Math 21a
Homework 14 Solutions
Spring, 2014
Due Monday, March 10th (MWF) or Tuesday, March 11th (TTh)
1.
(a) Consider the surface given by sin(x − y) − x − y + z = 0. Find a parametrization of the part of the surface that
has |x| ≤ 3 and |y| ≤ 3. Don’t forget to give bounds for your parameters.
Solution: We can rewrite the equation as z = x + y − sin(x − y). We let u = x, v = y to be the parameters.
Then r = hu, v, u + v − sin(u − v)i. The bounds are |u| ≤ 3, |v| ≤ 3.
(b) Look at the parametrization that you found in part (a). It is of the form r(u, v) = hu, v, f (u, v)i. Explain why it’s
not possible to find a parametrization of the surface x2 − y + z 2 = 1 that’s of the form r(u, v) = hu, v, f (u, v)i.
p
Solution: If we solve for z in the equation x2 − y + z 2 = 1, we get z = ± 1 + y − x2 . This means, given x and
y, there are two values of z for which the point (x, y, z) lies on the surface.
(c) Find a parametrization of x2 − y + z 2 = 1 that’s of the form r(u, v) = hu, f (u, v), vi. Don’t forget to give bounds
for your parameters.
Solution: This time, we solve for y
instead of z to obtain
the equation y = x2 + z 2 − 1. Letting x = u and z = v
2
2
gives us a parametrization r(u, v) = u, u + v − 1, v , where u and v are any real numbers.
z2
= 1. Explain why it’s not possible to find a parametrization of this surface that’s
4
of the form r(u, v) = hu, v, f (u, v)i, or of the form r(u, v) = hu, f (u, v), vi, or of the form r(u, v) = hf (u, v), u, vi.
(d) Consider the surface x2 − y 2 +
Solution:
We cannot parametrize the surface with x and y because if (x, y, z) lies on the surface, so does
(x, y, −z). By the same reason, we cannot parametrize the surface with x and z; or with y and z.
(e) Show (by substitution!) that every point that is swept out by the parametrization
Dp
E
p
r(u, v) =
1 + v 2 cos u, v, 2 1 + v 2 sin u
z2
= 1. (You might also use the ParametricSurfaces.nb applet below to verify that
4
this parametrization is in fact a good parametrization of the whole surface.)
lies on the surface x2 − y 2 +
Solution:
To verify that r(u, v) lies on the surface, we plug in (substitute) and compute:
2
2
p
z2
1 p
x2 − y 2 +
1 + v 2 cos u − v 2 +
=
2 1 + v 2 sin u
4
4
1
2
2
2
= (1 + v ) cos u − v + · 4(1 + v 2 ) sin2 u
4
= (1 + v 2 )(cos2 u + sin2 u) − v 2
= (1 + v 2 ) − v 2 = 1.
Thus every point on this parametrization is on the given surface.
√
(f) Show that every point that is swept out by the parametrization r(u, v) = hv cos u, v 2 − 1, 2v sin ui lies on the
2
√
z
surface x2 − y 2 +
= 1. Explain why, in spite of what you’ve just shown, r(u, v) = hv cos u, v 2 − 1, 2v sin ui is
4
z2
not a good parametrization of the surface x2 − y 2 +
= 1. (If you’re having trouble figuring out why, try using
4
the ParametricSurfaces.nb applet described below.)
Solution:
We substitute and compute as in the previous problem
p
2 1
z2
x2 − y 2 +
= (v cos u)2 −
v 2 − 1 + (2v sin u)2
4
4
1
= v 2 cos2 u − (v 2 − 1) + · 4v 2 sin2 u
4
= v 2 (cos2 u + sin2 u) − (v 2 − 1)
= v 2 − v 2 + 1 = 1.
Nevertheless, we are missing half the surface, for if (x, y, z) lies on the surface, so does (x, −y, z). But in the
parametrization of the y-coordinate, we take the positive square root and “miss” the point with negative square
root.
