C MATH*2080 SAMPLE F

Sample Final Exam Page 1 of 12
CALCULUS MATH*2080 SAMPLE FINAL EXAM
Prof. R.Gentry
Print Your Name
SIGNATURE
Student No.
Mark
This exam is worth 45% of your final grade.
In Part I
you will be asked to indicate a choice for the correct
procedure. Please indicate only one answer that you
believe is correct. If more than one answer choice is
indicated the problem will be considered wrong.
In Part II
you need not evaluate the given integral, simply indicate .if it is an
improper integral how it should be evaluated.
In part III
you should give complete answers in the space provided.
In part IV
you should circle the correct multiple choice. If more than one
answer is circled the problem will be marked wrong.
You may not utilize any calculator or computing device.
No cell phones or pagers may be used during the examination.
You may utilize a 3" by 5" facts card for formulas, etc., that has your
name and ID number and is turned in with the exam.
There are ___ pages including this cover page.
Sample Final Exam Page 2 of 12
4 pts
For each integral indicate in space proceeding the integral which of the following
methods should be used first to evaluate the integral. Write only one letter in the
space provided. Each problem is worth 0.5 points.
Part I
DO NOT EVALUATE THE INTEGRALS, IT WILL NOT INCREASE YOUR MARK.
CHOICES are denoted (a), (b), ...etc.
a.
A linear substitution u = ax + b
b.
A non-linear substitution u = a((xr + b
c.
The substitution x = a@sin(θ)
d.
The substitution x = a@tan(θ)
e.
Integration by Parts
f.
Partial Fractions
g.
Division
h.
i.
(1)
B
(3)
C
(5)
G
(7)
E
Apply a standard integration
rule. State the Rule.
None of the above
m
x(x 1/41) dx
9x
m x
m
m
2
dx
2 x3
65x x
cos2(x) dx
2
dx
(2)
D
(4)
E
(6)
F
(8)
H
2
x 9 4x dx
m
m
x2 e-0.5x dx
x 2
m x 4x 12
2
m
x2(9 - x3)2 dx
dx
Sample Final Exam Page 3 of 12
4 pts
Part II
In the box below each integral indicate if it is “improper”? If it is, indicate how it should be evaluated.
You do not need to complete the evaluations. Each problem is worth 0.5 points.
10
II-(1)
1
m x 1/3 2
1
: Improper
"
dx
II-(2)
m
x 2e xdx
0
9 Not Improper
: Improper
9 Not Improper
"
5
25 x
m
x
II-(3)
2
II-(4)
dx
m
sin(x) e cos(x) dx
0
1
9 Improper
10
II-(5)
m
1
: Not Improper
: Improper
2
dx
9 Not Improper
II-(6)
x
dx
m ln(x)
1
: Improper
9 Not Improper
x 5
dx
m x 2 4x 12
5
: Improper
9 Not Improper
0
e
II-(7)
9 Not Improper
0
1
65x x
: Improper
II-(8)
m
x 2 e xdx
"
: Improper
9 Not Improper
Sample Final Exam Page 4 of 12
27 pts Part III
Solve each problem in the space allotted.
2 pts
The concentration of a drug in an individual's blood was monitored following an injection at time
t = 0 and the data fit the curve C(t) = 25 t e-0.5t µg/l (t in hrs.). What is the individual’s average blood
concentration of this drug over the 24 hours following the injection?
Part III 1.
Average blood concentration is A =
And to evaluate
integration by parts. Set u = 25t and vN = e-0.5t. Then uN = 25 and v = -2e-0.5t
we use
(Note: 1/-0.5 = -2). So,
I 25t e-0.5t dt = -50 t e-0.5t - I -50 e-0.5t dt = -50 t e-0.5t - 100 e-0.5t = -50[ t + 2] e-0.5t
= -50[ t + 2] e-0.5t |t = 0t = 24 = -50@26 e-12 +100 = 100 [-13e-12 + 2] So A = 100 [2 - 13e-12 ]/24
If you use a calculator you will find that A . 4.16633 .
2 pts
Part III 2.
If a sequence {Xn} is generated by the rule that the next term is the sum of the present term and
twice the previous term, with X1 = 1 and X0 = 1, what is the value of x100? (I know, it is a big
number!)
x100 = ____________________________________
Model: Present term = Xn next term = Xn+1 and previous term = Xn-1.
