Math 2850-005, Fall 06 — Solutions for exam 2
1. Find the equation of the sphere with center (1, 1, −5) that is tangent to the plane z = 5.
The (minimum) distance from the point to the plane is 10, and hence the equation of the sphere is
(x − 1)2 + (y − 1)2 + (z + 5)2 = 102 .
2. Let ~a = h2, 1,
(a) − 21 ~a; (b) |~a|; (c) a unit vector that has the same direction as ~a.
−3i. 1Compute
1
3
(a) − 2 ~a = −1, − 2 , 2
p
√
(b) |~a| = 22 + 12 + (−3)2 = 14√
√
√ (c) The unit vector is ~a/|~a| = 2/ 14, 1/ 14, −3/ 14
3. Let ~a = 3~ı + 2~ and ~b = ~ı − ~k. (a) Draw ~a and ~b at the origin. (b) Find their lengths and the angle
between them.
p
√
√
√
(b) |~a| = 32 + 22 = 13, |~b| = 12 + (−1)2 = 2, ~a · ~b = 3 · 1 + 2 · 0 + 0 · (−2) = 3, and finally
√
∠(~a, ~b) = arccos(~a · ~b/(|~a| · |~b|)) = arccos(3/ 26).
~
~
~
4. Let ~a = 3~ı − 2~ and
(a) √
comp~a (~b); (b)
pb = ~ı + 3~. Compute
√
√ proj~a (b); (c) orth~a (b).
2
2
2
2
~
~
First of all, |~a| = 3 + (−2) = 13, |b| = 1 + 3 = 10, ~a · b = 3 · 1 + (−2) · 3 = −3.
√
(a) comp~a (~b) = ~a · ~b/|~a| = −3/ 13
9
6
~ı + 13
~
(b) proj~a (~b) = (~a · ~b/|~a2 |) ~a = − 13
33
16
(c) proj~a (~b) = b − proj~a (~b) = − 13~ı + 13~
5. Find a unit vector that is perpendicular to both ~a = h−1, 0, 2i and ~b = h3, 3, 0i.
One such vector (there are exactly two) is ~u = ~a × ~b/|~a × ~b|. We have
~a × ~b = h0 · 0 − 2 · 3, −(−1) · 0 + 2 · 3, (−1) · 3 − 0 · 3i = h−6, 6, −3i .
Also, |~a × ~b| =
√
36 + 36 + 9 = 9. Hence ~u = − 32 , 23 , − 31 .
6. Let P = (1, 2), Q = (0, 3), and R = (−1, 1). Find the area of triangle P QR.
The area of triangle P QR is 21 |P~Q × P~R|. (Here we consider all plane vectors as space vectors having
0 as the third coordinate.) We have
P~Q × P~R = h−1, 1, 0i × h−2, −1, 0i = h0, 0, (−1) · (−1) − (−2) · 1i = h0, 0, 3i .
Hence the area is 3/2.
7. Find both the vector equation and the parametric equation for the line through (3, −4, 0) perpendicular to the plane 2x − 2y + z = 2.
The normal vector of the plane is ~v = h2, −2, 1i, which is the direction vector of the line. Hence the
line is parametrized as ~r(t) = h2t + 3, −2t − 2, ti. That is, x = 2t + 3, y = −2t − 2, z = t.
8. Find an equation for the plane that is perpendicular to the line x = 2t − 3, y = 1 − t, z = t + 4 and
contains the point (1, 2, −2).
A direction vector of the line is ~v = h2, −1, 1i, which is a normal vector for the plane. Hence an
equation for the plane is 2(x − 1) − (y − 2) + (z + 2) = 0, or 2x − y + z = −2.
9. Find the traces of the surface x2 − 4y 2 = z 2 + z in the planes x = k, y = k and z = k. Identify the
type of surface.
The traces where x = k are ellipses 4y 2 +(z + 21 )2 = k 2 + 41 . Note that none of these ellipses degenerate
to a point, since k 2 + 41 > 0. The traces where y = k is a constant are hyperbolas x2 − (z + 21 )2 = 4k 2 − 41 .
The traces where z = k are hyperbolas x2 − 4y 2 = k 2 + k. Hence the surface is an elliptic hyperboloid
of one sheet, with central axis y = 0, z = − 21 .
10. (a) Given that the cartesian coordinates of P are (3, 3, −2) determine both the cylindrical and
spherical coordinates of P . (b) Given that the spherical coordinates of Q are (1, π/4, π/3) determine
both the cylindrical
and√
cartesian coordinates of Q.
√
(3, 3) is in the
θ = arctan(3/3) =√π/4. Hence the
(a) r = 9 + 9 = 3 2. Since
√
√
√ first quadrant,
cylindrical coordinates are√(3 2, π/4, −2. ρ = √ 18 + 4 = 22, and φ = arccos(−2/ 22). Hence the
spherical coordinates are ( 22, π/4, arccos(−2/ 22).
