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QUIZ 2 : MATH 251, Section 516
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Write up your result and detail your calculations if necessary
1. [25pts] Find an equation of the plane passing through the given point (-1, 3, -8) and parallel to the
plane 3x − 4y − 6z = 9.
2. [25pts] Classify the surface (identify it) and sketch it (be a bit rigorous) :
a/ 9x2 + y 2 − z 2 − 2y + 2 = 0,
b/ x2 + 4z 2 = y 2 .
3. [25pts] Find the unit tangent vector of the curve with parametrization r(t) = h2t, 3t2 , 4t3 i at t=1.
4. [25pts] Find the length of the curve r(t) = h2t, 3 sin t, 3 cos ti, 0 ≤ t ≤ π.
Answers :
1. Let P be the plane 3x − 4y − 6z = 9. The vector (3, −4, −6) 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 3x − 4y − 6z = d
for d ∈ R. We obtain d knowing that the plane passes through (−1, 3, −8) which means
3.(−1) − 4.3 − 6.(−8) = d, so d = 33. The equation of the wanted plane is 3x − 4y − 6z = 33 .
2. a/ 9x2 + y 2 − z 2 − 2y + 2 = 9x2 + (y − 1)2 − z 2 + 1 = 0 which is equivalent to −9x2 − (y − 1)2 + z 2 = 1.
There are 2 minus signs and Traces with vertical planes are hyperbolas. So, this is a hyperboloid
of 2 sheets whose axis is the z-axis.
b/ Traces with xz-plane are ellipses. This is a cone whose axis is the y-axis.
3. For all t, r0 (t) = h2, 6t, 12t2 i. This vector doesn’t vanish so at t = 1, the curve described by r
h2, 6, 12i
admits a tangent vector r0 (1) = h2, 6, 12i. So,the unit tangent vector at t = 1 is : √
=
22 + 62 + 122
2
6
12
1
3
6
√ , √ , √
= √ ,√ ,√
.
2 46 2 46 2 46
46 46 46
4. Denote ` the length of the curve between 0 and π. for any t, r0 (t) = h2, 3 cos t, −3 sin ti.
Z π
Z πp
Z π
√
√
0
2
2
`=
|r (t)|dt =
4 + 9 cos t + 9 sin tdt =
4 + 9dt = 13π .
0
0
0
of two sheets.png
Figure 1 – Hyperboloid of two sheets
Figure 2 – Cone