Calculus Stewart ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves
1. The component functions t , t 1 , and 5 t are all defined when t 1 0 t 5 , so the domain of r(t) is 1,5 . 2. The component functions
2
t
1 and 5 t
0
t 2 2 2 , sin t , and ln (9 t ) are all defined when t 2 and 9 t >0 t+2 3<t<3 , so the domain of r(t) is ( 3, 2 ) ( 2,3) . ln t 1/t 3. lim cos t=cos 0=1 , lim sin t=sin 0=0 , lim tln t=lim =lim =lim t=0 . Thus 2 + + + + 1/t + + t 0 t 0 t 0 t 0 t 0 1/t t 0 lim cos t,sin t,tln t =
t 0
+
lim cos t,lim sin t,lim tln t
t 0
+
= 1,0,0 .
t
0
+
t
0
+
e 1 e 4. lim =lim =1 [ using l'Hospital's Rule], t 0 t t 0 1 lim
t 0
t
t
1+t 1 =lim t t 0
1+t 1 t
1+t +1 1+t +1 1,
=lim
t 0
1 1 3 = , lim =3 . 1+t +1 2 t 0 1+t
1 ,3 . 2 t 1 1 1 5. lim t+3 =2 , lim 2 =lim = , 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
tan t t
=tan 1 .
2
, lim e =0 , lim
t 2t t
2t
ln t 1/t =lim =0 [ by l'Hospital's Rule]. t t 1 ,0,0 .
4
Thus lim
t
arctant,e ,
ln t t
=
2
7. The corresponding parametric equations for this curve are x=t +1 , y=t . We can make a table of values, or we can eliminate the parameter: t=y x=y +1 , with y R . By comparing different values of t , we find the direction in which t increases as indicated in the graph.
4
8. The corresponding parametric equations for this curve are x=t ,y=t . We can make a table of values, or we can eliminate the parameter: x=t
3
3
2
t= x
3
y=t =
2
( 3 x ) 2=x2/3 , with t
R
x R . By
1
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 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.
2 2 2 2 2 2
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 .
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 .
2
2
2
2
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 .
13. The parametric equations are x=t , y=t , z=t . These are positive for t 0 and 0 when t=0 . So the curve lies entirely in the first quadrant. The projection of the graph onto the xy plane is y=x , y>0 , a half parabola. On the xz plane z=x , z>0 , a half cubic, and the yz plane, y =z .
3 3 2 2
2
4
6
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 .
2
2
2
2
2
15. Taking r = 0,0,0 and r = 1,2,3 , we have from Equation 13.5.4
0 1
r(t)=(1 t) r +t r =(1 t) 0,0,0 +t 1,2,3 , 0 t
0 1
1 or r(t)= t,2t,3t , 0 t 1.
1.
2
Parametric equations are x=t , y=2t , z=3t , 0 t
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)=(1 t) r +t r =(1 t) 1,0,1 +t 2,3,1 , 0 t
0 1
1 or r(t)= 1+t,3t,1 , 0 t 1.
0 1
1.
Parametric equations are x=1+t , y=3t , z=1 , 0 t
0 1
17. Taking r = 1, 1,2 and r = 4,1,7 , we have r(t)=(1 t) r +t r =(1 t) 1, 1,2 +t 4,1,7 , 0 t or r(t)= 1+3t, 1+2t,2+5t , 0 t
0 1
1 1.
1 . Parametric equations are x=1+3t , y= 1+2t , z=2+5t , 0 t
0 1
18. Taking r = 2,4,0 and r = 6, 1,2 , we have r(t)=(1 t) r +t r = ( 1 t ) 0 t 0 t 1 or r(t)= 2+8t,4 5t,2t , 0 t 1.
2,4,0 +t 6, 1,2 ,
1 . Parametric equations are x= 2+8t , y=4 5t , z=2t ,
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. 20. x=t , y=t , z=e . At any point on the curve, y=x . So the curve lies on the parabolic cylinder 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. 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. 22. x=e cos 10t , y=e sin 10t , z=e . 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. 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. 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 , so the graph is III. the z axis. As t 0 , z 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 z =x +y . Since z=t , the curve is a spiral on this cone.
