CALCULUS 2210–001: HOMEWORK 2, STUDY GUIDE
13.5: #30. We have y = sin(x) for −π ≤ x ≤ π. So y 0 = cos(x) and y 00 = − sin(x),
whence
| sin(x)|
κ=
.
3/2
[1 + cos2 (x)]
Now 1 + cos2 (x) ≥ 1, and 0 ≤ | sin(x)| ≤ 1, so we have 0 ≤ κ ≤ 1. On the other hand,
κ = 1 for the x-values where sin(x) = ±1 and cos(x) = 0; i.e., for x = ± π2 . So κ is
maximized at x = ± π2 and, at those points, κ = 1.
13.5: #40. r(t) = cos3 (t)i + sin3 (t)j. So
v(t) = −3 cos2 (t) sin(t)i + 3 sin2 (t) cos(t)j
a(t) = −3 cos(t) cos2 (t) − 2 sin2 (t) i + 3 sin(t) 2 cos2 (t) − sin2 (t) j.
q
dS
= |v(t)| = 9 cos4 (t) sin2 (t) + 9 sin4 (t) cos2 (t)
dt
= 3 |sin(t) cos(t)| .
We are interested in this function near t = π/3. But cos(π/3) =
1
2
√
and sin(π/3) =
3
.
2
So
dS
π
= 3 sin(t) cos(t),
for t near
dt
3
2
d S
π
= −3 sin2 (t) − cos2 (t) ,
for t near .
2
dt
3
√
√
3 3
i
−
j, so
When t = π3 then, S 0 = 3 3/4 and aT = S 00 = −3/2. Also for this t, a = − 15
8
8
that |a| ≈ 1.9843. So a2N = |a|2 − a2T ≈ 3.0375 − (−3/2)2 = 1.6875. Therefore, aN ≈ 1.299.
13.5: #46. I will do this for ω = 1. In the more general case, the answer is my answer
when you replace t by ωt. (This is not at all obvious, and q
you should check it.)
a2 sin2 (t) + b2 cos2 (t). There-
First, v(t) = −a sin(t)i+b cos(t)j. Therefore, |v(t)| =
fore,
T(t) = q
−a sin(t)
a2
2
sin (t) +
b2
cos2 (t)
i+ q
b cos(t)
a2
2
sin (t) +
b2
cos2 (t)
j.
Differentiate, and write S for sin(t) and C for cos(t) to get,
T0 (t) =
−ab2 C 3 − (2a3 − ab2 )S 2 C
[a2 S 2 + b2 C 2 ]
3/2
i+
−a2 bS 3 − (2b3 − a2 b)SC 2
3/2
[a2 S 2 + b2 C 2 ]
j.
Its magnitude (or length) is,
q
2
2
[ab2 C 3 + (2a3 − ab2 )S 2 C] + [ab S 3 + (2b3 − a2 b)SC 2 ]
0
|T (t)| =
.
3/2
[a2 S 2 + b2 C 2 ]
1
Therefore, the vector N is the vector T0 (t) divided by its length. (Does not simplify much
beyond this.)
14.2: #32. The planes are −3x + 2y + z = 9 and 6x − 4y − 2z = 19. The first plane
is certainly perpendicular to n = h−3, 2, 1i, and the second to m = h6, −4, −2i. But
−2n = m, so n and m are parallel, and this means that the planes are also parallel. So
we need to only find a point on the first plane (e.g., P : (−3, 0, 0)), and find its distance
to the second plane via Example 6, p. 602. Here, x0 = −3, y0 = 0, and z0 = 0, whereas
A = 6, B = −4, C = −2, and D = 19. Therefore, the said distance is:
L=
|Ax0 + By0 + Cz0 − D|
37
√
= √ ≈ 4.944.
2
2
2
A +B +C
56
14.2: #48. Choose a point on the first plane, as in the previous exercise. Let us assume
that A 6= 0. Then we can choose (x0 , y0 , z0 ) = (D/A, 0, 0) which is indeed on the first
plane (i.e., Ax0 + By0 + Cz0 = D). Then, the distance from this point to the second plane
is:
|Ax0 + By0 + Cz0 − E|
|D − E|
√
L=
= √
.
A2 + B 2 + C 2
A2 + B 2 + C 2
If A = 0 but B 6= 0, then choose instead (x0 , y0 , z0 ) = (0, D/B, 0), and get the same
formula. If A = B = 0 but C 6= 0, then choose instead (x0 , y0 , z0 ) = (0, 0, D/C) and still
get the same formula. Since A2 + B 2 + C 2 6= 0 (definition of a plane), one of these must
be nonzero, and so the distance formula is always valid.
14.3: #2. Here, a = h3, 3, 1i, b = h−2, −1, 0i,
(a) a × b is
i
j
a × b = 3
3
−2 −1
(b) b + c = h−4, −4, −1i, so that
i
a × (b + c) = 3
−4
c = h−2, −3, −1i.
k 1 = i − 2j + 3k.
0
j
k 3
1 = 7i − j.
−4 −1 (c) a · (b × c) = (a × b) · c = 1, by using part (a).
(d) a × (b × c) = (a · c)b − (a · b)c = −18b + 9c = h−54, −45, −9i.
14.3: #10. Subtract, from all three points, the point (1, 2, 3) say. This gives you the
vectors, u = h2, −1, 2i and v = h3, 3, 3i. Now u × v = −9i + 3j + 12k. Thus, the area of
the triangle is half of the area of the parallelopiped described by u and v, which is
1√
1
81 + 9 + 144 ≈ 7.6485.
|u × v| =
2
2
14.3: #30. See the previous exercise for the reason (in numbers).
2