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Calculus III: Practice Midterm I ... The exam contains 6 problems. • Good luck! Question Points Score ... Calc III (Spring '13). Practice Midterm I.
Typology: Exercises
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Question Points Score 1 8 2 10 3 10 4 10 5 10 6 12 Total: 60
(a) (2 points) There is a vector v such that
v × 〈 1 , 1 , 1 〉 = 〈 1 , 2 , 3 〉.
Solution: False. (Because 〈 1 , 2 , 3 〉 is not perpendicular to 〈 1 , 1 , 1 〉.)
(b) (2 points) The sixth power of 2eiπ/^6 is a real number.
Solution: True. (Because the argument of the sixth power is π.)
(c) (2 points) The surface described by x^2 + y^2 − z^2 = 1 is a hyperbolic paraboloid.
Solution: False. (Because the traces are ellipses and hyperbolas, making it a hyperboloid.)
(d) (2 points) The plane 2x + 4y + 6z = 9 is perpendicular to the vector 〈 1 , 1 , − 1 〉.
Solution: False. (Because the normal direction to that plane is 〈 2 , 4 , 6 〉, which is not the same as 〈 1 , 1 , − 1 〉.)
Solution: There are many ways to do this problem. Let us denote the points by A, B, C, and D. Here are two strategies:
AC, and
If the points were on a plane, it would be zero (the paralellopiped will be flat), otherwise it would be non-zero.
First way: We first find the equation of the plane passing through A, B and C. This is the plane passing through 〈 1 , 1 , 0 〉 and with normal direction −→n =
We have −→ AB = 〈 1 , 1 , − 2 〉 − 〈 1 , 1 , 0 〉 = 〈 0 , 0 , − 2 〉 = − 2 k −→ AC = 〈 0 , 2 , − 1 〉 − 〈 1 , 1 , 0 〉 = 〈− 1 , 1 , − 1 〉 = − i + j − k −→n = (− 2 k) × (− i + j − k) = 2 k × i − 2 k × j = 2 j + 2 i.
The plane through A and perpendicular to −→n is
(〈x, y, z〉 − 〈 1 , 1 , 0 〉) · 〈 2 , 2 , 0 〉 = 0 2(x − 1) + 2(y − 1) = 0 2 x + 2y = 4 or equivalently x + y = 2.
The fourth point 〈 5 , − 3 , 0 〉 satisfies this equation, so it lies on this plane. In other words, the four points are coplanar. Second way: We know that the volume of the paralellopiped formed by
is
AC). We need to check whether this is zero. We have already computed −→ AB ×
AC = 2 i + 2 j = 〈 2 , 2 , 0 〉.
Now,
AD = 〈 5 , − 3 , 0 〉 − 〈 1 , 1 , 0 〉 = 〈 4 , − 4 , 0 〉. Hence the triple product is −−→ AD · (
In conclusion, the paralellopiped is flat and hence A, B, C, and D are coplanar.
x^3 = i.
Express your answers both in polar and Cartesian forms.
Solution: Taking the magnitudes, we get
|x|^3 = 1.
Since |x| is a positive real number, we conclude that |x| = 1. It remains to find the argument of x. Since arg i = π/2, we must have
3 arg x =
π 2
or
(π 2
or
(π 2
etc.
Therefore, arg x =
π 6
or
π 6
2 π 3
or
π 6
4 π 3
Hence, in polar form, the solutions to x^3 = i are
x = e
iπ 6 or e
5 iπ 6 or e
9 iπ 6 .
Since eiθ^ = cos θ + i sin θ, we get the Cartesian forms
x = cos(π/6) + i sin(π/6) or cos(5π/6) + i sin(5π/6) or cos(9π/6) + i sin(9π/6)
=
i or −
i or −i
Note: In class, we called r(cos θ + i sin θ) the polar form. Since this is equal to reiθ (which is more compact), it is more common to call reiθ^ the polar form. In the exam, you may use either one.
(a) (3 points) parallel?
Solution: For the vectors to be parallel, they must be proportional. By looking at the third coordinate, we see that the second vector must be twice the first. So we must have 2 〈 1 , a, 2 〉 = 〈a, 4 , 4 〉. By comparing the first and the second coordinates, we get
a = 2 and a = 2.
Hence a = 2.
(b) (3 points) perpendicular?
Solution: For the vectors to be perpendicular, the dot product must be zero:
〈 1 , a 2 〉 · 〈a, 4 , 4 〉 = a + 4a + 8 = 0,
which gives a = − 8 /5.
(c) (6 points) For which a, does the first vector 〈 1 , a, 2 〉 make an angle of π/4 with the vector j?
Solution: One way of doing this is to use the dot product. If the angle is π/4, we must have j · 〈 1 , a, 2 〉 = | j||〈 1 , a, 2 〉| cos(π/4). Simplifying, we get
a =
a^2 + 5 · (1/
2 a =
a^2 + 5.
Squaring, we get
2 a^2 = a^2 + 5 a^2 = 5.
So that a = ±
5 since that would make the dot product negative (which corresponds to an obtuse angle). Hence a =
