Math 222-004 – Spring 2014
Practice Final
1. Circle TRUE if the statement is always true and circle FALSE otherwise. If the statement is false,
explain why or give an example.
P
P
P
(a) If
an does not converge and
bn does not converge, then
an b n
does not converge
The sum
P
1/n diverges, but
P
True
False
True
False
True
False
True
False
1/n2 converges.
(b) The position vectors for the points (1, 4, −1) and (2, 1, 6) are perpendicular.
(c) The integral
Z
7
−7
cos
πx 14
(1 + x2 )3/2 dx is positive.
(d) If a sequence sn is convergent, then the terms sn tend to 0 as n tends
to ∞
Sequences can converge to any value, not just 0.
1
2. The marginal propensity to consume, c, is the percentage of every dollar that consumers will spend
(rather than save). Suppose that the U.S. government estimates that spending will decrease by 900
billion dollars over the next year. In order to compensate for this, the government passes a 780
billion dollar stimulus package. So the government will spend 780 billion dollars. Each dollar spent
is received by someone, who will spend c of that dollar. So in addition to 780 billion dollars, c780
billion dollars are also spent. Of the c780 billion dollars spent, each dollar is received, and c of each
of those dollars is spent, so an additional c2 780billion dollars is spent, and so on.
(a) If c = 0.5, what is the total contribution to spending from the stimulus package?
The total stimulus spending will be the total:
780 + 780(.5) + 780(.5)2 + 780(.5)3 + · · · =
∞
X
780(.5)n
n=0
This series is geometric, so we can write the total of the sum
∞
X
780(.5)n =
n=0
780
= 1560.
1 − .5
So the total spending will be 1.56 trillion dollars.
(b) Given your answer to the previous question, will the stimulus package be sufficient to overcome
the 900 billion dollar decrease in spending?
Yes, if c = .5, the stimulus package will make up for the decrease in spending.
(c) Suppose that the government measured c and chose the 780 billion dollar stimulus plan to
exactly compensate for the 900 billion dollar drop in spending. What value of c did they
obtain?
In order to get the exact value of c, we go back to the geometric series. Since
∞
X
780cn =
n=0
, solving for c, we get c = 1 −
780
900
= .1333.
2
780
= 900
1−c
3. Newton’s law of universal gravitation states that the force due to gravity on an object r meters
from the center of the earth changes at a rate proportional to the force times the inverse of the
distance.
(a) Write a differential equation for the force F on an object r meters from the center of the earth.
We are thinking of force as a function of r, so we have
F
dF
=k
dr
r
(b) At the surface of the earth (6, 378, 100 meters from the center), the force due to gravity on an
object of 10kg mass is 980 Newtons. Use this information to solve an initial value problem for
the force on the object. You will have one unknown constant remaining.
The differential equation above is separable, so we will have
F
dF
=k
dr
Zr
Z
kdr
dF
=
F
r
ln |F | = k ln |r| + C
F (r) = Crk
To solve the initial value problem, we plug in the initial distance r and the initial force:
F (6378100) = C(6378100)k = 980
980
⇒C =
(6378100)k
Then, the solution is F (r) = 980
k
r
6378100
(c) What happens to the force due to gravity on this object as r goes to infinity?
The force due to gravity as r tends to infinity depends on the sign of the constant of proportionality k. As objects get further from the earth, we expect that the gravity will decrease,
that is that k is negative. Therefore, as r → ∞, F (r) tends to 0.
3
4. Find the radius and interval of convergence for the following series.
∞
X
2n xn
(a)
n!n2
n=0
We will use the ratio test:
n+1 n+1 2 x
n+1
(n+1)!(n+1)2 n2
n!
= lim 2
lim n
n
2 x
n→∞ 2n (n + 1)! (n + 1)2 |x|
n→∞ n!n2
2
n2
|x| = 0
n→∞ (n + 1) (n + 1)2
= lim
The radius of convergence for this series is ∞, and the interval is (−∞, ∞).
(b)
∞
X
(−1)n xn
Again, we use the ratio test:
n ln n
n=2
(−1)n+1 xn+1 (n+1) ln(n+1) ln n
n ln n
|x| = |x| lim
lim (−1)n xn = lim
n→∞
n→∞ n→∞
(n
+
1)
ln(n
+
1)
ln(n
+ 1)
n ln n
= |x| lim
n→∞
1/n
= |x|
1/(n + 1)
Therefore the radius of convergence in 1, so we will need to
the end points ±1. At
P check
1
x = −1, we need to determine the convergence of the series
.
n ln n This series has positive
terms and n ln n < (n + 1) ln(n + 1), the terms are decreasing. Then we can use the integral
test:
Z b
Z ∞
dx
dx
= lim
b→∞
x
ln
x
x
ln x
2
2
b
= lim ln(ln x)|2 = ∞.
b→∞
The improper integral diverges, so by the integral test, the series diverges.
At the other endpoint, when x = 1, we have an alternating series. Since the terms are positive
and n ln n < (n + 1) ln(n + 1), by the alternating series test implies that the series converges.
Thus the interval of convergence is (−1, 1].
(c)
∞
X
xn
n2 5 n
n=1
By the ratio test,
xn+1
(n+1)2 5n+1 |x|
n2 5 n
= |x| lim
=
lim n
x
2
n+1
n→∞
n→∞ (n + 1) 5
5
n2 5n
This series converges for |x| < 5, so the radius of convergence is 5. Then, we need to test ±5.
P 1
P (−1)n
When x = 5, we get the series
n2 , which converges. When x = −5, the series is
n2 ,
which converges in absolute value and therefore converges.
4
5. When dealing with the positions of objects in a distant solar system, place the origin at the star at
the center.
(a) If two planets, P1 and P2 , are at points (10, 1, 5) and (−2, 4, 1). Write the position vectors p~1
and p~2 for the two planets.
Since these points are given with respect to the origin, the position vectors are




