Calculus II (MAC2312-02)
Test 2 (2015/06/25)
Name (PRINT):
Please show your work. An answer with no work receives no credit.
You may use the back of a page if you need more space for a problem.
You may not use any calculators.
Page
Points Score
2
26
3
20
4
20
5
14
6
20
Total:
100
Page 1 of 6
Calculus II (MAC2312-02)
Page 2 of 6
1. Circle True of False.
(a) (3 points) TRUE
False The differential equation
Solution: Note that
dz
dt
dz
= −ez+t is separable.
dt
= −ez+t = −ez et .
(b) (3 points) TRUE False The hydrostatic pressure at a point in a liquid is proportional
to depth and is independent of direction.
Solution: Hydrostatic pressure P is given by ρgd, where d is the depth of the point at
which we want the pressure. It is independent of direction (e.g., it is the same in x, y,
and z directions at any point).
Z
(c) (5 points) True
FALSE
0
3
dx
= (ln |x − 1|)30 = ln(2).
x−1
Solution: This is an improper integral, since x = 1 is a vertical asymptote in the
interval [0, 3]. So,
Z 3
Z t
Z 3
dx
dx
dx
:= lim−
+ lim+
,
t→1
s→1
0 x−1
0 x−1
s x−1
if both of the integrals on the right-hand side converge. But
Z t
dx
= lim− (ln |x − 1|)t0 = lim− ln |t − 1| = −∞.
lim−1
t→1
t→1
t→1
x
−
1
0
FALSE Let a and b be real numbers. If a ≤ b, then a2 ≤ b2 .
(d) (5 points) True
Solution: The implication only holds if 0 ≤ a ≤ b, since x2 is an increasing function for
nonnegative real numbers. On the other hand, −2 ≤ −1 but (−2)2 = 4 6≤ (−1)2 = 1.
(e) (5 points) TRUE
functions and 0 ≤ f (x) ≤ g(x)
Z ∞ False Suppose f and g are Zcontinuous
∞
for all x ≥ 1. If
g(x) dx is convergent, then
f (x) dx is convergent.
1
1
Solution: This is the comparison test for convergence of integrals of nonnegative functions. If the integral of the larger function converges, then so does that of the smaller
function.
2. (5 points) Find the volume of the solid resulting from rotating the region bounded by the curve
x2 + (y − 6)2 = 4 around the x axis.
Hint: A theorem of Pappus (which connects centroids and volumes of revolution ) may help.
Solution: The equation x2 + (y − 6)2 = 22 represents a circle with center at (0, 6) and
radius 2 (and area A = πr2 = π(2)2 = 4π). The centroid of a circle is at its center (due to
symmetry), which is at a distance d = 6 from the axis of rotation (the x axis). So, according
to a theorem of Pappus, the volume of the solid resulting from rotating the region bounded
by this circle around the x axis is equal to the distance traveled by the centroid of the circle
(2πd) times the area of the circle (A), V = 2πdA = 2π(6)(4π) = 48π 2 .
Calculus II (MAC2312-02)
Page 3 of 6
Z
3. Consider the definite integral I :=
1
2
e−x dx.
0
(a) (10 points) Use Simpson’s rule with 5 points to approximate I. There is no need to evaluate
the final sum.
Hint: The integral of a parabola with heights ystart , ymid , and yend at evenly-spaced points over an
interval of length ` is `( 16 ystart + 64 ymid + 16 yend ).
Solution: With 5 points we have n = 4 subintervals, so the points are ∆x =
apart.
1−0
4
=
1
4
2(1/4) −(0)2
2
2
2
2
(1e
+ 4e−(1/4) + 2e−(2/4) + 4e−(3/4) + 1e−(4/4) )
6
1
= (1 + 4e−1/16 + 2e−1/4 + 4e−9/16 + e−1 ) ≈ 0.715878
12
SI =
(b) (10 points) Estimate the error involved in approximating I using the trapezoidal rule with
5 points.
Hint: If |f 00 (x)| ≤ K for all a ≤ x ≤ b, then |ET | ≤
K(b−a)3
,
12n2
where n is the number of trapezoids.
2
2
2
Solution: The integrand is f (x) := e−x . So, f 0 (x) = −2xe−x and f 00 (x) = −2(e−x +
2
2
2
x(−2xe−x )) = −2(1−2x2 )e−x . To find a loose bound for |f 00 (x)| = |−2(1−2x2 )e−x | =
2
2|1−2x2 |e−x , note that |1−2x2 | ≤ |1|+|−2x2 | = 1+2x2 , which is at most 1+2(1)2 = 3
2
in [0, 1]. Also, e−x is at most e−(0) = 1 in that interval. So, |f 00 (x)| ≤ 2(3)(1) = 6 := K1
and
1
6(1 − 0)3
=
= 0.0625.
|ET | ≤
2
12(4)
16
We can find a tight bound for |g(x)| := |f 00 (x)| in [0, 1] by finding its global maximum
2
−x2
in that
The derivative g 0 (x) = f 000 (x)
is zero at x = 0 and
p interval. p
p = −4(2x
p − 3)xe
√
x = 3/2. But 3/2 is not in [0, 1] (since 3/2 > 2/2 = 1 = 1). So, there are no
critical points and hence no extrema (local or global) in the interior of the interval. At
the end points x = 0 and x = 1 we have g(0) = f 00 (0) = −2 and g(1) = f 00 (1) = 2/e.
