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Allen – Math 265 B2 Practice Exam #1 Name:
- Given that ~u = 〈− 1 , 3 〉, ~v = 〈 2 , 1 〉, and w~ = 〈 5 , 6 〉 are three non-collinear vectors, find scalars a and b so
that w~ = a~u + b~v.
We need 〈−a, 3 a〉 + 〈 2 b, b〉 = 〈 5 , 6 〉. This leads to the following system of equations: { −a + 2b = 5
3 a + b = 6
−a + 2b = 5
− 6 a − 2 b = − 12
Adding the two equations together yields − 7 a = −7. Thus
a = 1. Plugging back into −a + 2b = 5, we get −1 + 2b = 5 ⇒ 2 b = 6. Therefore, b = 3.
- Let ~u = 〈 1 , 4 , 2 〉, ~v = 〈− 2 , 3 , 1 〉, and w~ = 〈 4 , − 1 , − 1 〉. Perform the indicated operations.
(a) Compute the angle (in degrees) between ~u and ~v.
Use the formula ~u · ~v =
~u
~v
cos θ. ~u · ~v = 1(−2) + 4(3) + 2(1) = −2 + 12 + 2 = 12. ∣ ∣ ~u
2
~v
2
Thus, 12 =
14 cos θ ⇒ 12 = 7
6 cos θ ⇒
= cos θ. Therefore, θ ≈ 45. 5847
◦ .
(b) Find the volume of the parallelepiped determined by ~u, ~v, and w~.
We can use 2 methods.
~u · (~v × w~)
from part (b). This equals ∣ ∣ 1(−2) + 4(2) + 2(−10)
Another method we can use is to find
, which is
- Let ~u = 〈 3 , 4 〉 and ~v = 〈 5 , 12 〉. We can write ~u = m~ + ~n, where m~ is parallel to ~v and ~n is perpendicular
to ~v. Find m~ and ~n.
m ~ = pr~v~u =
~u · ~v
|~v| 2
~v. ~u · ~v = 3(5) + 12(4) = 15 + 48 = 63. |~v|
2 = 5
2
2 = 25 + 144 = 169.
Thus, m~ =
Then ~n = ~u − m~ = 〈 3 , 4 〉 −
- Find the equation of the plane that passes through the points P = (1, 4 , 6), Q = (− 2 , 5 , −1), and
R = (1, − 1 , 1).
If ~u is the vector from Q to P, then ~u = 〈(1 + 2), (4 − 5), (6 + 1)〉 = 〈 3 , − 1 , 7 〉. Similarly, if ~v is the vector
from Q to R, then ~v = 〈(1 + 2), (− 1 − 5), (1 + 1)〉 = 〈 3 , − 6 , 2 〉. To compute a normal to the plane, we
compute ~u × ~v.
~u × ~v =
~i ~j ~k
= (−2 + 42)~i − (6 − 21)~j + (−18 + 3)
k = 40~i = 15~j − 15
k = 〈 40 , 15 , − 15 〉.
This vector has the same direction as 〈 8 , 3 , − 3 〉. So the equation of the plane has the form
8 x + 3y − 3 z = D. Using the point P, we get 8(1) + 3(4) − 3(6) = D ⇒ D = 8 + 12 − 18 = 2. Therefore,
the equation of the plane is 8x + 3y − 3 z = 2.
- Find the center and radius of the sphere 4x 2 + 4y 2 + 4z 2 − 24 x + 56y − 40 z − 197 = 0.
4 x 2
- 4y 2
- 4z 2 − 24 x + 56y − 40 z = 197 ⇒ x 2
- y 2
- z 2 − 6 x + 14y − 10 z =
⇒ (x 2 − 6 x) + (y 2
⇒ (x 2 − 6 x + 9) + (y 2
- 14y + 49) + (z 2 − 10 z + 25) =
⇒ (x − 3) 2
⇒ (x − 3) 2
Center: (3, − 7 , 5). Radius:
- A particle P travels in the plane, and its vector position at time t is given by ~r(t) = (1 − t)~i +
1 − t
j.
(a) Find the Cartesian equation for the path of the particle.
From the vector equation, the parametric equations are given by x = 1 − t and y =
1 − t 2
. Solve
the first equation for t and get t = 1 − x. Plug into the second equation to get
y =
1 − (1 − x) 2 =
1 − (1 − 2 x + x 2 ) =
2 x − x 2
. Hence, the path of the particle is
y
2 = 2x − x
2 in Cartesian coordinates.
(b) Assuming that P starts moving at time t = 0, how far will it have traveled at t = 1/2 seconds?
