =. (. 2 2. =. 3. ) (. 2. 4t +2. /. / /. .. 3/2. (4t +1). /. / /. 19. r (t)=3 i+4cos t j4sin t k , r /. 2. (t)=2 k , r (t) = 1 +1 +(2t) = 4t +2 , r (t) r. 2. 3. /. r (t) r. ). 2. 2. =. 3. (t) = 2 +2 +0 = 8 =2 2 . Then. r (t) r. (t)=. (. / /. 18. r (t)=i+ j+2t k , r r (t) r. 2. =. 3. 2. 2. 2t +1. Then (t)=. / /. t. 20. =. 3. /. r (t) /. (t). 3. t. .. (2t +1). 2. 2. / /. 2. 2. (t) = 256+144cos t+144sin t = 400 =20 .. 4 . 25. =. 5. t. 3/2. /. /. /. 2. (t)=4sin t j4cos t k , r (t) = 9+16cos t+16sin t = 9+16 =5 ,. (t)=16 i+12cos t j12sin t k , r (t) r r (t) r. ). 3. 1. =. t. 20. r (t)= e cos te sin t,e cos t+e sin t,1 . The point ( 1,0,0 ) corresponds to t=0 , and /. /. 2. 2. 2. r (0)= 1,1,1 r (0) = 1 +1 +1 = 3 . / /. r. t. t. t. t. t. t. t. t. (t) = e cos te sin te cos te sin t,e cos te sin t+e cos t+e sin t,0 t. t. = 2e sin t,2e cos t,0 r /. r (0) r. / /. / /. /. (0)= 2,0,2 . r (0) r /. Then (0)=. r (0) r /. / /. r (0) /. 21. r (t)= 1,2t,3t. 2. /. (0). 3. (0)= 0,2,0 .. / /. 2 2. =. (. 3). 2. 3. =. 2 2 3 3. /. /. / /. /. (1)= 6,6,2 , so r (1) r. /. (1)=. r (1) r /. / /. r (1). (1). 3. 2. or. 2 6 . 9. . The point ( 1,1,1 ) corresponds to t=1 , and. r (1)= 1,2,3 r (1) = 1+4+9 = 14 . r r (1) r. 2. (0) = (2) +0 +2 = 8 =2 2 .. =. 76 14. 3. =. 1 7. / /. / /. (t)= 0,2,6t r. / /. (1)= 0,2,6 .. (1) = 36+36+4 = 76 . Then. 19 . 14. 6.

(56) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.3 Arc Length and Curvature. 22.. r(t)= t,4t r. / /. 3/2. ,t. (t)= 0,3t. /. / /. r (t) r /. r (t) r. 2. 3. /. ,. (. ,2 , r (t) = 1+36t+4t. (t)= 12t. / /. 1/2. 1/2. 1/2. 1/2. +6t ,2,3t. 1/2. / /. /. r (t). 36t +4t+9 t. =. 3. ,. . 2. (t). ). 1/2. 36t +4t+9 t. (t) = 36t+4+9t =. /. 2 3/2. 2. 1. r (t) r. (t)=. /. r (t)= 1,6t ,2t. 1/2. 2. 1 2 3/2. =. (1+36t+4t ). 36t +4t+9 1/2. 2 3/2. .. t (1+36t+4t ). The point ( 1,4,1 ) corresponds to t=1 , so the curvature at this point is (1)=. 36+4+9 3/2. (1+36+4) 3. /. 2. 23. f (x)=x , f (x)=3x , f. / /. f. (x)=6x , (x)=. / /. (. (x). /. 1+ f (x) /. 24. f (x)=cos x , f (x)=sin x , f (x)=. f. / /. (x) 2 3/2. /. 25. f (x)=4x (x)=. 5/2. f. /. / /. (x) 2 3/2. /. =. /. 1 ,y x. / /. =. 1 2. 2. .. 6 x. ( 1+9x4) 3/2. 3/2. (1+sin x). 3/2. 1/2. / /. (x)=15x. 1/2. 1+( f (x)) 26. y =. 2 3/2. ,f. 41 41. cos x. =. 1+(sin x). , f (x)=10x. 3/2. 7. (x)=cos x ,. cos x. =. 1+( f (x)). / /. )2. =. =. 15x. 3/2 2 3/2. 1+(10x ). =. , 15 x 3 3/2. (1+100x ). ,. x. 7.

