Calculus III, TEST 1 Solutions
1. Give a short answer to the following questions:
(a) What is a direction vector for the line with symmetric equations
x − 2 = − y−4
5 =
z+3
4 ?
Let t = x − 2 = − y−4
5 =
z+3
4 .
Then the line has parametric equations x = t + 2, y = −5t + 4, z = 4t − 3.
Hence a direction vector is given by h1, −5, 4i .
(b) What is a normal vector to the plane with equation z = 2x − y?
Equivalently, the plane has equation 2x − y − z = 0. Hence, a normal vector is given by h2, −1, −1i .
2. Find the parametric and symmetric equations for the line passing through the point (5, 1, 0) that is perpendicular to the plane 2x − y + z = 1 .
A normal vector h2, −1, 1i to the plane is a direction vector to the desired line. Hence the line has parametric
equations x = 2t + 5, y = −t + 1, z = t . Consequently, the symmetric equations are
x−5
2
=1−y =z .
3. Find the equation of the plane containing the lines with parametric equations x = 2t + 1, y = −3, z = −t + 2
and x = −t + 1, y = t − 3, z = 2 respectively.
The direction vectors for the
lines are u =
h2, 0, −1i and v = h−1, 1, 0i respectively. Hence a normal vector for
i
j k our plane is n = u × v = 2 0 −1 = i + j + 2k = h1, 1, 2i. Furthermore, notice that the point (1, −3, 2)
−1 1 0 is on both lines and hence, on the plane. Thus, the equation of the plane is x + y + 2z = 2 .
4. Find the equation of the plane consisting of all points equidistant from the two points (1, −1, 0) and (0, 2, 1).
Clearly such plane bisects the line segment defined by the two points. In particular, n = h1, −1, 0i − h0, 2, 1i =
h1, −3, −1i is a vector normal to the plane in question. In addition, the plane that we want contains the
midpoint
1 1 1
2, 2, 2
of the line segment defined by the two points.
Hence the equation of the plane is x − 3y − z = − 32 or 2x − 6y − 2z = −3 .
5. Let r(t) =< et , et cos t, et sin t >.
(a) Find r0 (t).
r0 (t) =<
d t d t
dt e , dt e
d t
cos t, dt
e sin t >=< et , et cos t−et sin t, et cos t+et sin t >= et h1, cos t − sin t, cos t + sin ti
(b) Reparametrize r with respect to arclength.
√ Rt
√
√
Rt
Rt q
2
2
s = 0 |r0 (u)| du = 0 eu 1 + (cos u − sin u) + (cos u + sin u) du = 3 0 eu du = et 3 − 3. Hence
et =
√
s+
√ 3
3
=
√
s 3+3
3
Consequently,
D √
r (s) = s 33+3 ,
and so t = ln s
√
s 3+3
3
√
3+3
.
3
√
cos ln s 33+3 ,
√
s 3+3
3
√
E
sin ln s 33+3
=
√
s 3+3
3
D
√
√
E
.
1, cos ln s 33+3 , sin ln s 33+3
(c) Find the unit tangent, unit normal and binormal vectors at the point (1, 1, 0).
T=
r0 (t)
|r0 (t)|
N=
T0 (t)
|T0 (t)|
=
=
et
1
√
3
et h1, cos t − sin t, cos t + sin ti =
1
√
h0,
3
1
√
3
√
− sin t−cos t, − sin t+cos ti
(− sin t−cos t)2 +(− sin t+cos t)2
The point (1, 1, 0) occurs
Thus B = T × N = √16 =
√1
2
√1
3
h1, cos t − sin t, cos t + sin ti,
h0, − sin t − cos t, − sin t + cos ti.
when t = 0. Hence T = √13 h1, 1, 1i and N = √12 h0, −1, 1i .
i
j k 1
1
1 1 1 = √6 (2i − j − k) and so B = √6 h2, −1, −1i .
0 −1 1 (d) Find the curvature, the normal plane and the osculating plane at the point (1, 1, 0).
|T0 |
The curvature is given by κ =
The normal plane has T =
|r0 |
√1
3
The osculating plane has B =
=
√
√2
√3
3
√
=
2
3
.
h1, 1, 1i as its normal vector and thus it is given by x + y + z = 2 .
√1
6
h2, −1, −1i as its normal vector and thus it is given by 2x − y − z = 1 .
6. Identify the following surfaces
(a) z = x2 + y 2 + 1
This is a circular paraboloid.
(b) 4x2 + z = 100
This is a parabolic cylinder.
7. At what point on the curve r (t) = t3 , 3t, t4 is the normal plane parallel to the plane 6x + 6y − 8z = 1?
Since r0 (t) gives a normal vector to the normal plane, we need to find the value(s) of t for which r0 (t) is parallel
to the normal vector h6, 6, −8i of the plane 6x + 6y − 8z = 1. Hence we need 3t2 , 3, 4t3 to be parallel to
h6, 6, −8i. In other words we need to find k and t, such that 3t2 = 6k, 3 = 6k, 4t3 = −8k. Solving the system
we obtain k =
1
2
and t = −1. Therefore the point that we seek is r (−1) = (−1, −3, 1).
8. Show that if r is a vector function such that r00 exists, then
d
[r (t) × r0 (t)] = r (t) × r00 (t) .
dt
By the product rule,
d
dt
[r (t) × r0 (t)] = r0 (t) × r0 (t) + r (t) × r00 (t).
|r0 (t)| |r0 (t)| sin 0 = 0 and so r0 (t) × r0 (t) = 0. Hence
d
dt
Now notice that |r0 (t) × r0 (t)| =
[r (t) × r0 (t)] = r0 (t) × r0 (t) + r (t) × r00 (t) =
0 + r (t) × r00 (t) = r (t) × r00 (t) which is what we wanted to show.