So, the global maximum of |f 00 (x)| is | − 2| = 2 := K. We could have seen this without
2
differentiating g(x) = f 00 (x) too. Both (1 − 2x2 ) and e−x are decreasing in [0, 1] (hence,
2
so is their product). So, the maximum of |f 00 (x)| = | − 2(1 − 2x2 )e−x | must happen at
one (or both) of the end points of the interval. In any case, this tight bound gives a
better error estimate:
|ET | ≤
1
2(1 − 0)3
=
≈ 0.020833.
2
12(4)
48
Calculus II (MAC2312-02)
Page 4 of 6
Z
4. (10 points) Find the values of p for which the integral
e
Solution:
Z ∞
e
∞
dx
converges.
x(ln x)p
Z t
dx
dx
:=
lim
t→∞ e x(ln x)p
x(ln x)p
Z t
1
= lim
(ln x)−p dx
t→∞ e x
Z t
(ln x)−p d(ln x) (substituting u := ln x “on the fly”)
= lim
t→∞ e

t
lim
(ln x)−p+1
,
p 6= 1
t→∞
−p+1
=
e
t
lim
p=1
t→∞ (ln(ln x))e ,
(
1
limt→∞ ((ln t)1−p − 1) ,
p 6= 1
1−p
=
∞,
p = 1.
Since limt→∞ ln t = ∞, the remaining limit is finite (and the given integral is convergent) only
if the power 1−p of lnZt is negative (so that ln t goes to the denominator). This gives 1−p < 0,
∞
1
1
1
dx
1−p
:=
lim
(ln
t)
−
1
=
(0
−
1)
=
.
i.e., p > 1, for which
x(ln x)p
1 − p t→∞
1−p
p−1
e
5. (10 points) Find the length of the curve y = ln(sec x) when 0 ≤ x ≤ π/4.
Hint: An antiderivative for sec x is ln | sec x + tan x|.
1
dy
= (sec x tan x)
= tan x.
dx
sec x
Solution: We have
Z
π/4
L :=
Z
0
π/4
ds
r
=
1+(
0
Z
π/4
=
dy 2
) dx
dx
p
1 + tan2 x dx
0
Z
π/4
=
√
sec2 x dx
0
Z
π/4
| sec x| dx
=
0
Z
π/4
sec x dx (since sec x > 0 for 0 ≤ x ≤ π/4)
√
√
= (ln | sec x + tan x|)π/4
=
ln
|1/(1/
2)
+
1|
−
ln
|1
+
0|
=
ln(1
+
2)
0
=
0
Calculus II (MAC2312-02)
Page 5 of 6
6. (8 points) Verify that the function y(t) = −t cos t − t is a solution of the initial-value problem
dy
t = y + t2 sin t, y(π) = 0.
dt
Solution: First, note that the given function satisfies the initial condition, y(π) = 0, since
y(π) = −π cos(π) − π = π − π = 0. To see if it also satisfied the rest of the differential
dy
= −(cos t + t(− sin t)) − 1 = − cos t + t sin t − 1. So, the left hand
equation, we find
dt
dy
side is t
= −t cos t + t2 sin t − t, which is identical to the right-hand side y + t2 sin t =
dt
−t cos t − t + t2 sin t.
7. (6 points) For what values of k does the function y(t) = cos kt satisfy the differential equation
d2 y
4 2 = −25y?
dt
d dy
d d(cos kt)
d(−k sin kt)
d2 y
=
=
= −k 2 cos kt. Substituting
Solution: We have 2 =
dt
dt dt
dt
dt
dt
in the differential equation we get
d2 y
= −25y
dt2
−4k 2 cos kt = −25 cos kt,
4
that is, (4k 2 −25) cos kt = 0, for which to hold for all t we need (4k 2 −25) = (2k−5)(2k+5) =
0, giving k = ±5/2.
Calculus II (MAC2312-02)
Page 6 of 6
8. (10 points) The arc of the parabola y = x2 from (1, 1) to (2, 4) is rotated about the y axis. Find
the area of the resulting surface.
dy
= 2x. So,
dx
Z 2 p
Z 2
Z 2 r
dy 2
x 1 + (2x)2 dx
S = 2π
x ds = 2π
x 1 + ( ) dx = 2π
dx
1
1
1
Z
2π 2 √
8x 1 + 4x2 dx
=
8 1
Z
2π 2
=
(1 + 4x2 )1/2 d(1 + 4x2 ) (substituting u := 4x2 “on the fly”)
8 1
√
π
π √
2π 2
(1 + 4x2 )3/2 |21 = (173/2 − 53/2 ) = (17 17 − 5 5)
=
8 3
6
6
Solution: We have
9. (10 points) Solve the initial-value problem
dL
= L2 ln t,
dt
L(1) = −8.
Hint: Integration by parts may help with one of the integrals.
Solution: This is a separable differential equation:
dL
= ln t dt
2
Z L
Z
dL
= ln t dt
L2
Z
Z
−2
L dL = ln t dt,
integrating the right-hand side by parts (with u := ln t and dv := dt, so that du =
v = t) we get,
−1
−L
Z
= uv −
Z
v du = t ln t −
−L−1 = t ln t − t + C
−1
= t ln t − t + C.
L
From the initial condition L(1) = −8, we have
−1
= (1) ln(1) − 1 + C,
−8
that is, C = 9/8. So, L(t) =
−1
1
=
.
t ln t − t + 9/8
t − t ln t − 9/8
t
dt
t
dt
and
t