Homework: 9/21 13.1 16.Draw the project ion of the curve on the three coordinate planes. Use these projecti ons to help sketch the curve. The three coordinate planes are the xy-, yz-, and xz-coordinate planes. We can look at each othem !y considerin" r#t$ % &t, t, t 2 ' two components at a time. ()ttin" them all to"ether, we see that the x- and y- coordinates will always !e on the plane y % x, whereas the z-coordinate will !e para!olic: 18.Find a vector equation and parametric equations for the line segment that joins P to Q. P(-! "! -"#! Q(-$! %! #
16. Draw the projection of the curve on the three coordinate planes. Use these projections to
help sketch the curve.
The three coordinate planes are the xy-, yz-, and xz-coordinate planes. We can look at each othem !y considerin" r#t$ % &t, t, t2' two components at a time.
()ttin" them all to"ether, we see that the x- and y- coordinates will always !e on the plane y % x,whereas the z-coordinate will !e para!olic:
18. Find a vector equation and parametric equations for the line segment that joins P to Q.
*n order to deine a line, we re+)ire its direction and a point on the line #or two points, !)t !asically the same$. We know that o)r line m)st "o thro)"h ( and , so let )s say that the pointon the line is (. or the direction, we look or the ector that "oes rom ( to namely (.
( % 0- #-1$, 3 2, 1 #-2$4 % 0-2, , 4. Thereore, we set )p a parameter 5t6 to help )s
deine this as direction: 0-1, 2, -24 7 t80-2, , 4 % 0-1 2t, 2 7 t, -2 7 t4 % r#t$. That is, westart at 0-1, 2, -24 and then span in the direction #or opposite direction$ o 0-2, , 4.
Howeer, the +)estion asked or the line segment , and so we m)st !o)nd o)r 5t6. ince we startat 0-1, 2, -24, and the se"ment has to reach and stop at 0-, 3, 14, !y o)r constr)ction we cantest/see/know that we sho)ld !o)nd t as ollows: ; t ; 1.
The parameterization, takin" each component one at a time, is:x#t$ % 1 2t y#t$ % 2 7 t z#t$ % -2 7 t
<=
>et?s start !y sayin" we?re "oin" to draw a line !etween points ( and , and it will !erepresented !y r#t$. @lso, let?s esta!lish that we?ll )se a aria!le 5t6 as a parameter to help )sdeine this line, and or simplicity, restrict it !etween and 1. Aeca)se we want r#t$ to e+)al r when t % , #at the start$ and r1 when t % 1, #at the end$ we constr)ct the "eneral e+)ation r#t$ %#1 t$r7 tr1. Then we set ( % r and % r1, and the res)lt is r#t$ % 0-1 -2t, 2 7 t, -2 7 t4,where ; t ; 1.
The parameterization, takin" each component at a time, is:x#t$ % 1 2t y#t$ % 2 7 t z#t$ % -2 7 t
28. &how that the curve with parametric equations ' sin t! ) cos t! * sin " t is the curve of
the intersection of the surfaces * '" and '" + )" . Use this fact to help sketch the curve.
irst let )s )nderstand that the s)races z % x2 and x2 7 y2 % 1 hae some intersection #whichmi"ht !e empty$. @ll we hae to do is show that the parametric e+)ations proided will containall those points o intersection. o or eery point o intersection, the z-coordinate will e+)al theal)e o the x-coordinate s+)ared, and the x-coordinate s+)ared s)mmed with the y-coordinates+)ared will e+)al 1. Bo o)r parametric e+)ations drien !y o)r parameter 5t6 satisy thisC
z % x2 z#t$ s. #x#t$$2 sin2 t % #sin t$2 Des, this is satisied.
x2 7 y2 % 1 #x#t$$2 7 #y#t$$2 s. 1 sin2 t 7 cos2 t % 1 Des, this is satisied.
Th)s we concl)de that the c)re is in act the intersection o the two s)races.
The simplest way to sketch this is to draw !oth s)races, and then ima"ine where they intersect:
#Eredit: math.)cla.ed) =onald Fiech$
#The intersection$
30. ,t what point does the heli' r (t# sin t! cos t! t intersect the sphere '" + )" + * " %/
We can )nderstand the ector )nction r#t$?s components separately: x % sin#t$, y % cos#t$, z % t.