Dynamical equation: Xn+1 = Xn + 2 Xn-1 or Xn+2 - Xn+ 1 - 2 Xn = 0.
Characteristic equation is λ2 - λ - 2 = 0. This factors as (λ -2)(λ +1) = 0, giving λ = -1 and λ = 2.
General form of the solution is thus Xn = c1 (-1)n + c2 2n.
X0 = c1 (-1)0 + c2 20 = c1 + c2
and
X1 = c1 (-1)1 + c2 21 = -c1 + 2c2
Setting X0 = 1 gives c1 = 1 - c2.
Substituting this into the equation 1 = X1 = -c1 + 2c2 gives 1 = -1 + 3 c2.
Hence, c2 = 2/3 and c1 = 1/3. Thus, Xn = 1/3 (-1)n + 2/3 2n.
Finally, X100 = 1/3 (-1)100 + 2/3 2100 = [1 + 2101 ]/3
Sample Final Exam Page 5 of 12
3 pts
Part III 3.
a.
Determine the general solution of the given equation:
yN + t y = 0
y(t) =
Use the formula for yN = a(t) y
b.
yO - 4y = 0
y(t) =
y = C e1 a(t) dt
a(t) = -t
C1 e2t + c2 e-2t
The characteristic equation λ2 - 4 GIVES λ = 2 AND λ = -2
c.
yON + 8yO + 12yN = 0
The characteristic equation is
y(t) = c1 e0t + c2 e-2t + c3 e-6t
λ3 + 8 λ2 + 12 λ = 0
Which factors as λ(λ2 + 8λ + 12) = λ(λ + 2)( λ + 6) = 0
3 pts
Part III 4.
Consider the equation yO + a yN + b y = 0 a and b parameters.
a.
What condition on a and b will ensure that the solution y(t) can be y = c1t ?
For this to happen λ = 0 must be a repeated root. Thus the characteristic equation must be λ2 = 0 and
hence we conclude that a = b = 0,
b.
What condition on a and b will ensure that lim y(t) = 4 ?
t$"
For the limit at infinity to equal infinity at least one of the eigenvalues must be positive, as terms
such as e-2t approach zero. Furthermore the solution can not oscillate so it can not be complex,
As λ = 0.5[ -a ± (a2 - 4b)½] we conclude that a > 0 and b 0.5 a2
c.
What condition on a and b will ensure that the solution y(t) will be positive for all t if y(0) = 5 ?
The solution can not have an oscillatory factor since these become negative. Hence b < 0.25a2
Beyond this, the problem becomes difficult. The answer depends on the coefficients A function
like y = 10e-2t - 5e-0.2t starts at (0,5) and then becomes negative before it approaches the t-axis. I will
not give this question!!!
Sample Final Exam Page 6 of 12
3 pts
Part III 5.
Compute the average of F(x,y) = x y over the circle of radius 1 centered at the origin. First sketch
this region to see how it looks and determine its boundary curves. Show all necessary work.
y
4
The circular region is
R = {(x,y) | -1 # x # 1 and -(1 - x2)½ # y # (1 - x2)½ }
The average A is the integral of F over R divided by the area
of R, which is 2π.
3
2
1
-4
-3
-2
-1
1
2
3
4
-1
=
x
-2
-3
=
=
-4
=0
Consequently the average value is just zero.
2 2x
3 pts
Part III 6.
Sketch the region of integration of
m m
F(x,y) dy dx and determine the equivalent iterated
0 1x
integral(s) with the order of integration reversed.
y
Note y = 1- x gives x = 1 -y and y = 2 - x gives x = 2 -y. Easy!
3
The region R = {(x, y) | 0 # x # 2 and 1-x # y # 2-x }
must be split into THREE regions to integrate first with respect
to x:
R1 = {(x,y) | 1 # y # 2 and 0 # x # 2 -y }
plus
R2 = {(x,y) | 0 # y # 1 and 1 - y # x # 2 - y }
R1
1
R2
-3
-2
1 R3 2
-1
-1
R3 = {(x,y) | -1 # y # 0 and 1 - y # x # 2 }
-2
Thus the integral is
F(x,y) dx dy +
2
F(x,y) dx dy +
F(x,y) dx dy
-3
3
x
Sample Final Exam Page 7 of 12
3 pts
Part III 7.
a.