(57) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.3 Arc Length and Curvature. (x)=. y. / /. (x). / 1+ ( y (x)). 2 3/2. =. 1. 1. 2. 2 3/2. =. 2 3/2. 1. (x ). 2. 2. 3/2. x. =. 2. 3/2. x. =. 2. 3/2. (x +1) x x (x +1) (x +1) (1+1/x ) (since x>0 ). To find the maximum curvature, we first find the critical numbers of (x) : 3 2 3/2 2 1/2 2 1/2 2 2 2 (x +1) x (x +1) (2x) / (x +1) [(x +1)3x ] 12x 2 (x)= = = ; 2 3 2 5/2 2 3/2 2 (x +1) (x +1) (x +1) 1 / 2 / (x)=0 12x =0 , so the only critical number in the domain is x= . Since (x)>0 for 2 1 1 1 / 0<x< and (x)<0 for x> , (x) attains its maximum at x= . Thus, the maximum 2 2 2 1 1 x ,ln =0 , (x) approaches 0 as x

(58) . curvature occurs at . Since lim 2 3/2 2 2 x

(59) (x +1) /. 27. Since y =y. / /. y. x. =e , the curvature is (x)=. / /. (x) 2 3/2. /. e. =. x. 2x 3/2. x. 2x 3/2. =e (1+e ). .. (1+e ) 1+(y (x)) To find the maximum curvature, we first find the critical numbers of (x) : /. x. 2x 3/2. (x)=e (1+e ) /. +e 2x. x. 3 2. 2x 5/2. (1+e ). 2x. (2e )=e. x. 2x. 2x. 1+e 3e. 2x 5/2. =e. (1+e ) 2x. (x)=0 when 12e =0 , so e =. 2x. 12e. x. 2x 5/2. .. (1+e ). 1 1 1 2x 2x or x= ln 2 . And since 12e >0 for x< ln 2 and 12e <0 2 2 2. 1 ln 2 , the maximum curvature is attained at the point 2 1 1 1 x 2x 3/2 ( ln 2 ) /2 ln 2,e = ln 2, . Since lim e (1+e ) =0, (x) approaches 0 as x

(60) . 2 2 2 x

(61). for x>. 28. We can take the parabola as having its vertex at the origin and opening upward, so the equation is 2. f (x)=ax ,a>0 . Then by Equation 11, (x)=. f. / / /. (x) 2 3/2. [1+( f (x)) ]. 2a. =. 2 3/2. [1+(2ax) ]. =. 2a 2 2 3/2. ,. (1+4a x ). 2. thus (0)=2a . We want (0)=4 , so a=2 and the equation is y=2x . 29. (a) C appears to be changing direction more quickly at P than Q , so we would expect the curvature to be greater at P . (b) First we sketch approximate osculating circles at P and Q . Using the axes scale as a guide, we measure the radius of the osculating circle at P to be approximately 0.8 units, thus 8.

(62) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.3 Arc Length and Curvature. 1 1 1 = 1.3 . Similarly, we estimate the radius of the osculating circle at Q to be 1.4 0.8 1 1 0.7 . units, so = 1.4. =. x. /. x. 30. y=xe y =e (1x) , y. / /. x. y. =e (x2) , and (x)=. x. / / / 2 3/2. =. [1+(y ) ] x. the curvature here is what we would expect. The graph of xe x. right of the origin. As x

(63) , the graph of xe approaches zero.. 4. /. 3. 31. y=x y =4x , y. / /. 2. =12x , and (x)=. e. x2. 2x. [1+e. 2 3/2. . The graph of. (1x) ]. is bending most sharply slightly to the. is asymptotic to the x axis, and so the curvature. y. / / / 2 3/2. [1+(y ) ]. 2. =. 12x. 6 3/2. . The appearance of the two. (1+16x ). 4. humps in this graph is perhaps a little surprising, but it is explained by the fact that y=x is very flat around the origin, and so here the curvature is zero.. 32. Notice that the curve a is highest for the same x values at which curve b is turning more sharply, and a is 0 or near 0 where b is nearly straight. So, a must be the graph of y= (x) , and b is the graph of y= f (x) . 9.

(64) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.3 Arc Length and Curvature. 33. Notice that the curve b has two inflection points at which the graph appears almost straight. We would expect the curvature to be 0 or nearly 0 at these values, but the curve a isn’t near 0 there. Thus, a must be the graph of y= f (x) rather than the graph of curvature, and b is the graph of y= (x) . 34. (a) The complete curve is given by 0 t 2 . Curvature appears to have a local (or absolute) maximum at 6 points. (Look at points where the curve appears to turn more sharply.). (b) Using a CAS, we find (after simplifying) (t)=. 3 2. 2. (5sin t+sin 5t). 3/2. . (To compute cross. (9cos 6t+2cos 4t+11) products in Maple, use the Linalg package and the crossprod(a,b) command; in Mathematica, use Cross.) The graph shows 6 local (or absolute) maximum points for 0 t 2 , as observed in part (a).. 35. Using a CAS, we find (after simplifying) (t)=. 6. 2. 4cos t12cos t+13 3/2. . (To compute cross. (1712cos t) products in Maple, use the Linalg package and the crossprod (a,b) command; in Mathematica, use Cross.) Curvature is largest at integer multiples of 2 .. 10.

