Quiz 4 Practice Problems
Practice problems are similar, both in difficulty and in scope, to the type of
problems you will see on the quiz. Problems marked with a ? are ‘‘for your
entertainment’’ and are not essential.
EXPECTED SKILLS:
• Be able to determine whether a curve is smooth.
• Be able to find the arc length of a smooth curve.
• Be able to reparametrize a smooth curve with respect to arc length.
• Be able to compute the unit tangent vector T of a curve.
• Given the arc length parametrization of a curve, know how to find basic
geometric data about the curve such as the unit tangent and at what
point on the curve will the arc length be a specific amount.
• Be able to find traces of quadric surfaces. In particular, be able to
recognize the resulting conic sections in the given plane.
• Given an equation for a quadric surface, be able to recognize the type
of surface and, in particular, its graph.
PRACTICE PROBLEMS:
For problems 1-3, determine whether the curve is smooth.
1. r(t) = ht3 , t4 , t5 i
2. r(t) = ht3 + t, t4 , t5 i
3. r(t) = hcos3 t, sin3 ti
1
For problems 4-8, find the length of the curve.
4. r(t) = h3 cos t, 3 sin ti, 0 ≤ t ≤ 2π
5. r(t) = h2 sin t, 5t, 2 cos ti, –10 ≤ t ≤ 10
6. r(t) = ht2 , sin t − t cos t, cos t + t sin ti, 0 ≤ t ≤ π
7. r(t) = i + t2 j + t3 k, 0 ≤ t ≤ 1
8. r(t) = 12ti + 8t3/2 j + 3t2 k, 0 ≤ t ≤ 1
For problems 9-10, reparametrize the curve with respect to arc
length measured from the point where t = 0 in the direction of
increasing t and then find T(s).
9. r(t) = 2ti + (1 − 3t)j + (5 + 4t)k
10. r(t) = e2t cos 2ti + 2j + e2t sin 2tk
11. Suppose you start at the point (0, 0, 3) and move 5 units along the
curve x = 3 sin t, y = 4t, z = 3 cos t, in the positive direction. Where
are you now?
For problems 12-14, find the unit tangent vector T(t) at the point
with the given value of the parameter t.
12. r(t) = h2 cos t, t, 2 sin ti, t = 0
13. r(t) = cos ti + 3tj + 2 sin 2tk, t = 0
14. r(t) = 2 sin ti + 2 cos tj + tan tk, t = π/4
• Section 11.7 Practice: http://www.math.drexel.edu/classes/math200/
201335/resources/Homework/Chapter11/Homework11.7.pdf
2
• Solutions: http://www.math.drexel.edu/classes/math200/201335/
resources/Homework/Chapter11/Homework11.7_Ans.pdf
? Reparametrize the curve
r(t) =
2
2t
−1 i+ 2
j
2
t +1
t +1
with respect to arc length measured from the point (1, 0) in the direction of
increasing t. Express the reparametrization in its simplest form. What can
you conclude about the curve?
? Determine the radius of the largest circle that can lie on the ellipsoid
x2 y 2 z 2
+ 2 + 2 =1
a2
b
c
(a > b > c).
3
SOLUTIONS
1. r0 (t) = h3t2 , 4t3 , 5t4 i ⇒ r0 (0) = h0, 0, 0i, and so r is not smooth.
2. r0 (t) = h3t2 + 1, 4t3 , 5t4 i. Note that the second and third components of
r0 (t) are 0 only when t = 0. Yet the first component of r0 (0) is 1 6= 0.
Therefore r0 (t) 6= 0 for all t, and so r is smooth.
3. r0 (t) = h3 cos2 t(– sin t), 3 sin2 t(cos t)i ⇒ r0 (0) = h0, 0i, and so r is not
smooth.
4.
2π
Z
||r0 (t)|| dt
L=
Z0 2π
||h–3 sin t, 3 cos ti|| dt
=
0
2π
Z
p
(–3 sin t)2 + (3 cos t)2 dt
=
Z0 2π p
=
9 sin2 t + 9 cos2 t dt
Z0 2π
3 dt = 6π
=
0
5.
Z
10
L=
||r0 (t)|| dt
Z–10
10
||h2 cos t, 5, –2 sin ti|| dt
=
Z–10
10
=
p
(2 cos t)2 + 52 + (–2 sin t)2 dt
Z–10
10 p
=
4 cos2 t + 25 + 4 sin2 t dt
–10
4
Z
10
=
Z–10
10
=
√
√
4 + 25 dt
√
29 dt = 20 29
–10
6.
Z
π
||r0 (t)|| dt
L=
Z0 π
||h2t, cos t − cos t + t sin t, – sin t + sin t + t cos ti|| dt
=
0
Z
π
||h2t, t sin t, t cos ti|| dt
=
0
Z
=
π
p
(2t)2 + (t sin t)2 + (t cos t)2 dt
Z0 π p
=
4t2 + t2 sin2 t + t2 cos2 t dt
Z0 π √
=
4t2 + t2 dt
0
Z π√
=
5t2 dt
Z0 π √
=
5|t| dt
0
Z π√
5 t dt
(since 0 ≤ t ≤ π)
=
0
√ 2 π √
5t 5 2
=
π
=
2 2
0
7.
