Download MATHEMATICS 3202

MATHEMATICS 3202
ASSIGNMENT 3 - SOLUTION
Due on Monday, February 1, 2016; Submitted to Department Exercise Box 33
Name
M.U.N. Number
1. Describe the projections of the circle r(t) = hsin t, 0, 4 + cos ti onto the coordinate planes.
Solution [p.728-Ex.12; SM-p.472]
The projection onto the xy-plane is traced by hsin t, 0, 4 + cos ti, which is the segment [−1, 1] on the x-axis. The
given circle is contained in the xz-plane, hence the projection on the sz-plane is the circle itself. We identify this
circle. Since x = sin t and z = 4 + cos t, we have x2 + (z − 4)2 = sin2 t + cos2 t = 1. This is the circle of radius 1
centered at h0, 0, 4i. The projection onto the yz-plane is traced by h0, 0, 4 + cos ti, which is the segment [3, 5] on
the z-axis.
2. i) Parametrize the intersection of the surfaces y 2 − z 2 = x − 2 and y 2 + z 2 = 9 using t = y as the parameter; ii)
find a parametrization of the curve in i) using trigonometric functions.
Solution [p.729-Ex.21,22; SM-pp.474-475]
p
i) Solve for z and x in terms of y. From y 2 + z 2 = 9 we get z = ± 9 − y 2 ; and
from y 2 − z 2 = x − 2 we have
√
x = 2 + y 2 − z 2 = 2y 2 − 7. Considering t = y as a parameter, we find z = ± 9 − t2 and x = 2t2 − 7, thereby
getting the vector parametrization
p
r(t) = h2t2 − 7, t, ± 9 − t2 i for t ∈ [−3, 3].
ii) The curve in i) is the intersection of the surfaces y 2 − z 2 = x − 2 and y 2 + z 2 = 9. Note that the circle
y 2 + z 2 = 9 is parametrized by y = 3 cos t, z = 3 sin t. Substituting in the first equation and using the identity
cos2 t − sin2 t = cos 2t yields
x = 2 + y 2 − z 2 = 2 + (3 cos t)2 − (3 sin t)2 = 2 + 9 cos 2t.
Below is the trigonometric parametrization: r(t) = h2 + 9 cos 2t, 3 cos t, 3 sin ti.
3. Use hyperbolic functions to parametrize the intersection of the surfaces x2 − y 2 = 4 and z = xy.
Solution [p.729-Ex.26; SM-p.477]
x = 2 cosh t & y = 2 sinh t satisfy the equation of the cylinder since x2 − y 2 = 4(cosh2 t − sinh2 t) = 4.
To find a parametrization of the curve of intersection, we substitute x = 2 cosh t & y = 2 sinh t in the equation of
the surface z = xy and solve for z. This gives z = 4 cosh t sinh t. We obtain r(t) = h2 cosh t, 2 sin tt, 4 sinh t cosh ti.
4. In i) and ii) below evaluate the derivative by using the appropriate Product Rule, where r1 (t) = ht2 , t3 , ti and
d 4
d
r2 (t) = he3t , e2t , et i: i) dt
(t r1 (t)); ii) dt
(r1 (t) · r2 (t))
assuming that r01 (2) = h2, 1, 0i and r02 (2) = h1, 4, 3i.
t=2
Solution [p.739-Ex.18,20; SM-p.489]
i)
d 4
dt4
(t r1 (t)) = t4 r01 (t) +
r1 (t) = t4 h2t, 3t2 , ti + (4t3 )ht2 , t3 , ti = h6t5 , 7t6 , 5t4 i.
dt
dt
ii)
d
d
d
(r1 (t) · r2 (t))
= ( (r1 (t))) · r2 (t)
+ ( r2 (t)) · r1 (t)
t=2
t=2
t=2
dt
dt
dt
2
3t 2t t 2 3
= h2t, 3t , 1i · he , e , e i
+ ht , t , ti · h3e3t , 2e2t , et i
t=2
6
4
t=2
2
= h2, 1, 0i · he , e , e i + h1, 4, 3i · h4, 8, 2i
= 2e6 + e4 + 4 + 32 + 6
= 2e6 + e4 + 42.
5. Compute
d
dt (r1 (t)
× r2 (t))
t=1
in two ways: i) Calculate r1 (t) × r2 (t) and differentiate; ii) Use the Product Rule.
Solution [p.739-Ex.22; SM-p.490]
i) since r1 (t) × r2 (t) = (et − 4t)i + (2t − t2 et )j + (2t2 − 1)k, it follows that
d
= (e − 4)i + (2 − 3e)j + 4k.
= (et − 4)i + (2 − 2tet − t2 et )j + (4t)k
(r1 (t) × r2 (t))
t=1
t=1
dt
ii) Using the Product Rule we see
d
d
d
= r1 (t) × r2 (t)
+ (r1 (t)) × r2 (t)
(r1 (t) × r2 (t))
t=1
t=1
t=1
dt
dt
dt
and so
d
= (e − 4)i + (2 − 3e)j + 4k.