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2t 2 2t 2 2t 2 2 2t 2 2 2 2 t t t 2 2 2 2 t 2
2
2
2
2
3
Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves
26. Here x =sin t=z and x +y =sin t+cos t=1 , so the curve is the intersection of the parabolic cylinder z=x with the circular cylinder x +y =1 .
2 2 2
2
2
2
2
2
2
27. r(t)= sin t,cos t,t
2
28. r(t)= t t +1,t,t
4 2
2
29. r(t)= t , t 1 , 5 t
2
30. We have the computer plot the parametric equations x=sin t , y=sin 2t , z=sin 3t , 0 t
2 . The
4
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, x +y =(1+cos 16t) cos t+(1+cos 16t) sin t =(1+cos 16t) =z , so the graph lies on the cone 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 2 2 2 2 2 2 2 2 2
x= 1 0.25cos 10t cos t , y= 1 0.25cos 10t sin t , z=0.5cos 10t . At any point on the graph,
5
2
2
Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves
x +y +z =(1 0.25cos 10t)cos t +(1 0.25cos 10t)sin t+0.25cos t =1 0.25cos 10t+0.25cos 10t=1 , 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
2 2 2 2 2 2 2 2
2
2
2
2
2
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 curve, we require y=1 3t=7 t= 2. But then z=1+( 2) = 7
3
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 . 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 36. The projection of the curve C of intersection onto the xy plane is the parabola y=x ,z=0 . Then we can choose the parameter x=t
2 2 2 22 2 2 2 2 2
y=t . Since C also lies on the surface +y z=4x , we have
2 2 4 4
2
2
2
z=4x +y =4t +(t ) . Then parametric equations for C are x=t , y=t , z=4t +t , and the corresponding vector function is r(t)=t i+t j+(4t +t ) k . 37.
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Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves
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 2 2
2
2
2 . Since C also lies on the surface z=x , we have
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 . 38.
2
x=t
y=t
2
4z =16 x 4y =16 t 4t
2
2
2
2
4
z=
4
1 t 2
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 39. For the particles to collide, we require r (t)=r (t)
2 2 1 2 2
t ,7t 12,t
2
2
= 4t 3,t ,5t 6 . Equating
2
2
components gives t =4t 3 , 7t 12=t , and t =5t 6 . From the first equation, t 4t+3=0 (t 3)(t 1)=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 ) . 40. The particles collide provided r (t)=r (t)
1 2
t,t ,t
2 3
= 1+2t,1+6t,1+14t . Equating components
7
Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves
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 and a value for s where r (t)=r (s)
1 2 2 3
2
3
t,t ,t
2 3
= 1+2s,1+6s,1+14s . Equating components, t=1+2s ,
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(2s 1)=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 real valued 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 t a 1 t a 2 t a 3 t a 1 t a 2 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 real valued 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 t a 1 t a 2 t a 2 t a 3 t a 3
lim
t a 1
u (t)+v (t) ,lim
1 1 t 2 a 2
u (t)+v (t) ,lim
2 2 t 3 a
u (t)+v (t)
3 3
= lim
t
u (t)+v (t),u (t)+v (t),u (t)+v (t) [using(1)backward
1 3
a
= lima u(t)+v(t) t (b) lim cu(t) =lim
t a t a
cu (t),cu (t),cu (t) = lim cu (t),lim cu (t),lim cu (t)
1 2 3 t a 1 t a 2 t a 3
= clim u (t),clim u (t),clim u (t) =c