(65) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.3 Arc Length and Curvature. /. /. /. 36. Here r(t)= f (t),g ( t ) , r (t)= f (t),g (t) , r. / /. / /. (t)= f. /. r (t) r. / /. (t),g. / /. /. (t) , r (t) = /. / /. 0,0,f (t) g. (t) =. 3. /. 2. /. 2. ( f (t)) +(g (t)) / /. (t) f. 3. /. 1/2. . 2. = (x yx y ). (t) g (t). /. 2 3/2. . 2. . 2 3/2. =[( f (t)) +(g (t)) ] =(x +y ). . .. ... /. 2. , and. . .. . ... = x yyx . Thus. . .. . ... x yyx. (t)=. . 2. .. . 2 3/2. (x +y ). .. t. ... t. t. t. t. 37. x=e cos t x=e (cos tsin t) x=e (sin tcos t)+e (cos tsin t)=2e sin t , .. t. ... t. t. t. t. y=e sin t y=e (cos t+sin t) y=e (sin t+cos t)+e (cos t+sin t)=2e cos t . Then . .. . ... (t) =. t. x yyx. ( x +y ) .. 2. .. 2. 3/2. =. t. t. t. e (cos tsin t)(2e cos t)e (cos t+sin t)(2e sin t). ([et(cos tsin t)]2+[et(cos t+sin t)]2) 3/2. 2t. 2. 2. 2e (cos tsin tcos t+sin tcos t+sin t). =. 2t. 2. 2. 2. 2. e (cos t2cos tsin t+sin t+cos t+2cos tsin t+sin t) 2t. 2t. 2e (1). =. 2t. e (1+1) .. 3. 2. 3/2. =. 2e 3t. 3/2. =. e (2). ... 3/2. 1 t. 2e 2. .. ... 38. x=1+t x=3t x=6t , y=t+t y=1+2t y=2 . Then . .. . ... (t) =. x yyx. ( x +y ) .. 2. .. 2. 3/2. 2. =. (3t )(2)(1+2t)(6t) 22. 2 3/2. (3t ) +(1+2t). 2. =. 6t 6t 4. 2. 3/2. (9t +4t +4t+1). 11.

(66) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.3 Arc Length and Curvature. 2. 6 t +t. =. 4. 2. 3/2. (9t +4t +4t+1). 39.. 2. 2. 2. 2. 2. 2. 1. 2. 2t,2t ,1 +(2t +1) 2. 2. 3. 2. 2. / /. =. 4. 2. , so. 2t +1. 4t +4t +1. 2,4t,0 [ by Theorem 14.2.3 [ET 13.2.3] #3]. 3. 2. 2. 8t +4t +2,8t +8t +4t,4t =2(2t +1). =(2t +1). T (t). 2t,2t ,1. .. T (t) =4t(2t +1). N(t) =. =. /. r (t). /. 2. 2t,2t ,1. r (t). corresponds to t=1 . T ( t ) =. 2 2 1 , , 3 3 3. T(1)=. 2. /. 2 1, ,1 3. =. 2. 2. T (t). 2. 2(2t +1) 2(2t +1). 2. 12t ,2t,2t 22. 2. 12t ,2t,2t. =. 2. 2. 12t ,2t,2t 2. (12t ) +(2t) +(2t). 4. 14t +4t +8t. 2. 2. 12t ,2t,2t. =. 1+2t . N(1)=. 2. 1 2 2 , , 3 3 3. . and B(1)=T(1) N(1)=. 4 2 4 1 , + 9 9 9 9. ,. 4 2 + 9 9. =. . 2 1 2 , , 3 3 3. .. t. 40. ( 1,0,1 ) corresponds to t=0 . r(t)=e 1,sin t,cos t , so t. /. t. t. r (t)=e 1,sin t,cos t +e 0,cos t,sin t =e 1,sin t+cos t,cos tsin t and t. /. T(t) =. r (t) /. r (t) =. T ( 0) =. =. e 1,sin t+cos t,cos tsin t t. 2. e. 2. 2. 2. 1+sin t+2sin tcos t+cos t+cos t2sin tcos t+sin t. 1,sin t+cos t,cos tsin t 3 1 1 1 , , 3 3 3. /. ,. . T (t)=. 1 0,cos tsin t,sin tcos t , so 3. 12.

