HW 4 - DUE 9/30
Please write neatly, and show all work. Caution: An answer with no
work is wrong!
Do the following problems from the book:
• §10.2: 3, 5, 12, 14, 17, 21, 24, 31 (no approximation necessary,
leave your answer exact), 33, 38, 49
• §10.3: 2, 4, 6, 11, 13, 24, 30, 37, 46, 54
Answers to selected excercises:
§10.2, 17. Find the unit tangent vector T(t) to the curve r(t) =
cos ti + 3tj + 2 sin 2tk at t = 0.
Solution. Differentiating we find r0 (t) = − sin ti + 3j + 4 sin 2tk. Thus
r0 (0) = 3j. This vector has length 3, so T(0) = r0 (0)/3 = j.
§10.2, 24. Find
equations for the tangent line to the curve
√ parametric
2
x = ln t, y = 2 t, z = t , at (0, 2, 1).
Solution. The given point
√ occurs for t = 1. Differentiating, we have
x0 (t) = 1/t, y 0 (t) = 1/ t, z 0 (t) = 2t, so that r0 (1) = h1, 1, 2i. This
means the line is given by (0, 2, 1) + th1, 1, 2i. Parametrically, we have
x = t, y = 2 + t, z = 1 + 2t.
§10.2, 31. The curves r1 (t) = ht, t2 , t3 i and r2 (t) = hsin t, sin 2t, ti
intersect at the origin. Find the angle of intersection.
Solution. We have r01 (t) = h1, 2t, 3t2 i and r02 (t) = hcos t, 2 cos 2t, 1i, so
that r01 (0) = h1, 0, 0i and r02 (0) = 1, 2, 1i. if θ is the desired angle, we
have
r01 (t) · r02 (0) = 1 = |r01 (0)||r02 (0)| cos θ.
1
2
Since |r01 (0)| √= 1 and |r02 (0)| =
θ = cos−1 (1/ 6).
√
√
6, we have cos θ = 1/ 6, so that
§10.3, 4. Find the length of r(t) = cos ti + sin tj + ln cos tk for
0 ≤ t ≤ π/4.
sin t
Solution. We have r0 (t) = − sin ti + cos tj − cos
k, so that
t
r
√
sin2 t p
2
|r0 (t)| = sin2 t + cos2 t +
=
1
+
tan
t
=
sec2 t.
cos2 t
Because sec t > 0 for 0 ≤ t ≤ π/4, we have |r0 (t)| = sec t. It follows
that the length is given by
Z π/4
π/4
√
√
sec t dt = ln | sec t + tan t| = ln( 2 + 1) − ln(1) = ln(1 + 2).
0
0
§10.3, 30. At what point does the curvature of y = ln x have maximum curvature? What happens to the curvature as x → ∞?
Solution. Differentiating, we have (ln x)0 = 1/x and (ln x)00 = −1/x2 .
Using formula 11 on page 711, we have that the curvature is given by
κ(x) =
|−1/x2 |
=
(1 + (1/x)2 )3/2
x2
1
x
=
.
x2 +1 3/2
(x2 + 1)3/2
x2
Differentiating, we have
1(x2 + 1)3/2 − x · 32 (x2 + 1)1/2 · 2x
(x2 + 1)3
(x2 + 1)3/2 − 3x2 (x2 + 1)1/2
=
(x2 + 1)3
x2 + 1 − 3x2
1 − 2x2
=
.
=
(x2 + 1)5/2
(x2 + 1)5/2
√
For x > 0, the latter is zero precisely when x = 1/ 2, which is therefore the only critical point of κ(x). Note that κ0 (x) → 1 as x → 0 and
0
κ0 (1)
√ < 0. Because κ goes from positive to negative as x goes through
1/ 2, this point is a local maximum.
√ Moreover, κ(x) → 0 as x → 0
and κ(x) → 0 as x → ∞. Thus 1/ 2 is the unique global maximum.
As κ(x) → 0 as x → ∞, the graph is ‘flattening out’.
κ0 (x) =
3
§10.3, 46. Find the vectors T, N, and B for r(t) = hcos t, sin t, ln cos ti
at the point (1, 0, 0).
Solution. We have r0 (t) = h− sin t, cos t, − tan ti, and
p
|r0 (t)| = sin2 t + cos2 t + tan2 t = sec t
(note that sec t > 0 for t < π/2, and the point we care about is t =
1 < π/2). Thus
1
h− sin t, cos t, − tan ti = h− sin t cos t, cos2 t, − sin ti.
T(t) =
sec t
This means we have
T0 (t) = h− cos2 t + sin2 t, −2 sin t cos t, − cos ti = h− cos 2t, − sin 2t, − cos ti.
Thus we have
p
cos2 2t + sin2 2t + cos2 t
√
= 1 + cos2 t.
|T0 (t)| =
and thus
1
1
T0 (t) = √
h− cos 2t, − sin 2t, − cos ti.
2
1 + cos t
1 + cos2 t
Plugging in t = 0, we have T(0) = h0, 1, 0i and N(0) = √12 h−1, 0, −1i.
Finally, in order to find B(0) we perform the cross product:
1
B(0) = T(0) × N(0) = h0, 1, 0i × √ h−1, 0, −1i
2
1
= √ h−1, 0, 1i.
2
N(t) = √
§10.3, 54. Show that the osculating plane at every point on the curve
r(t) = ht + 2, 1 − t, t2 /2i is the same plane. What can you conclude
about the curve?
Solution. Differentiating, we have r0 (t) = h1, −1, ti, so that
√
√
|r0 (t)| = 1 + 1 + t2 = 2 + t2 .
This means that
T(t) = √
1
h1, −1, ti.
2 + t2
4
Thus we have
1
1
T0 (t) = − (2 + t2 )−3/2 · 2th1, −1, ti + √
h0, 0, 1i
2
2 + t2
1
−t
h1, −1, ti + h0, 0, 1i
=√
2 + t2 2 + t2
1
−t
t
−t2
=√
,
,
+1
2 + t2 2 + t2 2 + t2 2 + t2
−t
1
t
2
=√
,
,
2 + t2 2 + t2 2 + t2 2 + t2
1
=
h−t, t, 2i.
(2 + t2 )3/2
Now we see that
p
√
2)
√
2(2
+
t
1
2
|T0 (t)| =
t2 + t2 + 4 =
=
,
2
3/2
2
3/2
(2 + t )
(2 + t )
2 + t2
so that
1
N(t) = p
h−t, t, 2i.
2(2 + t2 )
Now we compute a normal to the osculating plane by taking the cross
product T(t) × N(t) (aka B(t)):
1
1
T(t) × N(t) = √
h1, −1, ti × p
h−t, t, 2i
2 + t2
2(2 + t2 )
1
1
·p
=√
h1, −1, ti × h−t, t, 2i
2
2+t
2(2 + t2 )
1
=√
h−2 − t2 , −2 − t2 , 0i
2
2(2 + t )
1
= √ h−1, −1, 0i.
2
This means that the formula for the osculating plane at r(t) is
0 = h−1, −1, 0i · hx − (t + 2), y − (1 − t), z − (t2 /2)i,
which simpifies to
0 = −x + t + 2 − y + 1 − t = −x − y + 3,
evidently independent of t. In fact, what we’ve just checked is that r(t)
is constantly contained in the plane x + y = 3.