C Roettger, Spring 11
Math 265 Practice Exam 1 – Solutions
Problem 1 First find the direction of the line of intersection: it is given by
v = [1, −1, 5] × [1, 1, 0] = −5i + 5j + 2k.
Then find a point in the intersection of both planes: eg p = [1, 1, 1/5] works
(You can choose x = 1, get y = 1 from the second equation and then adjust
z using the first equation).
Problem 2 a) A normal vector to P is n = [3, −4, 1], so all planes parallel
to P will have equations
3x − 4y + z = d
with some constant d. Plug in the given point to find d = 3·5−4·0−22 = −7.
b) Find the multiple of n that gives a point within P . This is
16
n.
9 + 16 + 1
The distance of P from the origin is just the length of this multiple,
16
16
||n|| = √ .
26
26
Problem 3 a) For the angle θ, we know
cos θ =
u·v
−4
= √ .
||u|| ||v||
3 26
This gives θ ≈ 105.16o .
b) The area equals the length of the cross product,
−2
1
1 −2 2
2
i − k = −4i + 9j + 11k.
u × v = 4 −1 j + 4
3 3 −1 The length of u × v is
√
√
42 + 92 + 112 = 218.
1
Problem 4 a) v(t) = (2t + sin t)i + cos tj. The second component of v is
zero only for t = ±π/2. But for these values, the first component is nonzero.
So the curve is smooth for all values of t.
For b), we have done v already. Then
a = (2 + cos t)i − sin tj.
c) The speed in general is
p
√
(2t + sin t)2 + cos2 t = 4t2 + 4t sin t + 1,
√
which equals π 2 + 2π + 1 = π + 1.
d) The arc length between 0 and 2π is
Z 2π √
L=
4t2 + 4t sin t + 1 dt
0
Problem 5 Differentiate r for a), so
v(t) = 2i + 2tj − k,
a(t) = 2j.
b) First, the direction of the tangent is v(2) = 2i + 4j − k. Then the tangent
line is given by
r(2)+tv(2) = [7, 4, −2]+t[2, 4, −1] = (7+2t, 4+4t, −2−t] = (7+2t)i+4(1+t)j−(2+t)k.
Any of the last three expressions is a correct answer.
c) aT is the projection of a onto v,
aT =
4t
a.v
v=
[2, 2t, −1]
v.v
4 + 4t2 + 1
and then aN is just the difference
aN = a − aT =
1
[−8t, 10, 4t].
5 + 4t2
Note – the textbook also talks about scalar components of acceleration aT
and aN . These are just the lengths of aT and aN , respectively. You could
get the scalar aN slightly easier by
aN =
a×v
||v||
2
Problem 6 Using l’Hospital’s rule,
L = 2i − 4j + k = [2, −4, 1].
Problem 7 a) v = [10, 20, 30]+t[−12, 6, −4] just by integrating the constant
vector a.
b) The position at an arbitrary time is, integrating again,
r(t) = [1000, 0, 0] + t[10, 20, 30] + t2 [−6, 3, −2]
and for t = 4 this is r(4) = [944, 128, 88].
3