Chapter 3
Practice Test
Instructions.
Show your work and/or explain your answers. (Note: Concepts from
''DIFF GEOM'' sections included only in the last problem).
- Find the equation of the tangent plane to the level surface
at the point ( 2,2,1) .
Solution: Since x-y+z2 = 1, we let U( x,y,z) = x-y+z2. Thus,
The normal to the surface is n =
á 1,-1,2
ñ
and the equation is
1( x-2) -1( y-2) +2( z-1) = 0 |
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which results in z = -x/2 + y/2 +1.
- Find the level surface representation of the parametric surface
r( u,v) =
á vsin( u) ,v2,vcos( u)
ñ |
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Solution: Clearly, x = vsin( u) , y = v2, and z = vcos( u) . We can eliminate u by noticing that
x2+z2 = v2sin2( u) +v2cos2( u) = v2 |
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Since y = v2, we eliminate v by noticing that x2+z2 = y.
- Find the level surface representation of the parametric surface
r( u,v) =
á eucosh( v),eusinh( v) ,e-u
ñ |
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Solution: Clearly, x = eucosh( v) , y = eusinh( v) , and z = e-u. To eliminate v we notice that
x2-y2 = e2ucosh2( v) -e2usinh2( v) = e2u[ cosh2( v) -sinh2( v) ] = e2u |
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To eliminate u, we notice that z2 = e-2u, so that
( x2-y2) z2 = e2ue-2u = 1 |
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so that the surface is given by x2z2-y2z2 = 1.
- Find the parametric equation of the tangent plane to the parametric
surface
r( u,v) =
á vsin( u) ,v2,vcos( u)
ñ |
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at ( u,v) = ( p/4,1) .
Solution: ru =
á vcos( u),0,-vsin( u)
ñ and rv =
ásin( u) ,2v,cos( u)
ñ . Thus, we have
ru( p/4,1) = |
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1
Ö2
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,0, |
-1
Ö2
| ![](RANGLE.GIF) |
and rv( p/4,1) = | ![](LANGLE.GIF) |
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1
Ö2
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,2, |
1
Ö2
| ![](RANGLE.GIF) |
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Since r( p/4,1) =
á 1/Ö2,1,1/Ö2
ñ , the parameterization of the tangent plane is
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r( p/4,1) +ru( p/4,1) du+rv( p/4,1) dv |
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1
Ö2
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,1, |
-1
Ö2
| ![](RANGLE.GIF) |
+ |
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1
Ö2
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,0, |
-1
Ö2
| ![](RANGLE.GIF) |
du+ |
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1
Ö2
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,2, |
1
Ö2
| ![](RANGLE.GIF) |
dv |
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1
Ö2
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+ |
du
Ö2
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+ |
dv
Ö2
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,1+2dv, |
-1
Ö2
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- |
du
Ö2
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+ |
dv
Ö2
| ![](RANGLE.GIF) |
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- Find the image of the unit square under the coordinate transformation
Solution: Since x = u2-v2 and y = 2uv, we have
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1: x = 1-v2,y = 2vÞ x = 1- |
y2
4
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1: x = u2-1,y = 2uÞ x = |
y2
4
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-1 |
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Putting them all together yields the following region:
- Find the matrix of rotation through an angle of q = 45°.
Then use this to rotate the line v = u+1 through an angle of 45°.
Solution: The rotation matrix is
Rp/4 = |
é ê ê ê
ê ê ë
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ù ú ú ú
ú ú û
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= |
é ê ê ê
ê ê ë
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ù ú ú ú
ú ú û
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Thus, rotation of the vector
á u,v
ñ results in
T( u,v) = |
é ê ê ê
ê ê ë
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ù ú ú ú
ú ú û
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é ê
ë
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ù ú
û
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= |
é ê ê ê
ê ê ë
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ù ú ú ú
ú ú û
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That is, x = Ö2u/2 -
Ö2v/2 and y = Ö2u + Ö2v/2. Since v = u+1, we get
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1
2
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Ö2u- |
1
2
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Ö2( u+1) = - |
Ö2
2
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y = |
1
2
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Ö2u+ |
1
2
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Ö2( u+1) = Ö2u+ |
1
2
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Ö2 |
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Thus, v = u+1 is mapped to the line x = -Ö2/2.
- Convert the following into polar coordinates and solve for r:
Solution: Since x = rcos( q) and y = rsin( q) , we have
rsin( q) = 3rcos( q) +1, r = |
1
sinq-3cosq
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- Convert the following into polar coordinates and solve for r :
Solution: r2 = rcos( q) +rsin( q) implies that
- Sketch the graph of r = 4p-q in polar coordinates when q is in [ 0,4p] . Then find and sketch the tangent
vector to the curve when q = p
Solution: To sketch the curve, we use a table of values of r
versus q :
We plot the points in polar coordinates to obtain the curve.