(67) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.3 Arc Length and Curvature. 1 0,cos tsin t,sin tcos t 3. /. T (t). N(t) =. =. /. T (t). 1 3. 2. 2. 2. 2. 2. 0 +cos t2cos tsin t+sin t+sin t+2sin tcos t+cos t. 1 0,cos tsin t,sin tcos t . 2. = N ( 0) =. 0,. 1 1 , 2 2. and B ( 0 ) =T ( 0 ) N ( 0 ) =. 2 1 1 , , 6 6 6. . .. 41. ( 0, ,2 ) corresponds to t= . r(t)= 2sin 3t,t,2cos 3t /. T(t)=. r (t) /. 6cos 3t,1,6sin 3t. =. =. 1 6cos 3t,1,6sin 3t . 37. 2 2 r (t) 36cos 3t+1+36sin 3t 1 T( )= 6,1,0 is a normal vector for the normal plane, and so 6,1,0 is also normal. Thus an 37 equation for the plane is 6 ( x0 ) +1(y )+0(z+2)=0 or y6x= .. 1 / T (t)= 18sin 3t,0,18cos 3t T (t) = 37 /. 2. 2. 2. 2. 18 sin 3t+18 cos 3t 37. =. 18 37. /. N(t)=. T (t) /. = sin 3t,0,cos 3t . So N( )= 0,0,1 and. T (t) 1 1 B( )= 6,1,0 0,0,1 = 1,6,0 . Since B( ) is a normal to the osculating plane, so is 37 37 1,6,0 and an equation for the plane is 1(x0)+6(y )+0(z+2)=0 or x+6y=6 .. .. /. 2. /. 42. t=1 at ( 1,1,1 ) . r (t)= 1,2t,3t . r (1)= 1,2,3 is normal to the normal plane, so an equation for this plane is 1(x1)+2(y1)+3(z1)=0 , or x+2y+3z=6 . T(t)=. /. r (t) /. 1. =. 2. r (t). /. T (t) =. 1,2t,3t. 1+4t +9t 1. ( 1+4t2+9t 4) 3/2 2. 4. 6t(1+4t +9t ). 2. . Using the product rule on each term of T(t) gives. 4. . 1 3 2 4 1 3 (8t+36t ),2(1+4t +9t ) (8t+36t )2t, 2 2. 1 3 2 (8t+36t )3t 2 13.

(68) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.3 Arc Length and Curvature. 1. =. 2. 3. 4. 4t18t ,218t ,6t+12t. 4 3/2. (1+4t +9t ). 3. =. 2 3/2. 11,8,9 when t=1 .. (14). /. /. N(1) T (1) 11,8,9 and T(1) r (1)= 1,2,3 a normal vector to the osculating plane is 11,8,9 1,2,3 = 42,42,14 or equivalently 3,3,1 . An equation for the plane is 3(x1)3(y1)+(z1)=0 or 3x3y+z=1 . 43. The ellipse is given by the parametric equations x=2cos t , y=3sin t , so using the result from Exercise 36, . .. .. .. (t)=. x yx y . 2. 3/2. . 2. =. (2sin t)(3sin t)(3cos t)(2cos t) 2. 2 3/2. (4sin t+9cos t). =. 6 2. 2 3/2. (4sin t+9cos t). x +y. 6 2 9 = , so the radius of the osculating circle is 1/ (0)= and its center 27 9 2 2 5 5 6 3 2 81 +y = = = . is ,0 . Its equation is therefore x+ . At ( 0,3) , t= , and 2 2 4 2 2 8 4 4 5 So the radius of the osculating circle is and its center is 0, . Hence its equation is 3 3 5 2 16 2 x + y = . 3 9 At ( 2,0 ) , t=0 . Now (0)=. 44. y=. 1 2 / x y =x and y 2. / /. =1 , so Formula 11 gives (x)=. 1 2 3/2. . So the curvature at ( 0,0 ) is. (1+x ). 2. 2. (0)=1 and the osculating circle has radius 1 and center ( 0,1 ) , and hence equation x +(y1) =1 . The 1 1 1 1 curvature at 1, is (1)= . The tangent line to the parabola at 1, has = 2 3/2 2 2 2 2 (1+1 ) slope 1 , so the normal line has slope 1 . Thus the center of the osculating circle lies in the direction of the 14.

