# Math 217: Some solutions to Assignment 2 13.1 # 40: + y ```Math 217: Some solutions to Assignment 2
13.1 # 40:
For any fixed z, the cylinder x2 + y 2 = 4 is a circle, and can be parameterized by
x = 2 cos(t), y = 2 sin(t), 0 ≤ t &lt; 2π. So on the intersection with z = xy, we have
z = 2 cos(t) &middot; 2 sin(t). Thus the vector function
r(t) = h2 cos(t), 2 sin(t), 4 cos(t) sin(t)i,
0 ≤ t &lt; 2π
parameterizes the curve in question (though of course there are other possible correct
13.2 # 54:
Suppose r(t) &middot; r0 (t) = 0. Compute, using the product rule
d
d
|r(t)|2 = (r(t) &middot; r(t))
dt
dt
= r(t) &middot; r0 (t) + r0 (t) &middot; r(t) = 2r(t) &middot; r0 (t) = 0.
Thus |r(t)|2 = constant, call it R2 . That is another way of saying that the curve lies on a
sphere of radius R centred at the origin.
13.3 # 6
r(t) = h12t, 8t3/2 , 3t2 i , so r0 (t) = h12, 12t1/2 , 6ti, and so
p
p
|r0 (t)| = 6 4 + 4t + t2 = 6 (t + 2)2 = 6(t + 2)
so the length of the curve is
Z
1
0
Z
|r (t)|dt =
L=
0
1
6(t + 2)dt = 6(t2 /2 + 2t)|10 = 15.
0
13.3 # 32
For this question, it is easiest to use a formula from the text (not given in class) for
the curvature of a graph y = f (x):
κ=
|f 00 (x)|
.
[1 + (f 0 (x))2 ]3/2
Let us try f (x) = ax2 whose graph is, indeed, a parabola passing through the origin
(though there are other possibilities). Then
κ(x) =
|2a|
,
[1 + (2ax)2 ]3/2
1
and so the curvature at the origin is κ(0) = 2|a|. So if we choose a = 2, giving y = 2x2 ,
we have curvature 4 at the origin.
13.4 # 16
v0 (t) = a(t) = h2, 6t, 12t2 i → v(t) = h2t, 3t2 , 4t3 i + c.
Now h1, 0, 0i = v(0) = c. Then
r0 (t) = v(t) = h2t + 1, 3t2 , 4t3 i → r(t) = ht2 + t, t3 , t4 i + d.
Now h0, 1, −1i = r(0) = d, so
r(t) = ht2 + t, t3 + 1, t4 − 1i.
13.4 # 22
Suppose the particle’s path is described by the vector function r(t). If its speed is
constant, then |v(t)| = const, where v(t) = r0 (t). So
0=
d
d
|v(t)|2 = v(t) &middot; v(t) = 2v(t) &middot; v0 (t)
dt
dt
which says that the velocity is orthogonal to the acceleration.
(Sept. 26, 2013)
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