Winter 2012
Math 255
Problem Set 5
Section 14.3:
5) Reparametrize the curve
2
2t
r(t) = 2
−1 i+ 2
j
t +1
t +1
with respect to arc length measured from the point
(1, 0) in the direction of t. Express the reparametrization in the simplest form. What can you conclude
about the curve?
Solution
At (1, 0), t = 0.
0
r (t) =
|r0 (t)| =
=
=
2 − 2t2
−4t
i+ 2
j
(t2 + 1)2
(t + 1)2
4
1/2
1/2
4t + 8t2 + 4
2
4
2
=
t
+
2t
+
1
(t2 + 1)4
(t2 + 1)2
2
(t2 + 1)
2
2
(t + 1)
2
t2 + 1
Then, the arc length function s is
Z t
Z t
2
0
s=
|r (u)|du =
du = 2tan−1 (t)
2
0
0 u +1
=⇒ t = tan(s/2)
The reparametrization with respect to arc length be1
Winter 2012
Math 255
comes
r(t(s)) =
=
=
=
=
2
2tan(s/2)
−1 i+
2
tan (s/2) + 1
tan2 (s/2) + 1
2tan(s/2)
2
−
1
i
+
j
sec2 (s/2)
sec2 (s/2)
(2cos2 (s/2) − 1)i + 2sin(s/2)cos(s/2)j
1 + cos(s)
2
− 1 i + sin(s)j
2
cos(s)i + sin(s)j
with s ≥ 0. The curve is a circle.
6) At what point does the curve y = lnx have maximum
curvature? What happens to the curvature as
x → ∞?
Solution
curvature_lnx.pdf
The curvature
To find the maximum
of the curvature,
we should
1/x2
x
|f 00 (x)|
check
the critical points.
The end=
=
κ(x) =the endpoints and
2 ]3/2
[1 +
(f 0 (x))
[1 +
1/x2κ(x)
]3/2 →[x02 as
+ seen
1]3/2
points are
x→
0 and
x→∞
where
in the plot above. We can show this by using limits:
x
lim κ(x) = lim
= 0.
x→0
x→0 [x2 + 1]3/2
lim κ(x) = lim
x→∞
x→∞
x
[x2 + 1]3/2
= lim
x→∞ 3
2
1
[x2 + 1]1/2 2x
The critical points occur where
(x2 + 1)3/2 − x · 32 (x2 + 1)1/2 · 2x
κ (x) =
=0
[x2 + 1]3
0
2
= 0.
Winter 2012
Math 255
Then
(x2 + 1)3/2 − x · 23 (x2 + 1)1/2 · 2x
= 0,
[x2 + 1]3
=⇒ −2x2 + 1 = 0,
1
x = ±√ ,
2
1
x = √ ,
2
since ln(x) takes in only positive arguments. Then,
the maximum curvature
2
1
= √ ,
κ √
2
3 3
and κ(x) → 0 as x → ∞.
occurs at √12 , ln √12
7) Find the unit tangent, T, the unit normal, N, and
the binormal vector, B, to r(t) = het , et sint, et costi
at the point (1, 0, 1).
Solution
At (1, 0, 1), t = 0.
T(t) =
r0 (t) =
|r0 (t)| =
=
r0 (t)
|r0 (t)|
het , et sint + et cost, et cost − et sinti
[e2t + e2t (sint + cost)2 + e2t (cost − sint)2 ]1/2
√ t
3e
Then,
1
T(t) = √ h1, sint + cost, cost − sinti,
3
3
Winter 2012
Math 255
so,
1
T(0) = √ h1, 1, 1i,
3
N(t) =
T0 (t) =
|T0 (t)| =
=
T0 (t)
|T(t)|
1
√ h0, cost − sint, −sint − costi
3
1
√ [cos2 t − 2costsint + sin2 t + sin2 t + 2costsint + cos2 t]1/2
r3
2
3
Then,
1
N(t) = √ h0, cost − sint, −sint − costi,
2
so
1
N(0) = √ h0, 1, −1i
2
i j k
1 B(0) = T(0) × N(0) = √ 1 1 1
6 0 1 −1
4
= √1 h−2, 1, 1i
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Winter 2012
Math 255
Section 14.4
8) Use a computer to graph the path of the particle with
acceleration a(t) = ti + t2 j + cos2tk, initial velocity
v(0) = i + k and initial position r(0) = j.