(69) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.3 Arc Length and Curvature. . unit vector 1,. 1 2. 1 1 , 2 2 . +2 2. . The circle has radius 2 2 , so its center has position vector. 1 1 , 2 2. =. 1,. 5 2. 2. . So the equation of the circle is ( x+1 ) + y. 5 2. 2. =8 .. 45. The tangent vector is normal to the normal plane, and the vector 6,6,8 is normal to the given /. /. plane. But T(t) r (t) and 6,6,8 3,3,4 , so we need to find t such that r (t) 3,3,4 . 3. 4. /. 2. 3. r(t)= t ,3t,t r (t)= 3t ,3,4t 3,3,4 when t=1 . So the planes are parallel at the point r(1)= ( 1,3,1 ) . 46. To find the osculating plane, we first calculate the tangent and normal vectors. In Maple, we set x:=t^3; y:=3*t; and z:=t^4; and then calculate the components of the tangent vector 2. T(t) using the diff command. We find that T(t)=. 3t ,3,4t 6. 3. 4. . Differentiating the components of. 16t +9t +9 6. /. T(t) , we find that N(t)=. 5. 3. 2 4. 6t(8t 9),3(48t +18t ),36t (t +3). T (t). = . / 6 2 5 32 12 8 4 T (t) 144t(8t 9) +9(96t +36t ) +5,184t +31,104t +46,656t In Maple, we can calculate B(t)=T(t) N(t) using the linalg package. First we define T and N using T:=array(); and N:=array(); where f, g, h, F, G, and H are the components of T and N . Then we use the command B:=crossprod(T,N);. After normalization and 3. simplification, we find that B(t)=b 6t,2t ,3 , where 6. t. b= 2. 6. 2. 4. 16t +9t +9 5. 32. 12. 8. 4. 16t (8t 9) +(96t +36t ) +576t +3456t +5184t In Mathematica, we use the command Dt to differentiate the components of r(t) and subsequently T(t) , and then load the vector analysis package with the command << Calculus‘VectorAnalysis‘. After setting T={f,g,h} and N={F,G,H} , we use CrossProduct [T, N] to find B (before 15.

(70) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.3 Arc Length and Curvature. normalization). 3. Now B(t) is parallel to 6t,2t ,3 , so if B(t) is parallel to 1,1,1 for some t , then 6t=1 t= but 2. 47. =. 1 , 6. 3. 1 6. 1 . So there is no such osculating plane.. dT ds. =. dT/dt ds/dt. =. dT/dt ds/dt. dT/dt dT/dt. and N=. dT dt dT dt. , so N=. dT dt dT/dt dT = = by ds ds/dt ds dt. the Chain Rule. 48. For a plane curve, T= T cos i+ T sin j=cos i+sin j . Then dT d dT d = =(sin i+cos j) and ds ds d ds dT d d = sin i+cos j = . Hence for a plane curve, the curvature is ds ds ds = d /ds . 49. (a). d dB dB B=0 B ( B B) =0 2 ds ds ds. B =1 B B=1. (b) B=T N d B = d ( T N ) = d ( T N ) 1 = d ( T N ) ds/dt dt ds dt ds =. (T. /. ) (. N + T N. /. ). 1 /. r (t). =. /. T. 1 /. r (t) T T. / /. (. + T N. /. ). 1 /. r (t). =. T N. /. /. r (t). dB T ds (c) B=T N T N , BT and B N . So B , T and N form an orthogonal set of vectors in the. . 3. threedimensional space R . From parts (a) and (b), d B/ds is perpendicular to both B and T , so d B/ds is parallel to N . Therefore, d B/ds= (s)N , where (s) is a scalar. (d) Since B=T N , T N and both T and N are unit vectors, B is a unit vector mutually perpendicular to both T and N . For a plane curve, T and N always lie in the plane of the curve, so that B is a constant unit vector always perpendicular to the plane. Thus d B/ds=0 , but d B/ds= (s)N and N 0 , so (s)=0 . 50. N=B T 16.

(71) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.3 Arc Length and Curvature. dN d dB dT [by Theorem 14.2.3 [ET 13.2.3]#5] = ( B T ) = T+B. ds ds ds ds [by Formulas 3 and 1 ] = N T+B N [by Theorem 13.4.8 [ET 12.4.8 ]#2] = ( N T ) + ( B N ) But B N=B ( B T ) = ( B T ) B ( B B) T [ by Theorem 13.4.8 [ET 12.4.8]#6 ] =T dN/ds= (T N) T= T+ B . /. /. 51. (a) r =s T r. / /. =s. / /. /. /. T+s T =s. / /. T+s. dT / s =s ds. /. / /. / 2. T+ (s ) N by the first SerretFrenet. formula. (b) Using part (a), we have /. r r. / /. = (s / T) [s. / /. /. =. (s T) (s. = (s / s. / /. / 2. T+ (s ) N]. / /. /. / 2. T) + (s T) ( (s ) N) 3] / 3. / 3. / 3. )(T T)+ (s ) (T N)=0+ (s ) B= (s ) B. (c) Using part (a), we have r. / / /. / /. = [s = s = s. / /. / 2. / / /. T+s. / / /. = [s. /. / / /. / /. /. /. / 2. /. / /. / 2. T+ (s ) N] =s T+s T + (s ) N+2 s s N+ (s ) N / / / dT / / / 2 / / / / 2 dN / T+s s + (s ) N+2 s s N+ (s ) s ds ds 2. / /. /. /. / 2. /. s N+ (s ) N+2 s s. / 3. /. (s ) ]T+[3 s s. / /. /. / /. /. / 3. N+ (s ) ( T+ B) [by the second formula]. / 2. / 3. + (s ) ] N+ (s ) B. (d) Using parts (b) and (c) and the facts that B T=0 , B N=0 , and B B=1 , we get /. (r r /. / /. r r. ) r. / / /. / / 2. {. / 3. =. (s ) B [s. / / /. 2. / 3. /. (s ) ]T+[3 s s / 3. (s ) B / 3. =. / /. /. / 2. / 3. }. + (s ) ] N+ (s ) B. 2. / 3. (s ) (s ) / 3 2. =. (s ). 52. First we find the quantities required to compute : /. r (t)= asin t,acos t,b r. / /. (t)= acos t,asin t,0 r. / / /. (t)= asin t,acos t,0. 17.

