MATH 235 Midterm I(Spring 2015)
Instructor: Xiaowei Wang
Feb. 23rd, 9:20pm-11:20pm, 2015
Problem 1. Determine whether the following statements are true(no explanation is
required).
1. (F) projv u = proju v.
2. (F)If u is orthogonal to v and v is orthogonal to w, then u is orthogonal to w.
3. (F)The cross product of two nonzero vectors is a nonzero vector.
4. (T)If u × v = 0 and u · v = 0, then either u = 0 or v = 0 ( or both) .
5. (F)Any two nonparallel lines in R3 intersect.
6. (F)If the speed of an object is constant, then its velocity components are constant.
7. (F)It is not possible for a velocity vector to have a constant direction but a variable
magnitude for all t ≥ 0.
8. (F)If an object moves on a trajectory with constant speed S over a time interval
a ≤ t ≤ b, then the length of the trajectory is S(b − a).
9. (F)If the limits
lim
f (x, 0) and
(x,0)→(0,0)
lim
f (0, y) exist and equal L, then
(0,y)→(0,0)
lim
f (x, y) = L .
(x,y)→(0,0)
10. (T)If f is continuous at (a, b) then
lim
f (x, y) exists.
(x,y)→(0,0)
Problem 2. Let u = h4, 3i, v = h1, 1i, express u as the sum of u = p + n, where p is
parallel to v and n is orthogonal to v.
Solution. p = projv u =
u·v
v
|v|2
= 27 h1, 1i. n = u − p = 12 h1, −1i.
Problem 3. Determine the distance between the point P and the line ` through the
origin with
P (1, 1, −1); ` has the direction h1, 1, 0i.
Solution. Let v = h1, 1, 0i then
−−→ h1, 1, −1i · v
2
projv OP =
v = h1, 1, 0i
|v|2
2
−−→
−−→
−−→
−−→
and OP − projv OP = h0, 0, 1i and the distance is given by |OP − projv OP | = 1.
1
Problem 4. Find the area of the following triangle T . with the vertices of T are
O(0, 0, 0), P (0, 1, 1), and Q(2, 3, 0).
−−→
Solution. Notice that OP = h0, 1, 1i and
i
−−→ −−→ OP × OQ = 0
2
−−→
OQ = h2, 3, 0i. Then
j k
1 1 = −3i + 2j − 2k
3 0
1 −−→ −−→
1√
1p
area of T = |OP × OQ| =
(−3)2 + 22 + (−2)2 =
17 .
2
2
2
Problem 5. Find all vectors u satisfying the equation h1, 1, 1i × u = h0, 0, 1i.
Solution. Let u = hx, y, zi then
i j k
1 1
1 1
1 1
− j
h1, 1, 1i × u = 1 1 1 = i x z + k x y y z
x y z = i(z − y) + j(x − z) + k(y − x)
we want to solve z − y = 0, x − z = 0, and y − x = 1. There is no solution so there is
no such u.
Second approach, one notice that u · h1, 1, 1i =
6 0, they are NOT orthogonal which is
impossible by the definition of cross product.
Problem 6. Find the point (if it exists) at which the following planes and lines intersect.
x = 3; r(t) = ht, 2t, ti, for − ∞ < t < ∞ .
Solution. For the intersection point we must have ht, 2t, ti = h3, ∗, ∗i, this implies that
t = 3, hence the intersection point is given by h3, 6, 3i.
Problem 7. Let u = h1, t, t3 i and v = ht2 , −2t, 1i. Compute the derivatives of the
following functions
1. v(cos t)
Solution. (v(cos t))0 = − sin tv0 (cos t) = − sin th2 cos t, −2, 0i.
2. u(t2 ) × v(2t)
Solution.
2
i
j
k
1 t6 1
t
t2 t6 2
2
6
t
t = i
+ k 2
− j 2
u(t ) × v(2t) = 1
4t −4t
−4t 1 4t 1 4t2 −4t 1 = i(t2 + 4t7 ) − j(1 − 4t8 ) + k(−4t − 4t4 )
Hence (u(t2 ) × v(2t))0 = i(2t + 28t6 ) + j(32t7 ) + k(−4 − 16t3 ).
Problem 8. Given an acceleration vector, initial velocity hu0 , v0 , w0 i, and initial position
hx0 , y0 , z0 i, find the velocity and position vectors for t ≥ 0.
a(t) = hcos t, sin 2ti, hu0 , v0 i = h1, 1i, hx0 , y0 i = h0, 0i .
2
Solution.
t
Z
a(τ )dτ + hu0 , v0 i
v(t) =
0
t
Z
hcos τ, sin 2τ idτ + h1, 1i
=
0
= hsin t + 1,
3 − cos 2t
i.
2
and
t
Z
v(τ )dτ + hx0 , y0 i
r(t) =
0
t
3 − cos 2τ
idτ
2
0
6t − sin 2t
= h1 + t − cos t,
i.
4
Z
hsin τ + 1,
=
Problem 9. Find the length of the following curves:
1. r(t) = h4 cos t, 4 sin t, 3ti, for 0 ≤ t ≤ 4π.
Solution.
Z
4π
Z
0
4π
|r (t)|dt =
arc length =
0
Z
p
2
2
16 sin t + 16 cos t + 9dt = 5
4π
dt = 20π.
0
0
2. The complete cardioid r = 2 − 2 cos θ.
Solution.
Z
π
Z
2π
q
arc length = 2
+
=4
(1 − cos θ)2 + sin2 θdθ
0
0
√ Z π √
√ Z π√
= 4 2
1 − cos θdθ = −8 2
d( 1 + cos θ)
0
0
π
√ √
= −8 2 1 + cos θ = 16
p
r2
(r0 )2 dt
0
Problem 10. Find the unit tangent vector T and the curvature κ: r(t) = h4 sin t, 4 cos t, ti
at t = 1.
Solution. r0 (t) = h4 cos t, −4 sin t, 1i so
1
T (t) = √ h4 cos t, −4 sin t, 1i.
17
and
1
T 0 (t) = √ h−4 sin t, −4 cos t, 0i
17
and
κ(1) =
1
|r0 (1)|
0 T (1) = 4 .
17
3
Problem 11. Evaluate the following limits if they exists.
1.
x2 − 3xy
(x,y)→(6,2) x − 3y
lim
Solution.
x2 − 3xy
x(x − 3y)
=
lim
=
lim x = 6.
(x,y)→(6,2) x − 3y
(x,y)→(6,2) x − 3y
(x,y)→(6,2)
lim
2.
lim
(x,y)→(0,0)
3x − 2y
2x − 3y
Solution. Limit does not exist. Since if we let y = mx and let x → 0 then
3x − 2y
3 + 2m
=
.
2x − 3y
2 + 3m
This means (x, y) → (0, 0) along the line y = mx the limit depend on m.
Problem 12. Suppose an object moves on the surface of a sphere with |r(t)| constant
for all t. Show that r(t) and a(t) = r00 (t) satisfy r(t) · a(t) = −|v(t)|2 .
Proof. By assumption, we have r(t) · r(t) = const.. Taking derivative with respect to t
on both sides of the equation we have r(t) · r0 (t) = 0. Differentiate one more time of this
equation we obtain 0 = (r(t) · r0 (t))0 ) = r0 (t) · r0 (t) + r(t) · r00 (t). Our claim follows from
the fact that a(t) = r00 (t) and v(t) = r0 (t).
4