2. Identify and sketch the surface with the given vector equation.
(a) (Stewart 10.5 #4 ) r(u, v) = (2 sin u)i + (3 cos u)j + vk
Solution: The surface is a vertical elliptical cylinder. To see this, we note that the height z = v can be anything,
while x = 2 sin u and y = 3 cos u satisfy the equation x2 /4 + y 2 /9 = 1 (an ellipse in the xy-plane). The plot is
shown in Figure 1.
(b) (Stewart 10.5 #6 ) r(s, t) = hs sin 2t, s2 , s cos 2ti
Hint: What do the grid curves look like? You should be able to answer these questions without using Mathematica.
Solution: This is an elliptic paraboloid. When s is fixed, so is y = s2 . In this slice, (x, z) trace out a circle
centered at the origin of radius s. In fact, the equation of the paraboloid is y = s2 = x2 + z 2 . The plot is shown
in Figure 1.
Figure 1: The two surfaces from Problem 2: (a) an elliptical cylinder (left), and (b) an elliptic paraboloid (right)
3. (Stewart 10.5 #13-18 ) Match the equations and the graphs labeled I-VI that are given in the textbook. Give reasons
for your answers. Also determine which families of grid curves have u constant and which have v constant. You are
strongly encouraged to answer these questions without using Mathematica, since you may be asked to do such matching
on an exam.
Solution:
To qualitatively understand the shape of the surface, a useful idea is to consider the slices where one
coordinate is held constant.
13. If we fix z = v, then in the xy-plane, r(u) = hu cos v, u sin vi = uhcos v, sin vi is a line from the origin in direction
of hcos v, sin vi. As v changes, the direction of the line spins at constant rate. This matches up with Figure IV.
14. Since z = sin u, z is bounded (it ranges from −1 to 1). For a fixed u, in the xy-plane, r(v) = hu cos v, u sin vi
makes a circle of radius u (and we can think of this as depending on z = sin u). This matches up with Figure I.
15. If we fix v, then x = sin v is also fixed. In the yz-plane, (cos u sin 2v, sin u sin 2v) forms a circle of radius sin 2v.
This matches up with Figure II.
√
17. As x2/3 + y 2/3 + z 2/3 = 1, this looks like a “sphere” in terms of the cube roots ( 3 x = x1/3 , for example) of the
coordinates. This matches up with Figure III. By fixing v, we fix z, so the grid lines where v is constant are closed
curves lying in planes parallel to the xy-plane. This also suggests Figure III.
18. If we fix u, we are looking at the slice where z = u is fixed. This slice is a circle of radius 1−|u| as x2 +y 2 = (1−|u|)2 .
This matches up with Figure VI.
16. The equations are complicated, so perhaps, it is best to deduce that 16 corresponds to Figure V by process of
elimination. The curves where v is held constant are loops that go around the “tube”, while the curves where u
is held constant are curves that go down the length of the tube.
4. In this question, you are asked to plot parametric surfaces in R3 . You can do this manually in Mathematica (as in parts
(a) and (b)), or using the Mathematica applet you can download from the course homework page (see the instructions
before part (c)).
(a) (Stewart 10.5 #8 ) Use the Mathematica command
ParametricPlot3D[{u+v,u^2,v^2},{u,-1,1}, {v,-1,1}]
to plot the parametric surface given by r(u, v) = hu + v, u2 , v 2 i, −1 ≤ u ≤ 1, −1 ≤ v ≤ 1. Identify which of the
grid curves are the constant-u grid curves, and which are the constant-v grid curves.
Solution: Below the constant-u grid curves are drawn in black while the constant v curves are drawn in orange.
This choice is arbitrary as the parametrization is symmetric in u and v.
(b) (Stewart 10.5 #12 ) Modify the command above to plot the surface given by the parametric equation r(u, v) =
hu sin u cos v, u cos u cos v, u sin vi. Choose suitable bounds for u and v. Identify which of the grid curves are the
constant-u grid curves, and which are the constant-v grid curves.
Solution: Here we show the surface for 0 ≤ u ≤ 3π and 0 ≤ v ≤ 2π. When u is held constant, the (black) grid
curves are nearly circles centered at the origin, while if v is held constant, the (orange) grid curves are “ovals”
spiraling out (and slightly up or down).