Aeca)se where the c)re intersects the sphere their coordinates will !e e+)al, we can expect thati we s)!stit)te o)r x-, y-, and z-al)es rom r#t$ into the e+)ation or the sphere, the e+)ationsho)ld still hold:#sin#t$$2 7 #cos#t$$2 7 #t$2 % 3 1 7 t2 % 3
Then we know that r#t$ will intersect the sphere when t % -2, and t % 2. o to ind these points,we s)!stit)te these al)es into o)r ori"inal e+)ation:r#2$ % &sin#2$, cos#2$, 2'r#-2$ % &sin#-2$, cos#-2$, -2'#IA: the )nits or 5t6, or lack thereo, implies radians$
#40. and 42.$ Find a vector function that represents the curve of intersection of the two surfaces.
#IA: Do) may )se any parameterization that yo) wish these are the ones * elt were simplest.$
40. 0he c)linder '" + )" 1 and the surface * ').
<)r ector )nction will !e o the orm r#t$, so that means we?ll !e )sin" a parameter 5t6 as o)ronly aria!le. That is, we m)st express x, y, and z in terms o t. Aeca)se we know an ele"ant parameterization or x2 7 y2 % E in "eneral, let )s start rom there: x#t$ % 2cos#t$, y#t$ % 2sin#t$.
This satisies the e+)ation x2 7 y2 % G #eriy or yo)rsel$, and will reach all al)es o x and y iwe !o)nd ; t ; 2J. Then the parameterization or z ollows directly rom the e+)ation: z#t$ %x#t$8y#t$ % Gcos#t$sin#t$ % 2sin#2t$.
Th)s o)r ector )nction can !e written as r#t$ % &2cos#t$, 2sin#t$, 2sin#2t$', ; t ; 2J.
42. 0he para2oloid * 1'" + )" and the para2olic c)linder ) '".
<)r ector )nction will !e o the orm r#t$, so that means we?ll !e )sin" a parameter 5t6 as o)r
only aria!le. That is, we m)st express x, y, and z in terms o t. The second e+)ation seems tooer a startin" point, since that shows a simple relationship: x#t$ % t, y#t$ % t2. Then, we soleor z#t$ !y )sin" x#t$ and y#t$: z#t$ % Gt2 7 tG.
Th)s o)r ector )nction can !e written as r#t$ % &t, t2, Gt2 7 tG'.
#IA: we do not hae to !o)nd o)r parameter in this pro!lem, since a !o)nd was not implied inthe pro!lem or !y any o o)r )nctions$
47. 3f two o2jects travel through space along two different curves! it4s often important to knowwhether the) will collide. 0he curves might intersect! 2ut we need to know whether the o2jects
are in the same position at the same time. &uppose the trajectories of two particles are given 2)
the vector functions
r (t# t " ! 5t 6 "! t " r "(t# 1t 6 $! t " ! %t 6 7
irst, let )s )nderstand what?s happenin". Two particles are "oin" to ly thro)"h spaceindependently o each other. * we try to trace o)t their paths, we mi"ht ind that their pathsintersect. Howeer, this DOES NOT mean that they eer collid. @ter all, i yo)?re sittin" in achair in FKB 23, and someone else sits in the same chair 2 ho)rs later, yo) two do notnecessarily collide at that point. @ss)min" yo) let, yo) two happened to occ)py the same space
!)t at dierent times. o, when we look or points o intersection or these two ector )nctions,we m)st check to see whether the )nctions attain the same al)e at the same value of t.
o let?s analyze intersection points. When r1#t$ and r2#t$ intersect, their x-, y-, and z-coordinateswill !e e+)al. Th)s, we can e+)ate them to ind the al)es o t where this happens:x-coordinate: t2 % Gt t2 Gt 7 % #t $#t 1$ % t % 1 or
@t this point, we can check the other two coordinates and see which al)es o t are common toall three. Howeer, we may also pl)" in o)r o)nd al)es o t and see whether these work. *they do not, then we can concl)de that there are no al)es o t where the two intersect, since
there are only these two places where the x-coordinates are the same, at the same time t.
r1#1$ % $2, L#1$ 12, #1$2' % &1, -3, 1'r2#1$ % &G#1$ , #1$2, 3#1$ M' % &1, 1, -1'@ltho)"h they hae the same x-coordinate at t % 1, their other coordinates do not match )p.Th)s, the c)res do not intersect at t % 1. IextN
r1#$ % &#$2, L#$ 12, #$2' % &9, 9, 9'r2#$ % &G#$ , #$2, 3#$ M' % &9, 9, 9'TadaO @t the point t % , the two c)res hae the same coordinates, so they occ)py the same
space at the same time, and thereore they intersect #or are tan"ent$. Th)s, we can concl)de thatthe particles do collide.
