May 2016 - John Abbott Home Page

D
M
epartment of
athematics
Final Examination
16 May 2016
14h–17h
Mathematical Models
201-225-AB
Instructor: R.Masters, A.Panassenko
Student name:
Student number:
Instructor:
Instructions
1. Do not open this booklet before the examination begins.
2. Check that this booklet contains 6 pages, excluding this cover page and the formula sheet.
3. Write all of your solutions in this booklet and show all supporting work.
4. If the space provided is not sufficient, continue the solution on the opposite page.
Final Examination
201-225-AB
16 May 2016
(4) 1. Use Newton’s method to find the root of f (x) = x3 − 3x2 + 3 that is between 2 and 3. Give your answer
accurate to 3 decimals.
2. Find y 0 . Do not simplify your answer.
√
(3)
(a) y = arcsin( x)
(3)
(b) y = (arctan(ex ))3
(3)
(c) y = sec−1 (3x2 + 1)
(3)
(d) tan(x−1 y) = x2 ey
Hint: Solve for y 0 .
Z6
(4) 3. Use Trapezoidal Rule to approximate
sin cos(x) dx, using n = 6 (give your answer to 3 decimals)
3
y
(3) 4. Find the area of the region enclosed by y = x3 ,
the y-axis and y = −27. See figure at right.
0
x
f (x) = x3
y = −27
(4) 5. The volume and radius of a cylinder are increasing at a rate of 50π cm3 /s and 2 cm/s respectively. At what
rate is the height of the cylinder changing dh/dt, when the volume is 36π cm3 and the radius is 3 cm? (Recall:
V = πr2 h) Hint: use implicit differentiation.
6. Given f (x) =
(x − 2)(2x − 1)
(x + 1)2
f 0 (x) =
9(x − 1)
(x + 1)3
and f 00 (x) =
18(2 − x)
(x + 1)4
Find (if any):
(1) (a) The x and y intercept(s).
(1) (d) The local (relative) maxima and minima
(1) (b) The vertical and horizontal
asymptotes.
(1) (e) The inflection points.
(1) (c) The intervals on which f is increasing
or decreasing
(1) (f) Intervals of upward or downward
concavity.
(4) (g) Sketch the graph of f .
7. Evaluate the following limits.
cos(x − 2) + 2x − 5
(3)
(a) lim
x→2
x − 4 + 2ex−2
Page 1 of 6
Final Examination
16 May 2016
1
x−1
x
(3)
(b) lim
(3)
(c) lim x csc(x)
x→1
201-225-AB
x→0
(3) 8. If a resistor of R omhs is connected across a battery of E volts with internal resistor r omhs, then the power
E2R
. If E and r are constants such that E = 16V and
(in watts) in the external resistor is given by P =
(R + r)2
r = 8Ω, but R varies, what is the maximum possible value of P ? Hint: Find dP/dR.
(3) 9. Determine if y = ex sin(2x) + 4 is a solution to the differential equation y 00 − 2y 0 + 5y = 20.
y
10. Let R be the region bounded by the functions f (x) =
π
2 − 2 sin x, y = 0 and 0 ≤ x ≤ . Set up (but do
2
not evaluate) the integrals to find the volume of the
solid of revolution obtained by revolving R about;
2.0
f (x)
1.0
R
−1.0
0
1.0
2.0
3.0 x
−1.0
(2)
(a) the y−axis
(2)
(b) the x−axis
(2)
(c) the line y = −1
(4) 11. Solve the following differential equation for y.
dy
(ln y) dx
− xy = 0;
12. Integrate the following integrals.
Z
(4)
(a)
ex cos x dx
Z
(4)
(b)
(4)
(c)
(4)
(d)
Z
(4)
(4)
(4)
x3
dx
+1
x2
Z p
x2 − 4x + 7 dx
Z
ln(ln x)
dx
x
Z
sin5 x cos3 x dx
Z
x+5
dx
x3 + 2x2 − 3x
(e)
(f)
(g)
cos2 x dx
Page 2 of 6
√
with initial condition y( 2) = e
Final Examination
201-225-AB
16 May 2016
(4) 13. Solve the following first order linear differential equation for y.
dy
+ (sec x)y = cos x
with initial condition x = 0 when y = 5/2
dx
14. Given the function and its graph.
3.0
f (x)
2.0
(
−1,
f (x) =
1,
if −π ≤ x < 0
if 0 ≤ x < π
1.0
x
−π
0
π
−1.0
(2)
(a) Determine if the function is even or odd.
Show your work or Explain to obtain full marks
(4)
(b) Find the first three non-zero terms of the Fourier series for the function above and write the function
expansion.
