Download Solutions to Practice Exam 2

Math 21C
1 (15 pts.)
Brian Osserman
Practice Exam 2
Determine the radius and interval of convergence of the power series
∞
X
n(x − 2)n
3n+1
n=1
Solution:
.
First we use the root test to determine for which values of x the
series converges absolutely. The absolute value series is
∞
X
|an | =
n=1
∞
X
n|x − 2|n
n=1
3n+1
.
r
lim
p
n
n→∞
n|x − 2|n
n→∞
3n+1
√
|x − 2| n n
√
= lim
n→∞
3n3
|x − 2|
=
.
3
|an | =
By the root test, the series
P∞
|x − 2| < 3, and diverges if
|x−2|
3
n=1
lim
n
|an | converges if
|x−2|
3
< 1 or equivalently
> 1 or equivalently |x − 2| > 3. Thus the
radius of convergence of the power series is 3.
To nd the interval of convergence we have to test the endpoints, which are
x = −1, 5. At x = −1, the series is
∞
X
n(−3)n
n=1
3n+1
=
∞
X
n(−1)n
n=1
3
,
which diverges because the terms don't go to 0 as n goes to innity. At
x = 5, the series is
∞
∞
X
X
n3n
n
=
,
n+1
3
3
n=1
n=1
which also diverges because the terms don't go to 0. So the interval of
convergence is (−1, 5).
2 (20 pts.)
(a) Find the Taylor series for sin x at x =
π
2
. What is the radius and
interval of convergence?
Solution:
The derivatives of sin x cycle between cos x, − sin x, − cos x,
and sin x. If they are ± cos x, they vanish at x = π2 . If they are ± sin x,
they are ±1 at x = π2 . So the Taylor series is
1
π
1
π
1
π
1 − (x − )2 + (x − )4 + · · · + (−1)n
(x − )2n + . . . .
2
2
4!
2
(2n)!
2
We use the ratio test to nd for which x the series converges absolutely.
∞
X
|an | =
n=1
and
∞
X
n=1
1
π
|x − |2n ,
(2n)!
2
|x − π2 |2n+2 (2n)!
|x − π2 |2
|an+1 |
=
=
,
|an |
|x − π2 |2n+2 (2n + 2)!
(2n + 2)(2n + 1)
so
|x − π2 |2
1
|an+1 |
π
= lim
= |x− |2 lim
= 0.
n→∞ (2n + 2)(2n + 1)
n→∞ |an |
2 n→∞ (2n + 2)(2n + 1)
lim
Thus, by the ratio test the series converges absolutely for all x (the
radius of convergence is innite).
(b) If you approximate sin x using the Taylor polynomial of order 4 centered at x =
π
2
, what is a (reasonable) bound for the error when
|x − π2 | < 1?
Solution:
The fth derivative of sin x is cos x, and | cos x| ≤ 1 for all
x. Thus the error of the approximation at x = b is at most
1
|b
5!
− π2 |5 .
If we want an error bound which holds for all x with |x − π2 | < 1, we
nd the error is at most
1
5!
=
1
120
.
Note: this can be improved by observing that the Taylor polynomial
of degree 4 for sin x centered at x =
π
2
is also the Taylor polynomial
of degree 5, so we can similarly obtain a sharper error bound of 6!1 |x −
π 6
|
2
<
1
6!
.
2
3 (15 pts.)
Let u = h0, 1, 1i, v = h1, 1, 0i.
(a) Using projection, write u as a sum of a vector parallel to v and a vector
orthogonal to v.
Solution:
Note that u = projv u + (u − projv u), and by denition
projv u is parallel to v and u − projv u is orthogonal to v. Now,
u·v
v
|v|2
0·1+1·1+1·0
=
h1, 1, 0i
12 + 12 + 02
1
h1, 1, 0i
=
2
1 1
=
, ,0 .
2 2
Then u − projv u = h0, 1, 1i − 12 , 21 , 0 = − 21 , 21 , 1 , so we have
1 1
1 1
u=
, ,0 + − , ,1 ,
2 2
2 2
projv u =
where the rst is parallel to v and the second is orthogonal to it.
(b) What is the angle between u and v?