13.2
5. r (t# sin(t#i + "cos(t# j ! t :;1
5.a. &ketch the plane curve with the given vector equations.
We can test a ew points to "et an idea o what it looks like, !)t more likely we can reco"nize thee+)ation as a 5sort o6 circle. @lternatiely, we can directly see that it?s an ellipse. Howeer yo)"et there, here?s the sketch:
Ae care)l that yo) reco"nize the direction that the c)re is tracin". Aeca)se this is a parametricc)re, there exists a distinct direction !y which it traces its path, and this c)re is dierent romthe i")re we "et rom tracin" the c)re in the opposite direction. We test or direction !y tryin"a ew points o)t. @lso, it?s worth mentionin" that we achiee the )ll c)re on ; t ; 2J
5.c. &ketch the position vector r (t# and the tangent vector r 4(t# for the given value of t.
To sketch the position ector, we irst hae to eal)ate it:r#J/G$ % sin#J/G$i 7 2cos#J/G$ j #1/P2$i 7 P2 j
Ae care)l well lookin" or the tan"ent ector. They are not askin" )s to draw the position ector r?#t$. Th)s, we hae to place it, correctly, tan"ent to the c)re on the sketch.r?#J/G$ % cos#J/G$i 7 #-2$sin#J/G$ j #1/P2$i 7 -P2 j
24. Find the parametric equations for the tangent line to the curve with the given parametric
equations at the specified point.
x % et, y % tet, z % tetQ2 #1, , $
irst, we m)st reco"nize that they are not askin" )s to ind the parametric e+)ations or the)nction that will "ie )s the #direction o the$ tan"ent line at any point. They are askin" )s toind the parametric e+)ations that will descri!e the moement o a partic)lar tan"ent line, at a partic)lar point. o, they?re askin" )s to "ie the e+)ation o a line #that is, the tan"ent line$.
We can s)iciently descri!e a line with a point and a direction. Aeore we do either, let?s irstconsolidate the a!oe parametric e+)ations into a sin"le ector e+)ation:r#t$ % &et, tet, tetQ2'
Iow, we sho)ld o!tain the 5point6 that will descri!e the line. We are act)ally "ien a point, #1,, $. A)t the pro!lem with this is that i we?re "oin" to "o thro)"h the rest o this pro!lem witha sin"le )nction with common aria!le t, rather than three distinct )nctions, we hae todescri!e x % 1, y % , z % , in terms o t. >ookin" at the irst component, we see that this is onlyachiea!le when t % . o, we?re interested in the )nction at the point #1, , $, or when t % same thin".
<n to the direction. We can o!tain a )nction which will "ie )s the direction o the tan"ent lineat any point !y simply takin" the deriatie o o)r ori"inal:r#t$ % &et, tet, tetQ2' r?#t$ % &et, et 7 tet, etQ2 7 etQ2#tQ2$?t'
% &et, et 7 tet, etQ2 7 2t2etQ2'
o to ind the 5direction6 that will descri!e the line we?re interested in, we )se the point t % :r?#$ % &e, e 7 8e, eQ2 7 2#2$eQ2' &1, 1, 1'
#since this will "ie )s the tan"ent ector at the point t % $
Aeca)se we hae o)r point and o)r direction, we can now compose o)r parametric )nction #withthe idea that it will start at the point, and then span o)t in the direction$:v#t$ % r#$ 7 t8r?#$&1, , ' 7 t8&1, 1, 1'
<r, in parametric terms:x#t$ % 1 7 t, y#t$ % t, z#t$ % t
27. Find a vector equation for the tangent line to the curve of intersection of the c)linders
'" + )" "% and )" + * " "9 at the point ($! 1! "#.