Z
1
L=
Z0 1
=
||r0 (t)|| dt
||2tj + 3t2 k|| dt
Z0 1 p
=
(2t)2 + (3t2 )2 dt
0
5
Z
1
=
√
4t2 + 9t4 dt
0
Z
1
=
Z0 1
=
p
t2 (4 + 9t2 ) dt
Z
√
2
|t| 4 + 9t dt =
1
√
t 4 + 9t2 dt
0
0
since 0 ≤ t ≤ 1. Now let u = 4 + 9t2 ⇒ du = 18t dt ⇒
1
du
18
= t dt.
Hence
Z
L=
0
1
13
Z 13
√
√
1
8
1
1 2 3/2 2
t 4 + 9t dt =
u = 133/2 −
u du =
18 4
18 3
27
27
4
8.
Z
1
L=
Z0 1
=
||r0 (t)|| dt
||12i + 12t1/2 j + 6tk|| dt
Z0 1 q
=
122 + (12t1/2 )2 + (6t)2 dt
0
Z 1√
=
144 + 144t + 36t2 dt
0
Z 1p
=
36(t2 + 4t + 4) dt
Z0 1 p
=
6 (t + 2)2 dt
Z0 1
=
6|t + 2| dt
0
Z 1
=
6(t + 2) dt
(since 0 ≤ t ≤ 1 ⇒ t + 2 ≥ 0)
0
6
1
= 3t2 + 12t 0
= 3 + 12 = 15
9.
Z
t
s(t) =
||r0 (u)|| du
Z0 t
||2i − 3j + 4k|| du
=
0
Z tp
=
22 + (–3)2 + 42 du
0
Z t
√
4 + 9 + 16 dt
=
0
Z t√
√
=
29 dt = 29 t
0
Since s =
√
√
29 t we have that t = s/ 29 so that
3s
2s
4s
r(s) = √ i + 1 − √
j+ 5+ √
k.
29
29
29
Hence
3
4
2
T(s) = r0 (s) = √ i − √ j + √ k.
29
29
29
10.
Z
s(t) =
t
||r0 (u)|| du
0
7
t
Z
||(2e2u cos 2u − 2e2u sin 2u)i + (2e2u sin 2u + 2e2u cos 2u)k|| du
=
Z0 t p
=
(2e2u cos 2u − 2e2u sin 2u)2 + (2e2u sin 2u + 2e2u cos 2u)2 du
0
Z tp
=
8e4u sin2 2u + 8e4u cos2 2u du
Z0 t √
8e4u du
=
Z0 t √
=
8 e2u du
0
t
√
1 √ 2u = 2 2e = 2 e2t − 1
2
0
So
s=
√
s
2 e2t − 1 ⇒ e2t − 1 = √
2
s
2t
⇒e = √ +1
2
s
⇒ 2t = ln √ + 1
2
1
⇒ t = ln
2
s
√ +1 .
2
so that
r(s) =
s
s
s
s
√ + 1 cos ln √ + 1
i+2j+ √ + 1 sin ln √ + 1
k.
2
2
2
2
Because the derivative of the first component is
1
s
1
s
s
1
√ cos ln √ + 1 − √
√ + 1 sin ln √ + 1
√
2
2
2
2
2
s/ 2 + 1
8
and the derivative of the second component is similar, we have that
T(s) = r0 (s)
1
s
1
s
= √ cos ln √ + 1
− √ sin ln √ + 1
i+
2
2
2
2
1
s
1
s
+ √ sin ln √ + 1
+ √ cos ln √ + 1
k
2
2
2
2
(This is a rare example where the arc length parametrization is worse.)
11. Note that the reference point here is t0 = 0. We therefore want to find
the time t so that s(t) = 5. Since
Z
t
s(t) =
||r0 (u)|| du
Z0 t
||h3 cos u, 4, –3 sin ui|| du
=
0
Z tp
=
(3 cos u)2 + 42 + (–3 sin u)2 du
Z0 t p
=
9 cos2 u + 16 + 9 sin2 u du
Z0 t
√
=
9 + 16 du
0
Z t√
=
25 du
0
Z t
=
5 du = 5t.
0
9
Thus 5t = 5 ⇒ t = 1. Plugging this into the parametric equations of
our curve give (3 sin 1, 4, 3 cos 1).
p
12. r0 (t) = h–2 sin t, 1, 2 cos ti ⇒ ||r0 (t)|| = (2 sin t)2 + 12 + (2 cos t)2 =
√
√
√
4 sin2 t + 1 + 4 cos2 t = 4 + 1 = 5. Hence
h0, 1, 2i
r0 (0)
= √
T(0) = 0
=
||r (0)||
5
13. r0 (t) = – sin ti+3j+4 cos 2tk ⇒ ||r0 (t)|| =
√ 2
sin t + 9 + 16 cos2 2t. Hence
1 2
0, √ , √
5 5
.
p
(– sin t)2 + 32 + (4 cos 2t)2 =
r0 (0)
3j + 4k
3
4
=√
= j + k.
0
||r (0)||
5
5
9 + 16
p
14. r0 (t) = 2 cos ti − 2 sin tj + sec2 tk ⇒ (2 cos t)2 + (–2 sin t)2 + sec4 t =
√
√
4 cos2 t + 4 sin2 t + sec4 t = 4 + sec4 t. Hence
T(0) =
T
π 4
r0 (π/4)
= 0
=
||r (π/4)||
√
√
√
√
2
1 1
1
2 i − 2 j + 2k
2
2
√
= √ i− √ j+ √ k = i− j+ √ k.
2 2
4+4
2 2 2 2 2 2
2
10