(72) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.3 Arc Length and Curvature. /. =. r (t). 2. 2. 2. 2. 2. (asin t) +(acos t) +b = a +b. / / / r (t) r (t) = k j i asin t acos t b acos t asin t 0 2. =absin t iabcos t j+a k /. / /. r (t) r. 2. =. (t). ( r /(t) r / /(t)) r / / /(t) =. 2. 22. 2 2. (absin t) +(abcos t) +(a ) = a b +a. 4. 2. 2. (absin t)(asin t)+(abcos t)(acos t)+(a )(0)=a b. Then by Theorem 10, /. r (t) r. (t)=. / /. 2 2. (t). =. 3. /. r (t). a b +a. (. 2. 2. a +b. 4. =. 3. 2. a. 2. a +b. 2. 2. = 0,0,2 r r. / /. ) (. 2. 2. a +b. ). 3. a. =. a +b. which is a constant. From Exercise 51(d), the torsion is given by /. =. (r r. / /. / / /. ) r. /. r r. 2. =. / / 2. ab. (. 2 2. a b +a. 4. ). 2. =. b 2. 2. a +b. which is also a constant. 53. r=. t,. 1 2 1 3 t, t 2 3. /. / / / / / / ( r r ) r = = /. r r. 2. r = 1,t,t. / /. ,r. = 0,1,2t , r. 4. / / 2. 2. t +4t +1. =. / /. 4. 2. t +4t +1. 2. / /. = sinh t,cosh t,0 , r. / / /. = cosh t,sinh t,0 . 2. = cosh t,sinh t,cosh tsinh t = cosh t,sinh t,1 . /. =. 2. = t ,2t,1 . 2. /. /. /. 2. t ,2t,1 0,0,2. 54. r= sinh t,cosh t,t r = cosh t,sinh t,1 , r r r. / / /. r r r. / /. / 3. =. 2. cosh t,sinh t,1 cosh t,sinh t,1. 3. =. 2. cosh t+sinh t+1. ( cosh. 2. 2. t+sinh t+1. ). 3/2. =. 1 2. 2. cosh t+sinh t+1. =. 1 2. ,. 2cosh t. 18.

(73) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.3 Arc Length and Curvature. / / / / / / ( r r ) r = = /. cosh t,sinh t,1 cosh t,sinh t,0 2. / / 2. 2. 2. =. cosh t+sinh t+1 1 1 So at the point ( 0,1,0 ) , t=0 , and = and = . 2 2 r r. 2. cosh t+sinh t 2. =. 2cosh t. 1 2. 2cosh t. 55. For one helix, the vector equation is r(t)= 10cos t,10sin t,34t/(2 ) (measuring in angstroms), because the radius of each helix is 10 angstroms, and z increases by 34 angstroms for each increase of 8. 2 in t . Using the arc length formula, letting t go from 0 to 2.9 10 2 , we find the approximate length of each helix to be 8. L=. 2.9 10 2. . 8. 2.9 10 2. /. . r (t) dt=. 0. =. 2. 2. (10sin t) +(10cos t) +. 0. 100+. 8. 2.9 10 2. 2. 34 2. 34 2. t 0. 8. =2.9 10 2. 100+. 34 2. 2. dt 2. 2.07 1010 more than two meters!. 56. (a) For the function F(x)= P(1)=1 .. {. 0 if x<0 P(x) if 0<x<1 1 if x 1. /. /. to be continuous, we must have P(0)=0 and. /. For F to be continuous, we must have P (0)=P (1)=0 . The curvature of the curve y=F(x) at the point ( x,F(x) ) is (x)= P. / /. (0)=P. / /. F. ( 1+ F. / / /. (x). (x). 2 3/2. ). . For (x) to be continuous, we must have. (1)=0 . 5. 4. 3. 2. /. 4. 3. 2. Write P(x)=ax +bx +cx +dx +ex+ f . Then P (x)=5ax +4bx +3cx +2dx+e and P. / /. 3. 2. (x)=20ax +12bx +6cx+2d . Our six conditions are: P(0)=0 f=0 (1) P(1)=1 a+b+c+d+e+f=1 (2) /. P (0)=0 e=0 (3) /. P (1)=0 5a+4b+3c+2d+e=0 (4) P P. / / / /. (0)=0 d=0 (5) (1)=0 20a+12b+6c+2d=0 (6) 19.