This t)rns into a ery simple pro!lem with m)ltiple steps i we !reak down the +)estion:1$ ind a vector equation or the
2$ tangent line to the c)re o$ intersection o theG$ cylinders x2 7 y2 % 23 and y2 7 z2 % 2 at the point #, G, 2$
o we want to ind the ector e+)ation or the tan"ent line. A)t the tan"ent line is tan"ent to thec)re o the intersection. o we can?t ind 1$, or 2$, )ntil we ind $. imple irst +)estion then:Bescri!e the intersection o the cylinders x2 7 y2 % 23 and y2 7 z2 % 2 at the point #, G, 2$.
Do) may do this part howeer yo) want, !)t it?s )s)ally a "ood idea to represent x, y, and z in areasona!ly clean way. <ne s)ch way is x#t$ % 3cos#t$, y#t$ % 3sin#t$ #this satisies the irst
cylinder$. Ay the e+)ation o the second cylinder, we sole or z#t$ % P#2 23sin2
#t$$. Then o)r consolidated ector e+)ation or this intersection o two cylinders is:r#t$ % &3cos#t$, 3sin#t$, P#2 23sin2#t$$'
Iext step is to ind the tan"ent line, and we descri!e a line with a point and a direction. We haethe point tho)"h, #, G, 2$, so all we need is the direction. The )nction r?#t$ will "ie )s thedirection #and also ma"nit)de tho)"h that?s not important to )s$ o the tan"ent ectors !y nat)re
o dierentiation, so let?s ind that:r?#t$ % &-3sin#t$, 3cos#t$, -23sin#t$cos#t$/P#2 23sin2#t$$'
@h, !)t yo) see that o)r dierentiated )nction is in terms o t. Det, the point #, G, 2$ is in termso x-, y-, and z-coordinates. o, we hae to ind the appropriate t to "et an appropriate r?#t$:
* co)ld sole or t, and then * wo)ldn?t hae to consider the e+)ality rom the y-coordinate, !)tthis way is act)ally a little easier:r?#t$ % &-3sin#t$, 3cos#t$, -23sin#t$cos#t$/P#2 23sin2#t$$' % &-3#G/3$, 3#/3$, -23#G/3$#/3$/P#2 23#G/3$2$'
% &-G, , -12/P#2 1M$
% &-G, , -M'
Aeca)se we hae o)r point and o)r direction, we can now compose o)r ector )nction #with theidea that it will start at the point, and then span o)t in the direction$:v#t$ % r#$ 7 t8r?#$ % &, G, 2' 7 t8&-G, , -M'
% & Gt, G 7 t, 2 Mt'
32.a. Find the point of intersection of the tangent lines to the curve r (t# sin(:t#! "sin(:t#!
cos(:t# at the points where t 9 and t 9.%.
This +)estion can !e !roken down into irst 5indin"6 each o the tan"ent lines and then seein"where they intersect. o let?s ind the tan"ent lines. To descri!e a line, we need a point and itsdirection, and the dierentiated )nction r?#t$ will "ie )s a direction ector or the tan"ent line.The two tan"ent lines we?re analyzin" are deried rom the point t % , and t % .3. o, assem!lyline style:r#$ % &sin#J#$$, 2sin#J#$$, cos#J#$$' % &, , 1'r?#$ % &Jcos#J#$$, 2Jcos#J#$$, -Jsin#J#$$' % &J, 2J, 'r#.3$ % &sin#J#.3$$, 2sin#J#.3$$, cos#J#.3$$' % &1, 2, '
Ae ery care)l. Rach parametric )nction v#t$ and u#s$ operates completely apart rom the other,so * hae parameterized one with 5t6, and the other with 5s6 to make this distinction clearer. To
ind where they intersect, we hae to identiy all coordinates that !oth )nctions pass thro)"h.Ay e+)atin" each component:Jt % 1, 2Jt % 2, 1 % -Js t % 1/J, t % 1/J, -1/J % s
olin" or t and s is a little )nnecessary, !)t it does help to el)cidate the point that v#t$ and u#s$ !oth, in their own time #v#t$ when t % 1/J, u#s$ when s % -1/J$, "o thro)"h the point #1, 2, 1$.