(r1 (t) × r2 (t))
t=1
dt
6. Let v(s) = s2 i + 2sj + 9s−2 k. Evaluate
d
ds v(g(s))
at s = 4. assuming that g(4) = 3 and g 0 (4) = −9.
Solution [p.739-Ex.28; SM-p.492]
Applying the Chain Rule we have
d
2/3k, we get ds
v(s)
s=4
d
ds v(s)
= g 0 (s)v0 (g(s)), and
d
ds v(s)s=4
= −9v0 (3). Also since v0 (3) = 6i + 2j −
= −9v0 (3) = −54i − 18j + 6k.
7. In i)-ii)-iii)-iv), evaluate the integrals: i) 01 h(1 + s2 )−1 , s(1 + s2 )−1 i ds; ii)
R1
Rt
−2 −4 −5
2
1/2 hu , u , u i du; iv) 0 (3si + 6s j + 9k) ds.
R
R1
0
2
(te−t i + t(ln(1 + t2 ))j) dt; iii)
Solution [p.739-Ex.40,42,44,46; SM-pp.495-497]
i)
Z 1
Z 1
h(1 + s2 )−1 , s(1 + s2 )−1 i ds = h
0
(1 + s2 )−1 ds,
0
Z 1
s(1 + s2 )−1 dsi = hπ/4, 2−1 ln 2i.
0
ii)
Z 1
2
(te−t i + t(ln(1 + t2 ))j) dt = h2−1 (1 − e−1 ), −1/2 + ln 2i.
0
iii)
Z 1
Z 1
hu−2 , u−4 , u−5 i du = h
1/2
u−2 du,
1/2
Z 1
u−4 du,
1/2
Z 1
u−5 duih1, 7/3, 15/4i.
1/2
iv)
Z t
0
Z t
(3si + 6s2 j + 9k) ds = (
Z t
(3s)ds)i + (
0
0
(6s2 )dsj + (
Z t
0
9ds)k) = (3t2 /2)i + (2t3 )j + (9t)k.
8. In i)-ii), compute the length of the curve over the given interval: i) r(t) = 2ti − 3tk, 11 ≤ t ≤ 15; ii) r(t) =
(2t2 + 1)i + (2t2 − 1)j + t3 k, 0 ≤ t ≤ 2.
Solution [p.744-Ex.2,4; SM-p.509-510]
i) x(t) = 2t, y(t) = 0, z(t) = −3t gives s =
R 15
11
√
kr0 (t)kdt = 4 13.
ii) Since r0 (t) = h4t, 4t, 3t2 i, we use u = 32 + 9t2 to get
Z 2
s=
kr0 (t)k dt =
Z 2 p
t 32 + 9t2 dt =
0
0
Z 68
√ du
u
32
18
≈ 14.063.
9. Find an arc length parametrization of the cycloid with parametrization r(t) = ht − sin t, 1 − cos ti.
Solution [p.745-Ex.26; SM-p.517-518]
Step 1. Find the Rinverse of the arc length function. We are given r0 (t) = h1 − cos t, sin ti and kr0 k = 2 sin(t/2).
This gives s(t) = 0t kr0 (u)k du = 4(1 − cos t/2). Then solve s(t) for t: t = 2 arccos(1 − s/4).
Step 2. Reparametrize the curve using the t we just found. We will write this as t = 2 arccos(1 − s/4) and solving,
let us first note that
p
sin t = 8−1 (4 − s) 8s − s2 & cos t = 2((4 − s)/4)2 − 1.
So then, rewriting r(t) = ht − sin t, 1 − cos ti, and
p
r1 (t) = h2 arccos(1 − s/4) − 8−1 (4 − s) 8s − s2 , 2 − 8−1 (4 − s)2 i.
10. Express the arc length L of y = x3 for 0 ≤ x ≤ 8 as an integral in two ways, using the parametrizations r1 = ht, t3 i
and r2 = ht3 , t9 i. Do not evaluate the integrals, but use substitution to show that they yield the same result.
Solution [p.745-Ex.28; SM-p.518-519]
√
√
For r1 = ht, t3 i we have r01 = h1, 3t2 i and hence kr01 k = 1 + 9t4 . Similarly, we have kr02 k = 9t4 + 81t16 . The
length s may be computed using the two parametrizations by the following integrals (note that in the second
parametrization 0 ≤ t3 ≤ 8 hence 0 ≤ t ≤ 2)
Z 8
s=
0
kr01 k =
Z 8p
0
1 + 9t4 dt & s =
Z 2
0
kr02 k =
Z 2p
0
1 + 9t12 3t2 dt =
Z 8p
1 + 9u4 du.
0
This shows that the arc length is independent of the parametrization we choose for the curve.