Solution
Z
v(t) =
Z
a(t)dt =
t3
sin2t
t2
k + C1
[ti + t j + cos2tk]dt = i + j +
2
3
2
2
v(0) = i + k = C1
Then
v(t) =
t2
t3
sin2t
+1 i+ j+
+1 k
2
3
2
t2
t3
sin2t
v(t)dt =
+1 i+ j+
+ 1 k dt
2
3
2
3
t4
−cos2t
t
+t i+ j+
+ t k + C2
6
12
4
−1
k + C2
j=
4
1
j+
4
Z
r(t) =
=
r(0) =
C2 =
Z Thus,
r(t) =
4
t3
t
1 − cos2t
+t i+
+1 j+
+t k
6
12
4
which gives the path
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Winter 2012
Math 255
particle_path.pdf
9) Find the tangential and normal components of the
acceleration vector for the position vector
r(t) = ti + cos2 tj + sin2 tk.
Solution
The velocity vector is
v(t) = r0 (t) = i − sin(2t)j + sin(2t)k
a(t) = r00 (t) = −2cos(2t)j + 2cos(2t)k
a = aT T + aN N
v
2sin(2t)cos(2t) + 2sin(2t)cos(2t)
2sin(4t)
p
p
=
=
|v|
1 + 2sin2 (2t)
1 + 2sin2 (2t)



i
j
k

|r0 (t) × r00 (t)|
1
1 −sin(2t) sin(2t) p
=
=
|r0 (t)|
1 + 2sin2 (2t)  0 −2cos(2t) 2cos(2t) 
√
| − 2cos(2t)j − 2cos(2t)k|
2 2cos(2t)
p
=
=p
1 + 2sin2 (2t)
1 + 2sin2 (2t)
aT = a ·
aN
Section 15.1
9) Let f (x, y) = ln(x + y − 1).
(a) Evaluate f (1, 1).
(b) Evaluate f (e, 1).
(c) Find and sketch the domain of f .
(c) Find the range of f .
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Winter 2012
Math 255
Solution
(a) f (1, 1) = ln(1 + 1 − 1) = ln(1) = 0
fdomain_hw5.pdf
(b) f (e, 1) = ln(e + 1 − 1) = ln(e) = 1
(d) The range of f is the range of the natural log
which is −∞ < f (x, y) < ∞
(c) Domain of f is x + y − 1 > 0 or y > 1 − x.
10) Draw a contour map of the function
f (x, y) = y/(x2 + y 2 ) showing at least four level
curves.
contour_map_hw5.pdf
Solution
11) Describe the level surfaces of
(a) f (x, y, z) = x2 + 3y 2 + 5z 2
(b) f (x, y, z) = x2 − y 2
Solution
(a) The level suraces of f (x, y, z) = x2 + 3y 2 + 5z 2
are k = x2 + 3y 2 + 5z 2 , where k ≥ 0. They form
a family of ellipsoids.
(b) The level surfaces of f (x, y, z) = x2 − y 2 are
k = x2 − y 2 where k is any real number. The
level surfaces are a family of hyperbolic cylinders
with axis the z-axis.
Section 15.2
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Winter 2012
Math 255
12) Find the limit, if it exists, or show that the limit
does not exist for
(a)
lim
xycos(x − 2y)
(x,y)→(6,3)
x2 + sin2 y
(b)
lim
(x,y)→(0,0) 2x2 + y 2
Solution
(a)
xycos(x − 2y) = (6)(3)cos(6 − 2(3)) = 9
lim
(x,y)→(6,3)
(b) Approach along x-axis. =⇒ y = 0. Then,
x2
1
x2 + sin2 y
= lim 2 =
lim
2
2
x→0 2x
2
(x,y)→(0,0) 2x + y
Approach along y-axis. =⇒ x = 0. Then,
x2 + sin2 y
sin2 y
2sinycosy
lim
=
lim
=
lim
y→0 y 2
y→0
2y
(x,y)→(0,0) 2x2 + y 2
sin2y
2cos2y
= lim
= lim
=1
y→0 2y
y→0
2
The function has two different limits along two different lines, thus the limit does not exist.
13) Use polar coordinates to find the limit
lim
(x2 + y 2 )ln(x2 + y 2 ).
(x,y)→(0,0)
[Note that if (r, θ) are the polar coordinates of the
point (x, y) with r ≥ 0, then r → 0+ as
(x, y) → (0, 0).]
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Winter 2012
Math 255
Solution
lim
(x2 + y 2 )ln(x2 + y 2 ) = lim+ r2 lnr2 = 2 lim+ r2 lnr
r→0
(x,y)→(0,0)
r→0
lnr
1/r
=
2
lim
r→0 1/r 2
r→0+ −2/r 3
= − lim+ r2 = 0.
= 2 lim+
r→0
14) Use spherical coordinates to find
xyz
.
lim
(x,y,z)→(0,0,0) x2 + y 2 + z 2
Solution
xyz
ρ3 sin2 φcosφsinθcosθ
lim
= lim+
ρ→0
ρ2
(x,y,z)→(0,0,0) x2 + y 2 + z 2
= lim+ ρsin2 φcosφsinθcosθ = 0
ρ→0
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