The Jacobian in polar coordinates yields the tangent vector:
Since r( p) = 3p and r' (q) = -1, we have
v = |
é ê
ë
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ù ú
û
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é ê
ë
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ù ú
û
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= |
é ê
ë
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ù ú
û
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That is, the tangent vector is v =
á 1,-3p
ñ , which is shown in the plot above.
- Find the Jacobian determinant and area differential of the coordinate
transformation
Solution: Since x = u-v and y = u2+v2, the Jacobian is
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¶x
¶u
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¶y
¶v
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- |
¶x
¶v
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¶y
¶u
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Thus, the area differential is dA = | 2u+2v| dudv.
- What is the unit surface normal for the surface x2+y2+z2 = 2x? What is the unit surface normal for the surface in cylindrical coordinates?
Solution: Writing x2+y2+z2-2x = 0 leads us to define U = x2+y2+z2-2x. Thus, ÑU =
á2x-2,2y,2z
ñ and
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( 2x-2) 2+4y2+4z2
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| ![](rootin5.gif)
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4x2-8x+4+4y2+4z2
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4( x2-2x+y2+z2) +4
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4( 0) +4
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Thus, the unit surface normal is
n = |
ÑU
|| ÑU||
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=
áx-1,y,z
ñ |
| (1) |
In cylindrical coordinates, U = r2+z2-2rcos( q)
and the gradient is given by
where er =
á cos( q) ,sin(q) ,0
ñ and eq =
á-sin( q) ,cos( q) ,0
ñ .
Thus, in cylindrical coordinates, we have
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( 2r-2cos( q) )
á cos( q) ,sin( q) ,0
ñ + |
1
r
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( 2rsin( q) )
á -sin(q) ,cos( q) ,0
ñ +2z
á0,0,1
ñ |
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á 2rcos( q) -2cos2( q) ,2rsin( q) -2cos( q) sin( q) ,0
ñ +
á -2sin2( q) ,2sin( q) cos( q),0
ñ +
á 0,0,2z
ñ |
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á 2rcos( q) -2( cos2( q) +sin2( q) ) ,2rsin( q) -2cos( q) sin( q) +2cos( q) sin( q) ,2z
ñ |
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á 2rcos( q) -2,2rsin( q),2z
ñ |
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That is, ÑU =
á 2rcos( q) -2,2rsin( q) ,2z
ñ , and as shown above, ||ÑU|| = 2, so that
n = |
ÑU
|| ÑU||
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=
á rcos( q) -1,rsin( q) ,z
ñ |
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Notice that you could have obtained the same result by simply applying
cylindrical coordinates to (1).
- Find the pullback of the surface x2+y2+z2 = 2x into
spherical coordinates, and then use the result to construct a
parameterization of the surface.
Solution: In spherical coordinates, r2 = x2+y2+z2
, r = rsin( f) and x = rcos( q) = rsin( f) cos( q) . Thus, the
surface becomes
r2 = rsin( f) cos( q) or r = sin( f) cos( q) for r ¹ 0 |
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Since also y = rsin( f) sin( q) and
z = rcos( f) , the parameterization of the surface is
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á rsin( f) cos( q) ,rsin( f) sin( q) ,rcos( f)
ñ |
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á sin2( f) cos2( q),sin2( f) cos( q) sin( q) ,sin( f) cos( f) cos(q)
ñ |
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- Use the fundamental form of the plane in polar coordinates to find
the length of the polar curve r = e-q/4 , q in [0,2p] .
Solution: The fundamental form of the plane in polar coordinates is
given by
which leads to an arclength formula of
L = |
ó õ
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b
a
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ds
dq
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dq = |
ó õ
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2p
0
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![](roottall.gif) |
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dq |
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Substituting r = e-q/4 and r¢ = -e-q/4/4 leads to
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ó õ
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2p
0
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![](roottall.gif) |
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dq |
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ó õ
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2p
0
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e-q/4 |
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dq |
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e-q/4
-1/4
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ê ê
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2p
0
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- Find the fundamental form of the surface r( u,v) =
á vsin( u) ,vcos( u) ,v
ñ .