Consider the surface z = x3 - 75x + y2 - 4y.
What are the critical (x, y) points of this surface?
zx = 3x2 - 75 and zy = 2 y - 4 are defined everywhere so the only critical points are determined from the
solutions of zx = zy = 0. These are y = 2 and x = ±5. Critical points are (5,2) and (-5,2).
b.
Determine if the stationary points correspond to local maximum or minim values. I.e.,
Classify the critical points using the Second Partial Derivative Test.
As A = zxx = 6x
B = zxy = 0 and C = zyy = 2, the discriminate D = B2 - AC = -12x.
At the point (5,2), D = -60 < 0 and A > 0 so the surface has a local minimum.
At the point (-5,2), D = 60 > 0 and by the second partial derivative test so the surface has a saddle point and
neither a local minimum nor a local maximum.
2pts
Part III 8.
What is the equation of the tangent plane to the surface z = x sin(πy) at the point (2, 1)?
Standard form of Tangent Plane: z = z0 + zx(x0,y0) (x - x0) + zx(x0,y0) (y - y0)
zx = sin(πy) and zy = πx cos(πy)
_________________________________________
z(2,1) = 2sin(π) = 0
zx(2,1) = sin(π) = 0 and zy(2,1) = π2 cos(π) = -2 π
So the tangent plane is z = 0 + 0 (x - 2) + -2 π(y - 1) or simply z = -2 π(y - 1)
Sample Final Exam Page 8 of 12
6 pts
Part III 9.
Consider the non-linear dynamical system XN = X - XY2 YN = 3Y - 2XY.
a.
Determine all equilibriums of this system. Equilibriums are where XN = 0 and YN = 0. Factoring we see that
XN = X - XY2 = X(1 - Y2) = 0 Y X = 0 or Y = 1 or Y = -1
YN = 3Y - 2XY = Y(3 - 2X) = 0 Y Y = 0 or X = 3/2.
Therefore the equilibrium points are: (0,0), (3/2,1) and (3/2,-1).
b.
Determine the linear system obtained by linearizing these equations about one of the equilibriums with a
positive Y-value. Show all necessary work. The point is (3/2, 1)
Linear Model xN = 0x - 3y
yN = -2x + 0 y
We first linearize F(X,Y) = X - XY2 and G(X,Y) = 3Y - 2XY about (3/2, 1)
FX(X,Y) = 1 - Y2 FY(X,Y) = - 2XY
GX(X,Y) = - 2Y
FX(3/2, 1) = 1 - 12 = 0 FY(3/2, 1) = - 2 @3/2 @1 = -3
c.
GY(X,Y) = 3 - 2X
GX(3/2, 1) = - 2@1 = -2
What is the solution of the linear system you found in part b?
Differentiating xN = -3y gives xO = -3 yN Substitute yN = -2x to get xO = -3@(-2)x
This equation has the characteristic equation λ2 - 6 = 0 or λ = ±
Its solution is thus
+ c2
x(t) = c1
Since y = -xN/3 this gives
2 pts
GY(3/2, 1) = 3 - 2@3/2 =
0
Part III 10.
y(t) = c1
/3
- c2
or xO - 6x = 0.
.
/3
Use the Lagrange method to find the maximum of the function F(x,y) = xy for
points lying on the ellipse x2 + 4y2 = 1.
The constraint function is g(x) = x2 + 4y2 - 1. The Lagrange equations are Fx = λgx, Fy = λgy and g = 0.
These become y = 2λx x = 8λy and x2 + 4y2 - 1 = 0
The first two equations give y = 16λ2 y. Hence, either y = 0 or λ = 1/4 or λ = -1/4.
Case 1. If y = 0 the constraint equation gives x = ±1. Two critical points (1,0) and (-1,0): F(±1, 0) = 0.
Case 2.