Page 3 of 6
Final Examination
y − y0 ≈
yrms
f 0 (x
0 )(x
201-225-AB
− x0 )
v
u ZT
u
u1
y 2 dx
=t
T
Zb
f (x1 )
; x2 = x1 − 0
f (x1 )
1
; VC =
C
16 May 2016
Z
f (x) dx = F (b) − F (a)
;
udv = uv −
;
vdu
a
Z
Z
i dt
Z
; s=
Z
v dt
; v=
Z
a dt
; q=
i dt
0
Zb
b−a
2n
h
i
f (x0 ) + 2f (x1 ) + 2f (x2 ) + 2f (x3 ) + · · · + 2f (xn−1 ) + f (xn )
b−a
3n
h
i
f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + · · · + 4f (xn−1 ) + f (xn )
f (x) dx ≈
a
Zb
f (x) dx ≈
a
Zb
V =
π
2
[outer radius] − [inner radius]
2
Zb
dx
;
2π [radius] × [height] dx
V =
a
a
csc(x) =
1
sin(x)
;
1
cos(x)
sec(x) =
sin2 (x) + cos2 (x) = 1
Z
Z
√
cot(x) =
; 1 + tan2 (x) = sec2 (x)
cos(2x) = cos2 (x) − sin2 (x)
Z
;
;
√
Z
secn−2 (x) dx
a2 + x2 → sub x = a tan(θ)
y = e−
R
P (x)dx
Z
;
1 − cos(2x)
2
sin2 (x) =
csc(x) dx = ln csc(x) − cot(x) + C
1
n−2
sec (x) dx =
secn−2 (x) tan(x) +
n−1
n−1
−→
;
Z
n
;
sin(x)
cos(x)
sec(x) dx = ln sec(x) + tan(x) + C
;
a2 − x2 → sub x = a sin(θ)
tan(x) =
Z
;
cot(x) dx = ln sin(x) + C
;
cot(x) =
cos(x)
sin(x)
1 + cot2 (x) = csc2 (x)
;
sin(2x) = 2 sin(x) cos(x)
tan(x) dx = ln sec(x) + C
dy
+ P (x)y = Q(x)
dx
1
tan(x)
R
Q(x) e
P (x)dx
;
;
cos2 (x) =
1 + cos(2x)
2
(for n > 2)
√
x2 − a2 → sub x = a sec(θ)
dx
f (x) = a0 + a1 cos(x) + a2 cos(2x) + · · · + an cos(nx) + · · · + b1 sin(x) + b2 sin(2x) + · · · + bn sin(nx) + · · ·
1
a0 =
2π
Zπ
−π
f (x) dx
;
1
an =
π
Zπ
f (x) cos(nx) dx
1
; bn =
π
−π
Zπ
−π
Page 4 of 6
f (x) sin(nx) dx
Final Examination
201-225-AB
Answers
1. 2.532
1
2. (a) y 0 = p
2 x(1 − x)
3ex arctan2 (ex )
1 + e2x
6x
p
(c) y 0 =
2
(3x + 1) (3x2 + 1)2 − 1
(b) y 0 =
(d) y 0 =
2xey + x−2 y sec2 (x−1 y)
sec2 (x−1 y)x−1 − x2 ey
3. -0.350
4. 60.75
5. 2/9
6. (a) x-int = 2, x -int = 1/2 and y-int = 2
(b) V.A. x = −1 ; H.A. y = 2
(c) Increasing: −∞ < x < −1 and 1 < x < ∞
Decreasing: −1 < x < 1
(d) Local min: x = 1
(e) I.P. x = 2
(f) C.U. ] − ∞, −1[∪] − 1, 2[
C.D. ]2, ∞[
7. (a) 2/3
(b) e
(c) 1
8. 8 watts
9. yes
π/2
Z
x(2 − 2 sin x) dx
10. (a) Shell: V = 2π
Z
0
π/2
(2 − 2 sin x)2 dx
(b) Disk: V = π
0
Z
(c) Washer: V = π
π/2
[(3 − 2 sin x)2 − 1] dx
0
√
11. y = e
x2 −1
1
1
x + sin 2x + C
2
4
1
(b) ex (cos x + sin x) + C
2
1
(c) (x2 − ln(x2 + 1)) + C
2
12. (a)
Page 5 of 6
16 May 2016
Final Examination
201-225-AB
p
1
3
(d) (x − 2) x2 − 4x + 7 + ln
2
2
√
x2 − 4x + 7 + x − 2
√
3
!
(e) ln x(ln(ln x) − 1) + C
sin6 x sin8 x
−
+C
6
8
5
1
3
(g) − ln x + ln(x + 3) + ln(x − 1) + C
3
6
2
(f)
13.
16 May 2016
x − cos x + 7/2
sec x + tan x
14. (a) f (x) is odd
(b) coefficients: b1 = 4/π; b3 = 4/ 3π; b5 = 4/5π
∞
4X 1
function: f (x) =
sin(nx)
π
2n − 1
n=1
Page 6 of 6
+C