Then use it to compute the arclength of
v = sin |
æ è
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u
Ö2
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ö ø
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, u in |
é ë
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0, |
p
4
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ù û
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Solution: ru =
á vcos( u),-vsin( u) ,0
ñ and rv =
ásin( u) ,cos( u) ,1
ñ . Thus,
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ru·ru = v2cos2( u)+v2sin2( u) = v2 |
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ru·rv = vcos( u) sin( u) -vsin( u) cos( u) = 0 |
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rv·rv = sin2( u) +cos2( u) +1 = 2 |
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Thus, the fundamental form is
( ds) 2 = v2( du) 2+2( dv) 2 |
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and correspondingly, the arclength integral is
L = |
ó õ
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p/4
0
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![](roottall.gif)
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du |
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Since dv/du = cos( u/Ö2) /Ö2, the arclength is
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ó õ
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p/4
0
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sin2 |
æ è
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u
Ö2
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ö ø
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+ |
2
( Ö2) 2
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cos2 |
æ è
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u
Ö2
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ö ø
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du |
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ó õ
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p/4
0
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![](roottall.gif)
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sin2 |
æ è
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u
Ö2
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ö ø
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+cos2 |
æ è
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u
Ö2
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ö ø
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du |
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- DIFF GEOM: For the right circular cone, r( r,q) =
á rcos( q) ,rsin( q) ,r
ñ , do the following:
- Show that curves of the form q = k for k constant are
geodesics on the cone. What are these curves?
- Find the fundamental form of the cone and calculate the shortest
distance between the points with coordinates ( r,q) = ( 1,p) and ( r,q) = ( 3,p) .
- What are the principal curvatures of the surface?
- What are the mean and Gaussian curvatures of the surface? Do you
obtain the same Gaussian curvature if you use the theorem Egregium?
Solution: (a) Curves of the form q = k are
parameterized by
r( t) =
á r( t) cos(k), r( t) sin( k),
r( t)
ñ |
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for some unknown function r( t) , and the acceleration is r'' ( t) =
á r'' ( t) cos(k) ,
r'' ( t) sin( k) , r'' ( t)
ñ . However, rr =
á cos(q) ,sin(q) ,1
ñ and rq =
á -rsin(q), rcos(q) ,0
ñ . Clearly, we have
r'' · rr = r'' cos2( k) +r''
sin2( k) +r'' = 2r''
and r'' · rq = 0 |
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If r'' = 0, then r(t) = pt + c for constants
p and c. Thus, q = k constant and r = mt+b for m, b constant
parameterizes a geodesic, which implies that q = k for k constant is
a geodesic. These are straight lines on the cone that pass through the
origin.
(b) Since rr =
á cos( q) ,sin( q) ,1
ñ and rq =
á -rsin( q) ,rcos( q),0
ñ , we have g11 = rr·rr = 2 and
g22 = rq·rq = r2 with g12 = 0
(the parameterization is orthogonal). Thus, the fundamental form is
However, on q = k, the differential dq = 0, so that we have
Since r ranges from 1 to 3, this implies that
(c) Since rr×rq =
á -rcosq,-rsinq,r
ñ , the surface normal is
n = |
rr×rq
|| rr×rq||
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= |
1
Ö2
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á-cos( q) ,-sin( q) ,1
ñ |
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Since rrr = 0, rrq
= á -sin(q), cos(q), 0 ñ and rqq =
á -r cos(q) , -r sin(q) ,0
ñ, the normal curvature of the
cone is given by
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rrr·n
|| rr|| 2
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cos2( x) + |
rrq·n
|| rr|| || rq||
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sin( 2x) + |
rqq·n
|| rq|| 2
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sin2( x) |
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á -sin( q) ,cos( q) ,0
ñ ·
á -cos( q),-sin( q) ,1
ñ
rÖ2
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sin( 2x) + |
á -rcos( q) ,-rsin(q) ,0
ñ ·
á -cos( q) ,-sin( q) ,1
ñ
r2Ö2
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sin2( x) |
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Thus, the principal curvatures occur when x = 0 and when x = p/2,
and are correspondingly
(d) The mean curvature is
and the Gaussian curvature is 0 since k1
= 0. Using the theorem
Egregium, we obtain
K = |
-1
2Ög
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é ë
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¶
¶q
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æ è
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g11,q
Ög
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ö ø
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+ |
¶
¶r
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æ è
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g22,r
Ög
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ö ø
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ù û
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where g11 = 2, g22 = r2, and
g = g11g22 = 2 r2. Thus,
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-1
2rÖ2
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é ë
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¶
¶q
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æ è
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0
Ög
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ö ø
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+ |
¶
¶r
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æ è
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2r
rÖ2
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ö ø
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ù û
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-1
2rÖ2
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é ë
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¶
¶q
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(0) + |
¶
¶r
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æ è
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2
Ö2
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ö ø
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ù û
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Thus, the cone is a surface that can be ''made with a piece of paper''
without stretching or tearing the paper.