L =
0
[x′(t)]^2 + [y′(t)]^2 dt. x ′ (t) = −1, and so [x ′ (t)] 2 = 1.
y ′ (t) =
(1 − t 2 ) − 1 / 2 (− 2 t) = −
t √ 1 − t 2
, and so [y ′ (t)] 2 =
t
2
1 − t 2
Hence, [x ′ (t)] 2
t
2
1 − t 2
1 − t
2
1 − t 2
t
2
1 − t 2
1 − t 2
. Therefore,
L =
0
1 − t 2
dt =
0
1 − t^2
dt =
[
sin
− 1 t
] 1 / 2
0
= sin
− 1 (1/2) − sin
− 1 (0) =
π
π
- A cannon sits atop a cliff that is 176 feet tall. It is aimed at an angle of 53. 1301
◦ above the horizontal,
and it shoots a projectile at the speed of 25 feet per second.
(a) Find the initial velocity vector, ~v(0) of the projectile.
We are given that
~v(0)
= 25, and the angle between ~v(0) and the ground is 53. 1301
◦
. Let x be
the horizontal component of the velocity and y be the vertical component of the velocity. Then,
x = 25 cos(53.1301) = 15, and y = 25 sin(53.1301) = 20. Therefore, ~v(0) = 15
i + 20
j.
(b) Using part(a), the fact that ~a(t) = − 32
j, and assuming ~r(0) = 176
j, find the velocity, ~v(t), and the
position, ~r(t), of the projectile for any time t.
Taking the antiderivative of ~a(t) yields ~v(t) = C 1 ~i + (− 32 t + C 2 )~j. At t = 0, v(0) = C 1 ~i + C 2 ~j. But
from part (a), this should be 15~i+20~j. Hence, C 1 = 15, C 2 = 20, and thus ~v(t) = 15~i+(− 32 t+20)~j.
Now, we antidifferentiate again to get ~r(t) = (15t + C 3 )~i + (− 16 t 2
~r(0) = C 3 ~i + C 4 ~j. But this should be 176~j. Therefore, C 3 = 0, C 4 = 176, and thus
~r(t) = 15t~i + (− 16 t 2
(c) What is the maximum height of the projectile?
To answer this question, we set the vertical component of velocity equal to zero. So, − 32 t + 20 =
0 ⇒ 32 t = 20 ⇒ t =
= 0.625 seconds. Next we plug this time in for the vertical component
of position and get −16 (5/8)
2
Therefore, the maximum height is
= 182.25 feet.
(d) At what time does the projectile hit the ground?
This happens when the vertical component of the position is zero. That is, when
− 16 t
2
2 − 5 t − 44 = 0. You can use the quadratic formula to find t, but luckily
4 t 2 − 5 t − 44 factors into (4t + 11)(t − 4). Thus, 4t + 11 = 0 or t − 4 = 0, which implies t = −
or t = 4. Since t = −
is preposterous, we must have t = 4. So the projectile hits the ground 4
seconds after it is shot.
(e) With what speed does the projectile come smashing into the basin below the cliff?
First, we need to find the velocity at t = 4. ~v(4) = 15
i + (−32(4) + 20)
j = 15
i − 108
j. Then, the
speed is
~v(4)
2
11 , 889 ≈ 109 .037 feet per second.
- Let ~r(t) = t~i + t
j +
t
k.
(a) Find T~ (1).
~v(t) = ~i + 2t~j + 2t
k. Then ~v(1) = ~i + 2~j + 2~k. Therefore,
T^ ~ (1) =
i + 2
j + 2
k √ 1 + 4 + 4
i + 2
j + 2
k √ 9
~i +
~j +
~k =
(b) Find N~ (1).
~a(t) = 2~j + 4t~k. Then ~a(1) = 2~j + 4~k. Therefore,
aT (1) = T~ (1) · ~a(1) =
Note that |~a(1)|
2 = 2 2
- 4 2 = 20. Using the Pythagorean Identity |~a(1)|
2 = a 2 T (1) + a 2 N (1), we have
aN (1) =
2
Then, since ~a(1) = aT (1) T~ (1) + aN (1) N~ (1), we can compute N (1) as follows:
N^ ~ (1) =
~a(1) − aT (1) T~ (1)
aN (1)
(c) Find the curvature, κ at t = 1.
κ(1) =
∣~v(1) × ~a(1)
~v(1)
3
. But aN (1) =
∣~v(1) × ~a(1)
~v(1)
. Thus, we can use our calculation for aN (1) in
part (b) to get κ =
aN (1) ∣ ∣ ~v(1)
2
2
(d) Find the binormal B~(1).
B(1) =
T (1) ×
N (1) =
i
j
k
1 / 3 2 / 3 2 / 3
− 2 / 3 − 1 / 3 2 / 3