Solution:
u·v
) = cos−1 ( √02 +112 +12 √2 ) = cos−1 ( 12 ) =
The angle is cos−1 ( |u||v|
60◦ .
3
4 (15 pts.)
Let M be the plane passing through the three points P = (2, 0, 0), Q =
(0, 2, 0), and R = (0, 0, 2).
(a) What is the equation for M ?
Solution:
−→
−→
The vectors P Q = h−2, 2, 0i and P R = h−2, 0, 2i both
lie on the plane, so we can nd a normal vector by taking their cross
product. We have
i j k
−→ −→ P Q× P R = −2 2 0 = (4−0)i+(0−(−4))j+(0−(−4))k = h4, 4, 4i .
−2 0 2 Since M contains (2, 0, 0), we can write its equation as 4(x − 2) + 4y +
4z = 0, or 4x + 4y + 4z = 8, or x + y + z = 2.
(b) What is the distance from the origin to M ?
Solution:
The distance from the origin O to M is the length of the
−→
−→
−→
projection of P O onto the normal vector n = P Q × P R. This length
is
−→
| − 2 · 4 + 0 · 4 + 0 · 4|
8
2
|P O · n|
√
=
= √ =√ .
|n|
42 + 42 + 42
4 3
3
4
5 (10 pts.)
Find a parametric representation for the line which is the intersection of
the two planes x + 2z = 1 and x + y − z = 0.
Solution:
We rst nd a point which is in both planes by solving the
equations simultaneously. The rst equation gives x = 1 − 2z , so if we
substitute into the second we get (1 − 2z) + y − z = 0, or y = 3z − 1. If we
put z = 0 (for example), we get x = 1 and y = −1, so the point (1, −1, 0)
lies on both planes.
Next, the planes have normal vectors h1, 0, 2i and h1, 1, −1i, so the line of
intersection is perpendicular to both vectors, which means it is parallel to
i j k h1, 0, 2i × h1, 1, −1i = 1 0 2 1 1 −1
= (0 · (−1) − 2 · 1)i + (2 · 1 − 1 · (−1))j + (1 · 1 − 0 · 1)k
= −2i + 3j + k.
So a parametric representation for the line is
r(t) = h1, −1, 0i + t h−2, 3, 1i ,
or
x = 1 − 2t, y = −1 + 3t, z = t.
5
6 (10 pts.)
A particle moves with position at time t given by
r(t) = t2 i + (cos 3t)j + (sin 3t)k.
Find the velocity, speed, and direction of the particle at all times t.
Solution:
The velocity v(t) is r0 (t) = 2ti − 3(sin 3t)j + 3(cos 3t)k. The
speed is |v(t)| =
v
|v(t)|
7 (15 pts.)
=
√ 2t
i
4t2 +9
−
√
4t2 + 9 sin2 3t + 9 cos2 3t =
3(sin 3t)
√
j
4t2 +9
+
√
4t2 + 9. The direction is
3(cos 3t)
√
k.
4t2 +9
Two projectiles are red at the same time. One is red from (1, 0, 0) with
initial velocity h5, 10, 15i, while the other is red from (0, 1, 0) with initial velocity h10, 10, 10i. At what time are the projectiles closest together?
What is their closest distance?
Solution:
The motion equations for the two projectiles are
1
1
r1 (t) = i + t(5i + 10j + 15k) − gt2 k = (1 + 5t)i + 10tj + (15t − gt2 )k
2
2
and
1
1
r2 (t) = j + t(10i + 10j + 10k) − gt2 k = 10ti + (1 + 10t)j + (10t − gt2 )k,
2
2
and their distance at time t is
|r1 −r2 | = |(1−5t)i+(−1)j+(5t)k| =
p
(1 − 5t)2 + (−1)2 + (5t)2 =
√
2 − 10t + 50t2 .
This function is minimized when its square 2 − 10t + 50t2 is minimized,
which (taking the derivative) is when −10 + 100t = 0, or t =
1
10
. At this
time, the distance between the projectiles is
√
r
2 − 10t + 50t2 =
1
1
2 − 10( ) + 50(
)=
10
100
6
r
1
2−1+ =
2
r
3
.
2