If λ = 1/4 then y = x/2, which when substituted into the constraint equation gives x2 + 4x2/4 = 1 or
x2 = 1/2. This has two roots, x = +(1/2)½ and x = -(1/2)½ . Evaluating the function F we get
F((1/2)½, (1/2)½/2) = (1/2)½(1/2)½ /2 = 1/4 and F( -(1/2)½ , -(1/25)½ /2) = -(1/2)½ @ (-(1/2)½ /2) = 1/4.
Case 3. If λ = -1/4 then y = -x/2 and again substituting this into the constraint equation gives x2 + 4x2/4 = 1 or
x2 = 1/2. This has two roots, x = +(1/2)½ and x = -(1/2)½ , but this time the y-values have the
opposite sign and hence evaluating the function we find
F((1/2)½, -(1/2)½/2) = (1/2)½(-(1/2)½ /2) = -1/4 and F( -(1/2)½ , (1/2)½ /2) = -(1/2)½ @ (1/2)½ /2 = -1/4.
So, the maximum value of F(x,y) subject to the constraint is 1/4, which occurs at two critical points.
Sample Final Exam Page 9 of 12
10 pts Section IV
Part IV 1.
Multiple choice. Circle on this exam your choice for the correct answer.
The problem will be marked wrong if more than one answer is circled.
Each problem has only one correct answer and is worth one point.
lim
x$0
a.
0
b.
4
±
c.
1/2
d.
-1/2
e.
1/4
x@ln(1x)
equals
cos(2x)1
"
Part IV 2.
m
cos(3x)((e-2x dx =
0
a.
4
±
b.
2/13
c.
5/13
d.
-1/13
e.
1/13
Part IV 3.
To evaluate
4
m x@ 259x2
a.
x = 5 sin(3θ)
b.
x = (3/5) tan(θ)
c.
x = tan(5θ/3)
d.
x = (5/3) sin(θ)
±
e.
x = (5/3) tan(θ)
dx the correct substitution would be
Sample Final Exam Page 10 of 12
Part IV 4.
The solution of the differential equation yN = 3y - 6 for which y(0) = 4 is
a.
y(t) = 4 e3t
b.
y(t) = e3t + 3
c.
y(t) = 2e-3t + 2
±
d.
y(t) = 2e3t + 2
e.
y(t) = 5 e-3t - 9
Part IV 5.
The general solution of the equation yO + 2yN + 5y = 0 is
±
a.
y(t) = e-t [C1 cos(2t) + C2 sin(2t)]
b.
y(t) = [C1 + C2 t]e-t
c.
y(t) = C1 e-2t + C2 e2t
d.
y(t) = C1 cos(2t) + C2 sin(2t)
e.
y(t) = C1 e-2t + C2 et
If F(x,y) = cos(πxy) + x2y, then LF(1, 0.5) equals
Part IV 6.
a.
-1/4
b.
(0.5π + 1, 0.5π + 1)
c.
(-π + 1, -0.5π + 1)
d.
0.5π(x + y)cos(0.5πxy) + 2xy + x2
±
e.
(-0.5π + 1, -π + 1)
f.
[ln(t) - 1/t]/t - [sin(t) - e-sin(t)]cos(t)
Sample Final Exam Page 11 of 12
Part IV 7.
If F = x2y + exy, which of the following is false?
a.
Fx = y(2x + exy)
b.
Fy = x(x + exy)
c.
Fxy = Fyx
d.
Fyy = x2exy
±
e.
Fxx = y2exy
Part IV 8.
If u = (3, 1) and v = (-3, 2), which of the following is false
a.
uCv = -7
b.
**v** = SQRT(13)
±
c.
u is perpendicular to (1, 3)
d.
v is parallel to (18, -12)
e.
v + u = (0, 3)
Part IV 9.
If F(x,y) = xe-2y and d = (3,-1) then the directional derivative DdF(2,0) equals
a.
(-1,4)
b.
7
c.
1
10
±
d.
e.
5
10
Sample Final Exam Page 12 of 12
Part IV 10.
The equation of the plane normal to the vector (3,-1,2) that contains the point (1,-1,3) is
a.
x - 1 + y+1 + z -3 = -6
b.
x - 1 - y+1 + 2z -3 = 0
c.
3x - y + 2z = -6
d.
3(x - 1) - (y+1) + 2(z -3) = -6
±
e.
3(x - 1) - (y+1) + 2(z -3) = 0