(74) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.3 Arc Length and Curvature. From (1), (3), and (5), we have d=e= f =0 . Thus (2), (4) and (6) become (7) a+b+c=1 , (8) 5a+4b+3c=0 , and (9) 10a+6b+3c=0 . Subtracting (8) from (9) gives (10) 5a+2b=0 . Multiplying (7) by 3 and subtracting from (8) gives (11) 2a+b=3 . Multiplying (11) by 2 and subtracting from 5. 4. 3. (10) gives a=6 . By (10), b=15 . By (7), c=10 . Thus, P(x)=6x 15x +10x .. (b). 20.

(75) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.4 Motion in Space: Velocity and Acceleration. 1. (a) If r(t)=x(t) i+y ( t ) j+z(t) k is the position vector of the particle at time t , then the average velocity over the time interval 0,1 is r(1)r(0) ( 4.5 i+6.0 j+3.0 k ) ( 2.7 i+9.8 j+3.7 k ) v = = =1.8 i3.8 j0.7 k . Similarly, over the other ave 1 10 intervals we have 0.5,1 :v = r(1)r(0.5) = (4.5 i+6.0 j+3.0 k)(3.5 i+7.2 j+3.3 k) ave 10.5 0.5 = 2.0 i2.4 j0.6 k r(2)r(1) (7.3 i+7.8 j+2.7 k)(4.5 i+6.0 j+3.0 k) 1,2 :v = ave = 21 1 = 2.8 i+1.8 j0.3 k 1,1.5 :v = r(1.5)r(1) = (5.9 i+6.4 j+2.8 k)(4.5 i+6.0 j+3.0 k) ave 1.51 0.5 = 2.8 i+0.8 j0.4 k (b) We can estimate the velocity at t=1 by averaging the average velocities over the time intervals 1 0.5,1 and 1,1.5 : v(1) (2 i2.4 j0.6 k)+(2.8 i+0.8 j0.4 k) =2.4 i0.8 j0.5 k . Then the speed 2 2. 2. 2. is v(1) (2.4) +(0.8) +(0.5) 2.58 . 2. (a) The average velocity over 2 t 2.4 is r(2.4)r(2) =2.5 r(2.4)r(2) , so we sketch a vector in the same direction but 2.5 times the length of 2.42 r(2.4)r(2) .. r(2)r(1.5) =2 r(2)r(1.5) , so we sketch a vector in the 21.5 same direction but twice the length of r(2)r(1.5) . r(2+h)r(2) (c) Using Equation 2 we have v(2)=lim . h h 0 (b) The average velocity over 1.5 t 2 is. 1.

(76) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.4 Motion in Space: Velocity and Acceleration. (d) v(2) is tangent to the curve at r(2) and points in the direction of increasing t . Its length is the speed of the particle at t=2 . We can estimate the speed by averaging the lengths of the vectors found in parts (a) and (b) which represent the average speed over 2 t 2.4 and 1.5 t 2 respectively. Using the axes scale as a guide, we estimate the vectors to have lengths 2.8 and 2.7 . Thus, we 1 estimate the speed at t=2 to be v(2) (2.8+2.7)=2.75 and we draw the velocity vector v(2) with 2 this length.. 2. 3. r(t)= t 1,t At t=1 : /. v(t)=r (t)= 2t,1 , v(1)= 2,1 a(t)=r. / /. (t)= 2,0 , a(1)= 2,0 2. v(t) = 4t +1. 4. r(t)= 2t,4 t At t=1 : /. v(t)=r (t)= 1,2/ t a(t)=r. / /. (t)= 0,1/t. , v(1)= 1,2 3/2. , a(1)= 0,1. v(t) = 1+4/t. 2.