t a 1 t 2 a 3 2 t a 3
lim u (t),lim u (t),lim u (t)
t a 1 t a 2 t a 3
=clim
t a
u (t),u (t),u (t) =clim u(t)
1 t a
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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 t a 1 t a 2 t 1 a 3
lim v (t),lim v (t),lim v (t)
t 2 a 1 t 2 a 2 t a 3 3
= lim u (t)
t t t a a a 1 1 1
lim v (t) + lim u (t)
t t 1 a a 2 t 2 t 3 a 2 3
lim v (t) + lim u (t)
t 3 a t a
lim v (t)
t a 3
=lim u (t)v (t)+lim u (t)v (t)+lim u (t)v (t)
a 3
=lim (d)
u (t)v (t)+u (t)v (t)+u (t)v (t) =lim u(t) v(t)
1 2 t a
lim u(t) lim v(t) = lim u (t),lim u (t),lim u (t)
t a t a t a 1 t a 2 t a 3
lim v (t),lim v (t),lim v (t)
t a 1 t a 2 t a 3
=
lim u (t)
t 3 a 2
lim v (t)
t 1 a 3
lim u (t)
t 1 a 3
lim v (t) ,
t 3 a 2
lim u (t)
t a
lim v (t)
t a
lim u (t)
t a
lim v (t) ,
t a
lim u (t)
t a 1
lim v (t)
t a 2
lim u (t)
t a 2
lim v (t)
t a 1
= lim
t a
u (t)v (t) u (t)v (t) ,lim
2 3 3 2 t a
u (t)v (t) u (t)v (t) ,
3 1 1 3
lim
t a
u (t)v (t) u (t)v (t)
1 2 2 1
=lim
t 1 a
u (t)v (t) u (t)v (t),u ( t ) v (t) u (t)v (t), 2 3 3 2 3 1 1 3
2 2 1
u (t)v (t) u (t)v (t) =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
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= 0,4 . Since z=sin 1.5t , z is maximized where sin 1.5t=1 1.5t= , , or 2 2 9 2 5 3 7 11 7 t= , , or 3 . z is minimized where sin 1.5t= 1 1.5t= , , or t= , , or 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 self intersections 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 computer drawn graph of the curve from above, as well as views from the front and from the right side.
10
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.
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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 b=lim r(t)= lim f (t),lim g(t),lim h(t)
t t a a 2 t a t a t a 3 t a
1 2 3
. If lim r(t)=b , then lim r(t) exists, so by (1),
t a t a t a 1
. By the definition of equal vectors we have lim f (t)=b ,
lim g(t)=b and lim h(t)=b . But these are limits of real valued functions, so by the definition of limits, for every >0 there exists whenever 0< t a <
2
1
>0 ,
2
>0 ,
3
3
>0 so f (t) b < /3 whenever 0< t a <
1 3
1
, g(t) b < /3
2
, and h(t) b < /3 whenever 0< t a <
1 2 3
. Letting = minimum of
{
=
1 2 3
, ,
}
1
, we have f (t) b + g(t) b + h(t) b < /3+ /3+ /3= whenever 0< t a < . But f (t) b ,g(t) b ,h(t) b
1 2 2 2 2 3 3 2
r(t) b =
( f (t) b ) +(g(t) b ) +(h(t) b )
1 2 3
[ f (t) b ] +
1
2
[g(t) b ] +
2
2
[h(t) b ]
3
2
= f (t) b + g(t) b + h(t) b r(t) b
1 2
. Thus for every >0 there exists >0 such that
3
f (t) b + g(t) b + h(t) b < whenever 0< t a < . Conversely, if for every >0 ,
2 2 2
there exists >0 such that r(t) b < whenever 0< t a < , then f (t) b ,g(t) b ,h(t) b
1 2 2 2 1 2 3
<
2 2 2 3
[ f (t) b ] +[g(t) b ] +[h(t) b ] <
1 2 3
[ f (t) b ] +[g(t) b ] +[h(t) b ] <
1
whenever 0< t a < . But each term on the left side of this
2
inequality is positive so [ f (t) b ] <
, [g(t) b ] <
2
2
2
and [h(t) b ] <
3
2
2
whenever 0< t a < , or
1 2
taking the square root of both sides in each of the above we have f (t) b < , g(t) b < and h(t) b < whenever 0< t a < . And by definition of limits of real valued functions we have
3
12
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 t a 2 t a 3 t a t a t a t a
, so
lim r(t)= b ,b ,b
t a
1 2 3
=b .
13