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QUIZ 2 : MATH 251, Section 516
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Write up your result, detail your calculations if necessary and BOX your final answer
1. [25pts] Give an equation of the plane passing through (2, −1, 3) and parallel to the plane x+2y −4z =
36.
2. [25pts] Classify the surface (identify it) and sketch it (be a bit rigorous) :
a/ x2 + 4x + 2y + z 2 − 10z + 29 = 0,
b/ 4x2 + y 2 = z 2 .
π
3. [25pts] Find the unit tangent of the curve with parametrization r(t) = t~i + 2 sin t~j + 2 cos t~k at t = .
4
t
t
t
~
~
~
4. [25pts] Find the length of the curve r(t) = e i + e sin tj + e cos tk, 0 ≤ t ≤ 2π.
Answers :
1. Let P be the plane x + 2y − 4z = 36. The vector h1, 2, −4i is a normal vector to the plane P but is
also orthogonal to any parallel planes to P. So the wanted plane is of the form x + 2y − 4z = d for
d ∈ R. We obtain d knowing that the plane passes through (2, −1, 3) which means
2.1 − 1.2 − 4.3 = −12, so d = −12. The equation of the wanted plane is x + 2y − 4z + 12 = 0 .
2. a/ x2 + 4x + 2y + +z 2 − 10z + 29 = (x + 2)2 − 4 + 2y + (z − 5)2 − 25 + 29 = 0 which is equivalent to
(x + 2)2 + (z − 5)2 = −2y. The trace with the yz-plane or the xz-plane are parabolas. the trade with
planes y = k for k < 0 are circles. So, this is a circular paraboloid whose axis is the y-axis.
b/ Traces with the planes z = k, k > 0 are ellipses. This is a cone whose axis is the z-axis.
π
3. For all t, r0 (t) = h1, 2 cos t, −3 sin ti. This vector doesn’t vanish so at t = , the curve descri4
√
√
π
bed by r admits a tangent vector r0 (0) = h1, 2, − 2i. So, the unit tangent vector at t =
is :
4
*
+
√
√
√
√
1
2
2
h1, 2, − 2i
q
√ , √ , −√
=
.
√ 2 √ 2
5 5
5
12 + 2 + 2
4. Denote ` the length of the curve between 0 and 2π. For any t, r0 (t) = het , et sin t + et cos t, et cos t −
et sin ti.
Z
2π
0
Z
|r (t)|dt =
`=
0
0
2π
Z
p
2t
2
2
e (1 + (cos t + sin t) + (cos t − sin t) )dt =
0
2π
3et dt = 3(e2π − 1) .
Figure 1 – Paraboloid
Figure 2 – Cone