MATH 2413 HW 2.7
Due Friday, October 2
(1) A 13-ft ladder is leaning against a vertical wall when someone pulls the foot of the
ladder away from the wall at a rate of 0.5 ft/s. How fast is the top of the ladder
sliding down the wall when the foot of the l
MATH 2413 HW 3.1
Due Monday, October 5
(1) Make a rough sketch of the function. y = 2(1 ex )
Find the limit.
(2) lim e3x
x
(3) lim+ e5/(3x)
x3
e3x e3x
x e3x + e3x
(4) lim
3 + 2x
x 5 10x
(5) lim
(6) lim 5x 7x
x
(7)
lim
x(/2)+
etan
2
2x
(8) lim e
x
x
2
3
MATH 2413 HW 2.6
Due Monday, September 28
For problems (1) - (6), find dy/dx by implicit differentiation.
(1) y 5 + x2 y 3 = 1 + x4 y
(2) y sin(x2 ) = x sin(y 2 )
(3)
x + y = 1 + x2 y 2
(4) x sin y + y sin x = 1
(5) tan(x y) =
y
1 + x2
(6) y 3 + x3 y 2 =
12.4: Double Integral Applications Name:
Important Formulas:
ZZ
Mass: M =
(x, y) dA
D
ZZ
ZZ
Moments: Mx =
y(x, y) dA and My =
D
x(x, y) dA
D
Center of Mass: (
x, y) where x = My /M and y = Mx /M
ZZ
ZZ
2
Moments of inertia: Ix =
y (x, y) dA, Iy =
x2 (x
Name:
MATH 2414: Third Exam Solutions
October 31, 2014
PLEASE TURN OFF ALL ELECTRONIC DEVICES. THE USE OF
CALCULATORS, NOTES, BOOKS AS WELL AS ANY OTHER
ASSISTANCE ARE NOT PERMITTED DURING THIS EXAMINATION.
Be sure to answer all of the questions. Please c
Name:
MATH 2414: Quiz 1 Solutions
September 5, 2014
Z
1. (4 points) Calculate
1
Solution: Let u =
2
e1/x
dx.
x2
1
1
, then du = 2 dx. Therefore
x
x
Z 1/2
Z 2 1/x
e
eu du
dx =
2
x
1
1
1/2
= eu 1
= (e1/2 e)
= e e1/2 .
Points earned:
Z
2. (8 points) Find
ar
Name:
MATH 2414: Second Exam Solutions
October 10, 2014
PLEASE TURN OFF ALL ELECTRONIC DEVICES. THE USE OF
CALCULATORS, NOTES, BOOKS AS WELL AS ANY OTHER
ASSISTANCE ARE NOT PERMITTED DURING THIS EXAMINATION.
Be sure to answer all of the questions. Please
Name:
MATH 2414: Quiz 4 Solutions
October 3, 2014
1. (8 points) Find the volume
of the solid obtained by rotating the region enclosed by the
graphs of y = x and y = x about the line y = 1.
Solution: Since x =
x implies x = 0 or x = 1, we have the followi
Name:
MATH 2414: Quiz 6 Solutions
October 24, 2014
1. (8 points) Determine whether the series is convergent or divergent. If it is convergent,
then find its sum.
X
2
n2 1
n=2
Solution: The form of the partial fraction decomposition is
n2
2
2
A
B
=
=
+
.
1
Name:
MATH 2414: Fourth Exam
December 3, 2014
PLEASE TURN OFF ALL ELECTRONIC DEVICES. THE USE OF
CALCULATORS, NOTES, BOOKS AS WELL AS ANY OTHER
ASSISTANCE ARE NOT PERMITTED DURING THIS EXAMINATION.
Be sure to answer all of the questions. Please check that
Name:
MATH 2414: First Exam Solutions
September 19, 2014
PLEASE TURN OFF ALL ELECTRONIC DEVICES. THE USE OF
CALCULATORS, NOTES, BOOKS AS WELL AS ANY OTHER
ASSISTANCE ARE NOT PERMITTED DURING THIS EXAMINATION.
Be sure to answer all of the questions. Please
Name:
MATH 2414: Quiz 7 Solutions
November 7, 2014
1. (5 points) Find a power series representation for the function and determine the interval
of convergence.
2
f (x) =
3x
Solution: Recall that
X
1
=
un ,
1 u n=0
where |u| < 1. Factor
of equation ():
2
3
Name:
MATH 2414: Quiz 3 Solutions
September 26, 2014
1. (6 points) Find the area of the region enclosed by the graphs of y = 12 x2 and y =
x2 6.
Solution: Since 12 x2 = x2 6 implies x = 3 or x = 3 we have the following
picture:
15 y
y = x2 6
y = 12 x2
10
Name:
MATH 2414: Quiz 5 Solutions
October 17, 2014
1. (5 points) Determine whether the sequence converges or diverges. If it converges, find
the limit.
n2
an =
n3 + 4n
Solution: Divide the numerator and the denominator by the highest power of n
that occu
Name:
MATH 2414: Quiz 9 Solutions
November 24, 2014
1. (6 points) Find
dy
d2 y
and
. For which values of t is the curve concave upward?
dx
dx2
x = t2 + 1,
y = t2 + t
Solution: The curve will be concave upward when
dy
=
dx
and
d2 y
=
dx2
d dy
dt dx
dx
dt
=
Name:
MATH 2414: Quiz 8 Solutions
November 14, 2014
1. (7 points) Find the Taylor series for f (x) centered at the given value of a.
f (x) = e2x ,
a=3
Solution: Begin by computing the derivatives of f and their values at a = 3:
f (x)
= e2x
f (2)
= e6
f 0
MATH 2414: Quiz 2 Solutions
September 12, 2014
Z
1. (10 points) Find
x2 9
dx.
x3
Solution: Let x = 3 sec . Then dx = 3 sec tan d and
x2 9 = 9 sec2 9 = 9 tan2 = 3 tan .
Therefore
Z
x2 9
dx =
x3
=
=
=
=
=
Z
3 tan
3 sec tan d
(3 sec )3
Z
1
tan2
d
3
sec2