Math 2210–001, Midterm 1, Study Guide
Summer 2003
√
(1) (4 points each) Consider the following two 2-dimensional vectors: u = h−2, yi, v = h1, 2i.
(a) Find the projection of u onto v. √
p
√
Solution: Clearly, u · v = −2 + 2 y. Also, |u| = 4 + y 2 , and |v| = 3. Therefore,
√
√
√
−2 + 2
−2 2 + 2
u·v
−2 + 2y
v
=
i
+
j.
v
=
projv (u) =
|v|2
3
3
3
(b) If you knew that u and v are perpendicular, then find
√
√ the value of y.
Solution: Perpendicular means u · v = 0; i.e., −2 + 2 y = 0, and this means that y = 2.
(c) If you knew that they are parallel, then find the value of y.
√
Solution: Parallel means that v = cu for a scalar c. Solve to obtain, h−2, yi
√ the first
√ = ch1, 2i. Inspect
coordinates to see√that c = −2; therefore, the second coordinates tell us that c 2 = y, so y = −2 2. In other
words, u = h−2, 2 2i is parallel to v.
(2) (4 points each) Consider a particle that moves according to the following law:
π
r(t) = sin(2t)i + sin(4t)j,
0≤t≤ .
2
Here is the plot. The picture starts at the point of intersection and ends there.
(a) On the preceding figure, mark and label the position of the particle at the following times: t = π8 , t = π4 , t = 3π
8 ,
.
and t = 3π
4
(b) Identify the direction of motion on the figure, and compute the velocity and acceleration of this particle at any
time t.
Solution: v(t) = r0 (t) = 2 cos(2t)i + 4 cos(4t)j, and a(t) = r00 (t) = −4 sin(2t)i − 16 sin(4t)j.
1
2
(c) Find the length of this knot.
Solution: From its definition,
Z πp
Z π
|v(u)| du =
4 cos2 (2t) + 16 cos2 (4t) dt
S=
0
0
Z πp
cos2 (2t) + 4 cos2 (4t) dt.
=2
0
(d) Compute and plot the unit tangent vector at time t =
3π
8 .
Solution: In general, T(t) = v(t)/ |v(t)| . Therefore, here,
2 cos(4t)
cos(2t)
i+ p
j.
T(t) = p
cos2 (2t) + 4 cos2 (4t)
cos2 (2t) + 4 cos2 (4t)
At time t =
3π
8 ,
we have
2 cos(3π/2)
cos(3π/4)
i+ p
j.
T= p
2
2
2
cos (3π/4) + 4 cos (3π/2)
cos (3π/4) + 4 cos2 (3π/2)
√
Now cos(3π/4) = −1/ 2 ≈ −0.707 (radians). Moreover, cos(3π/2) = 0. Therefore, at t = 3π/8,
T = −i.
π
4
(e) Compute the curvature when t = and when t = 3π
8 .
(Hint: If you want, you may use one of the following formulas, whichever applies.)
|y 00 |
,
κ
=
.
3/2
3/2
[(x0 )2 + (y 0 )2 ]
[1 + (y 0 )2 ]
Solution: We can use only the first formula in the hint, since the parametrization of x is not x = t. In fact,
x = sin(2t) and y = sin(4t), so that x0 = 2 cos(2t), x00 = −4 sin(2t), y 0 = 4 cos(4t), and y 00 = −16 sin(4t). Plug
in to get:
|−32 cos(2t) sin(4t) + 16 sin(2t) cos(4t)|
.
κ(t) =
3/2
[4 cos2 (2t) + 16 cos2 (4t)]
√
In particular, upon using the fact that sin(3π/4) = cos(3π/4) = 1/ 2, we obtain
π |−32 cos(π/2) sin(π) + 16 sin(π/2) cos(π)|
=
κ
4
[4 cos2 (π/2) + 16 cos2 (π)]3/2
16
1
= 3/2 = , whereas
4
16
3π
|−32 cos(3π/4) sin(3π/2) + 16 sin(3π/4) cos(3π/2)|
κ
=
3/2
8
[4 cos2 (3π/4) + 16 cos2 (3π/2)]
√
32/ 2
= 3/2 = 8.
2
κ=
|x0 y 00 − y 0 x00 |