10
−2
p~1 =  1  and p~2 =  4 
5
1
(b) Solar systems organize themselves into planes. Write an expression for plane containing both
planets and the star.
To find an expression for the plane, we need a point and a normal vector. The star at the
center of the solar system clearly lies in the plane, so we can take the origin as our base point.
The normal vector will be the cross product of the position vectors for p~1 and p~2 . Therefore,
~i ~j ~k 10 1 5 = ~i(1 − 20) − ~j(10 + 10) + ~k(40 + 2)
−2 4 1 

−19
p~1 × p~2 =  −20  .
42
The expression for the plane is then


−19
~x ·  −20  = 0 or − 19x − 20y + 42z = 0
42
(c) Comets travel in enormously large elliptic orbits in the universe. On the scale of solar systems,
the trajectory of a comet looks like a line. If the trajectory of a comet near the solar system
described in this problem is given by




4
1
~c(t) =  2  + t  1  ,
1
−2
does the comet ever lie in the same plane as the planets and star?
To decide if the comet hits the plane of the solar system, we simply plug ~c(t) into the equation
for the plane:

 
 

−19
4+t
−19
~c(t) ·  −20  =  2 + t  ·  −20 
42
1 − 2t
42
= −19(4 + t) − 20(2 + t) + 42(1 − 2t)
−19t − 20t − 84t − 76 − 20 + 42 = 0
−123t = 74 ⇒ t =
5
−74
123
Therefore, the comet passes through the point
418 172 272
,
,
123 123 123
in the plane containing the solar system.
6
6. Given a function f (t), we can find the average value of the function over the interval [0, x] using
the following expression:
Z
1 x
A(x) =
f (t)dt.
x 0
(a) Find a differential equation satisfied by A(x).
Differential equations are equations satisfied by derivatives, so let’s take a derivative:
Z x
1
1
′
f (t)dt
A (x) = f (x) − 2
x
x 0
Notice that the second term has exactly A(x), se we have
A′ (x) =
f (x)
1
− A(x).
x
x
(b) What kind of differential equation did you find in (a)?
The equation certainly isn’t separable, but it does have A(x) times x1 , so it’s first order linear.
(c) Given that A(0) = 1, solve your differential equation. How does your answer compare to the
given A(x)?
Since the differential equation is first order linear, we use an integrating factor. In particular,
1
f (x)
A(x) =
x
x
f
(x)
1 d
A(x)eln x =
eln x dx
x
d
(xA(x)) = f (x)
dx
Z
A′ (x) +
x
f (t)dt
xA(x) − 0A(0) =
0
A(x) =
1
x
Z
x
f (t)dt
0
You’ll notice that we get exactly the same answer as before! You should be a little freaked
out by this, because as x goes to 0, that factor in front becomes division by zero. However,
for some choices of f (t) this actually works out, such as f (t) = t.
7
7. A badger, in Peru for a study-abroad semester, discovers a pond hidden high in the Andes mountains. The pond has perfectly circular
p horizontal cross-sections, and its radii r at different depths
y are given by the function r(y) = 36 − y 2 . At the surface, the pond has a radius of 6 meters.
(a) Sketch the pond, clearly labeling your variables. Which direction should you slice, and what
shape do you see?
p
We can see from the formula r(y) = (36 − y 2 ) that the maximum depth of the pond is 6 m.
We are given that the slices are circles, so we get
p
Volume of slice = πr2 ∆y = π( 36 − y 2 )2 ∆y.
y=0
r
∆y
y=6
(b) Set up an integral that gives the total volume of the pond. Evaluate the integral to find the
total volume of the pond.
Since the range of y is (0, 6), using our slice from a), we have:
Total Volume = π
Z
6
36 − y 2 dy
0
6
y3 = 36y − 3 0
63
= π 36(6) −
3
8