(77) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.4 Motion in Space: Velocity and Acceleration. t. t. 5. r(t)=e i+e j t. t. t. t. v(t)=e ie j , a(t)=e i+e j At t=0 : v(0)=i j , a(0)=i+ j 2t. 2t. v(t) = e +e. t. =e. 4t. e +1. t. t. ln x. Since x=e , t=ln x and y=e =e. =1/x , and x>0 , y>0 .. 6. r(t)=sin t i+2cos t j 3 = i j 2 6 1 a(t)=sin t i2cos t j , a = i 3 j 6 2. v(t)=cos t i2sin t j , v. 2. 2. 2. v(t) = cos t+4sin t = 1+3sin t 2. 2. 2. 2. And x +y /4=sin t+cos t=1 , an ellipse.. 3.

(78) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.4 Motion in Space: Velocity and Acceleration. 7. r(t)=sin t i+t j+cos t k v(t)=cos t i+ jsin t k , v(0)=i+ j a(t)=sin t icos t k , a(0)=k 2. 2. v(t) = cos t+1+sin t = 2 2. 2. Since x +z =1 , y=t , the path of the particle is a helix about the y axis.. 2. 3. 8. r(t)=t i+t j+t k 2. v(t)=i+2t j+3t k , v(1)=i+2 j+3 ka(t)=2 j+6t k , a(1)=2 j+6 k 2. 4. v(t) = 1+4t +9t The path is a ‘‘twisted cubic’’ (see Example 14.1.7 ).. 2. 3 2. /. /. 2. 9. r(t)= t +1,t ,t 1 v(t)=r (t)= 2t,3t ,2t , a(t)=v (t)= 2,6t,2 , 2. 22. 2. 4. 2. v(t) = (2t) +(3t ) +(2t) = 9t +8t = t /. 2. 9t +8 . /. 10. r(t)= 2cos t,3t,2sin t v(t)=r (t)= 2sin t,3,2cos t , a(t)=v (t)= 2cos t,0,2sin t , 2. 2. v(t) = 4sin t+9+4cos t = 13 . 4.

(79) Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.4 Motion in Space: Velocity and Acceleration. t. t. t. /. t. t. t. /. 11. r(t)= 2 t i+e j+e k v(t)=r (t)= 2 i+e je k , a(t)=v (t)=e j+e k , 2t. 2t. v(t) = 2+e +e. t. t 2. t. t. = (e +e ) =e +e .. 2. /. 12. r(t)=t i+ln t j+t k v(t)=r (t)=2t i+t t. 1. /. j+k , a(t)=v (t)=2 it. t. /. 2. 2. 2. j , v(t) = 4t +t +1 .. t. t. 13. r(t)=e cos t,sin t,t v(t)=r (t)=e cos t,sin t,t +e sin t,cos t,1 =e cos tsin t,sin t+cos t,t+1 a(t) =v /(t)=et cos tsin tsin tcos t,sin t+cos t+cos tsin t,t+1+1 t. =e 2sin t,2cos t,t+2 v(t) =et cos 2t+sin 2t2cos tsin t+sin 2t+cos 2t+2sin tcos t+t 2+2t+1 t. =e. 2. t +2t+3 2. /. 14. r(t)=tsin t i+tcos t j+t k v(t)=r (t)=(sin t+tcos t) i+(cos ttsin t) j+2t k , /. a(t)=v (t)=(2cos ttsin t) i+(2sin ttcos t) j+2 k , 2. 2. 2. 2. 2. 2. 2. 2. v(t) = (sin t+2tsin tcos t+t cos t)+(cos t2tsin tcos t+t sin t)+4t = 5t +1 . 15. a(t)=k v(t)= a(t)dt= kdt=t k+c and i j=v(0)=0 k+c , so c =i j and v(t)=i j+t k . 1. 1. 1. 1 2 1 2 r(t)= v(t)dt= ( i j+t k ) dt=t it j+ t k+c . But 0=r(0)=0+c , so c =0 and r(t)=t it j+ t k . 2 2 2 2 2 16. a(t)=10 k v(t)= (10k)dt=10t k+c , and i+ jk=v(0)=0+c , so c =i+ jk and 1. 1. 1. v(t)=i+ j(10t+1) k . r(t)= i+ j(10t+1) k dt=t i+t j(5t +t) k+c . But 2 i+3 j=r(0)=0+c , so c =2 i+3 j and 2. 2. 2. 2. 2. r(t)=(t+2) i+(t+3) j(5t +t) k . 17. (a) a(t)=i+2 j+2t k v(t)= (i+2 j+2t k)dt=t i+2t j+t k+c , and 0=v(0)=0+c , so c =0 and 2. 1. v(t)=i+2t j+t k . r(t)= (t i+2t j+t k)dt= 2. r(t)= 1+. 2. 1. 1. 1 2 2 1 3 t i+t j+ t k+c . But i+k=r(0)=0+c , so c =i+k and 2 2 2 2 3. 1 2 1 3 2 t i+t j+ 1+ t k . 2 3 5.

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