We can !r)te orce this proo !y splittin" r into components #o which there may or may not !e, !)t since cross prod)ct or )s has only !een deined or -dimensional spaces, this is okay$and then act)ally indin" the deriaties and the cross prod)cts. Howeer, there?s a aster way.
>et?s start with the let side o the e+)ation we want to proe. Ay theorem part 3, we know thatd/dtu#t$ V v#t$U % u?#t$ V v#t$ 7 u#t$ V v?#t$. o, i we let u#t$ % r#t$, and v#t$ % r?#t$, we "et:r?#t$ V r?#t$ 7 r#t$ V r??#t$
Howeer, we also know that i yo) take the cross prod)ct o any ector with itsel, the res)lt is 0 #the ector, not the n)m!er$. o then we "et:r#t$ V r??#t$, which is the ri"ht side o the e+)ation we want to proe.
ast track:d/dtr#t$ V r?#t$Ur?#t$ V r?#t$ 7 r#t$ V r??#t$ Ay theorem part 3Ur#t$ V r??#t$ u#t$ V u#t$ % 0 or any ector u#t$U
53. 3f r (t# ? 0 ! show that d;dt@r (t#@ (;@r (t#@# r (t# A r 4(t#.
<BintC @r (t#@" r (t# A r (t#>
Waste not want not:r#t$2 % r#t$ X r#t$ #1$
o, it doesn?t seem that the let side o #1$ is similar to either side o the e+)ality o what we
want to proe, !)t nor is the ri"ht side. A)t i we had to choose, * mean, it seems like it?d take aew strai"htorward manip)lations to "et r#t$2 to look like d/dtr#t$ namely, s+)are root, andthen dierentiation:r#t$2 % r#t$ X r#t$r#t$ % P#r#t$ X r#t$$d/dtr#t$ % d/dtP#r#t$ X r#t$$#IA: We can take the s+)are root o !oth sides !eca)se the dot prod)ct o any ector with itselis nonne"atie, and the let hand side is in 5a!sol)te al)es6.$
o the let side is all taken care o. Iow we hae to "et d/dtP#r#t$ X r#t$$ to t)rn into the
expression #1/r#t$$ r#t$ X r?#t$ somehowY. Well, let?s )se o)r r)les o dierentiation #o dot prod)cts and chain r)le$ to work it o)t:d/dtP#r#t$ X r#t$$ % #1/2$ #r#t$ X r#t$$-1/2 d/dt#r#t$ X r#t$$
o this is ery promisin". *t seems the only part that doesn?t match )p with the what we want isthat #r#t$ X r#t$$-1/2 doesn?t seem to e+)al #1/r#t$$Y or does itC =emem!er that r#t$2 % r#t$ X r#t$,so then #r#t$ X r#t$$1/2 sho)ld e+)al r#t$, and #r#t$ X r#t$$-1/2 % r#t$-1 % 1/r#t$.
Hooray we?re doneN
ast track:
r#t$2 % r#t$ X r#t$ #Zien$r#t$ % P#r#t$ X r#t$$ #+)are root [)stiied !y nonne"atiity o sel dot prod)ct$d/dtr#t$ % d/dtP#r#t$ X r#t$$OOOd/dtP#r#t$ X r#t$$ #Eonsider [)st the ri"ht hand side$#1/2$ #r#t$ X r#t$$-1/2 d/dt#r#t$ X r#t$$ #Ehain r)le$#r#t$ X r#t$$-1/2 r#t$ X r?#t$ #Ehain r)le$
#1/r#t$$ r#t$ X r?#t$ #Ay the "ien$OOOd/dtr#t$ % #1/r#t$$ r#t$ X r?#t$ #)!stit)tion$
<=
r#t$2 % r#t$ X r#t$ #Zien$d/dt#r#t$2$ % d/dt#r#t$ X r#t$$2r#t$ d/dtr#t$ % 2#r#t$ X r?#t$$ #Ehain r)le$r#t$d/dtr#t$ % r#t$ X r?#t$d/dtr#t$ % #1/r#t$$ #r#t$Xr?#t$$
This way is m)ch aster, !)t it takes a little int)ition to see how it?ll work o)t.
#Bisclaimer: this sol)tion set is s)