√
1 · sin(π/3) = √ 3/2. Hence the √
cylindrical coordinates are
√ (b) z = 1 · cos(π/3)
√ = 1/2 and r = √
( 3/2, π/4, 1/2).√x = √3/2 · cos(π/4) = 6/4 and y = 3/2 · sin(π/4) = 6/4. Hence the cartesian
coordinates are ( 6/4, 6/4, 1/2).
11. (a) Sketch and describe the surface whose equation in spherical coordinates is φ = π/3. (b) Sketch
and describe the surface whose equation in cylindrical coordinates is r = 2 cos θ.
(a) The surface is the upper half of the cone centered along the positive z-axis, whose vertex lies at
the origin, and whose generator makes an angle of π/3 with the central axis.
π/3
(b) If we multiply both sides of the equation by r we obtain the equation r2 = 2r cos θ, or x2 +y 2 = 2x,
or finally (x − 1)2 + y 2 = 1. This determines a right circular cylinder of radius 1 whose central axis is
the vertical line x = 1, y = 0.
12. Sketch the curve with vector equation ht cos(t), t sin(t), ti. Indicate the direction of motion.
For positive t, in cylindrical coordinates the curve is parametrized as r = θ = z = t. Hence for positive
t the curve spirals counterclockwise up the cone r = z, with constant angular speed. For negative t the
curve is the reflection thru the origin.
5.0
2.5
5.0
−5.0
2.5
−2.5
0.0
0.0
0.0
−2.5
−5.0
2.5
−2.5
−5.0
5.0
13. Let ~r(t) = 2t~ı − t2~. (a) Determine an equation for the tangent line at t = 1. (b) Sketch the curve
and the tangent line.
(a) ~r0 (t) = 2~ı − 2t~, ~r(1) = 2~ı −~, and ~r0 (1) = 2~ı − 2~. Hence the tangent line is parametrized by the
equations x = 2t + 2, y = −2t − 1.
2
1
−1
0
1
2
3
4
5
0
−1
−2
−3
−4
14. Find ~r(t) if ~r0 (t) = h2t, et , −1i and ~r(0) = h1, 1, 0i.
Z t
~r(t) = h1, 1, 0i +
~r0 (u) du = t2 + 1, et , −t .
0
15. Let ~r(t) = t2 , 2t, ln t . Compute (a) the unit tangent vector; (b) the unit normal vector; (c)
curvature.
p
First of all ~v = ~r0 = h2t, 2, 1/ti, v = |~v | = 4t2p+ 4 + 1/t2 = 2t + 1/t = (2t2 + 1)/t, ~a = ~v 0 =
2, 0, −1/t2 , ~v × ~a = −2/t2 , 4/t, −4 , and |~v × ~a| = 4/t4 + 16/t2 + 16 = 2(2t2 + 1)/t2 .
(a) T~ = ~v /v = 2t2 /(2t2 + 1), 2t/(2t2 + 1), 1/(2t2 + 1)
(b) κ = |~v × ~a|/v 2 = 2/(2t2 + 1)
16. Let ~r(t) = hsin(t), 2 cos(t)i. Compute (a) velocity; (b) acceleration; (c) speed.
(a) ~v = ~r0 = hcos(t), −2 sin(t)i
(b) ~a = ~v 0 = h−
q sin(t), −2 cos(t)i
q
(c) v = |~v | =
cos2 (t) + 4 sin2 (t) =
1 + 3 sin2 (t).
17. Show that if a particle moves with constant speed then its velocity is always perpendicular to its
acceleration.
If v is constant, then so is v 2 = ~v · ~v . Hence 0 = (~v · ~v )0 = ~a · ~v + ~v ·~a = 2~v ·~a. This tells us that ~v ⊥ ~a.
18. (a) Sketch the graph of the equation f (x, y) = x2 − y. (b) Draw its contour map.
(b) The contour x2 − y = k is a parabola opening up from the vertex (0, −k).
3
2
y
10.0
1
7.5
0
−3
−2
−1
0
1
2
5.0
3
x
−1
2.5
−3
−3
−2
y
−2
00
1
2
3
−2
0.0 −1
−1
x
1
−2.5
2
3
−3
2
19. Compute the
√ partial derivatives fx and fy of each the following functions. (a) f (x, y) = sin(x y).
(b) f (x, y) = y/ x.
(a) fx = cos(x2 y) · 2xy, and fy = √
cos(x2 y) · x2 .
1
−3/2
, and fy = 1/ x.
(b) fx = − 2 y/x
20. The contour map below shows the contours k = 1, 2, . . . , 8 (reading left to right). Use the map to
estimate (a) f (−1, 0); (b) fx (−1, 0); (c) fy (−1, 0).
(a) f (−1, 0) ≈ 3.2.
(b) f (0, 0) ≈ 3.5. Hence fx (−1, 0) ≈ 0.3/1.
(c) f (−1, 1) ≈ 3.9. Hence fy (−1, 0) ≈ 0.7/1.
3
2
y
1
0
−3
−2
0
−1
1
2
x
−1
−2
−3
3