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DAWSON COLLEGE
Mathematics Department
Final Examination
Engineering Mathematics II (201-942-DW)
May 16, 2011
Instructor: N. Sabetghadam
Time: 3 Hours
Name:
ID:
Instructions:
- Print your name and ID in the provided space.
- Solve the problems in the space provided for each question and show all your work clearly.
- A Formula sheet is attached.
- Scientific non-programmable calculators are permitted.
- This examination booklet must be returned intact.
This examination consists of 12 questions. Please ensure that you have a complete
examination booklet before starting.
1
1.(5 marks) Evaluate the following limit.
lim xโโ 1
x^2 โ 1 3 x + 3
Solution: (^) xlimโโ 1 x^2 โ 1 3 x + 3 = (^) xlimโโ 1 (x โ 1)(x + 1) 3(x + 1) = (^) xlimโโ 1 x โ 1 3
2.(5 marks) Find the derivative of the following function by using the โdefinition of derivativeโ.
f (x) = 8x^2 โ 5 x + 1
Solution: lim hโ 0
f (x + h) โ f (x) h = lim hโ 0
8(x + h)^2 โ 5(x + h) + 1 โ (8x^2 โ 5 x + 1) h
lim hโ 0
8 x^2 + 16xh + 8h^2 โ 5 x โ 5 h + 1 โ 8 x^2 + 5x โ 1 h = lim hโ 0
16 xh + 8h^2 โ 5 h h = lim hโ 0
h(16x + 8h โ 5) h
hlimโ 0 16 x^ + 8h^ โ^ 5 = 16x^ โ^5 โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ 3.(5 marks) Find the โsecond derivativeโ of the given function.
f (r) = r(2r + 1)^3
Solution: f โฒ(r) = (2r + 1)^3 + 6r(2r + 1)^2 f โฒโฒ(r) = 6(2r + 1)^2 + 6(2r + 1)^2 + 24r(2r + 1)
โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ 4.(5 marks) Evaluate the derivative of the given function at the given point:
y^2 + 2x^2 = 6 at (1, โ2)
Solution: 2 yyโฒ^ + 4x = 0 โโ yโฒ^ = โ 4 x 2 y
5.(10 marks) Given the function f (x) = x^3 โ 3 x;
(a) Find all relative maximum and relative minimum.
Solution: f โฒ(x) = 3x^2 โ 3 = 0 โโ x^2 = 1 โโ x = 1 or x = โ 1 If x = 1 then y = โ2, and if x = โ1 then y = 2. So the points (1, โ2) and (โ 1 , 2) are relative maximum and minimum.
(b) Find the points of inflection.
Solution: f โฒโฒ(x) = 6x = 0 โโ x = 0 and f (0) = 0. So (0, 0) is the point of reflection.
(c) Sketch the graph of f (x).
Solution:
y
s
s s
s
s
10.(25 marks) Evaluate the following integrals:
(a)
( x
2 2
x^2 )dx
Solution:
( x
2 2
x^2 )dx =
( x
2 2
3 6 โ 2 xโ^1 + C
(b)
x^3 (x^4 + 1)^4 dx
Solution:
x^3 (x^4 + 1)^4 dx =
4 x^3 (x^4 + 1)^4 dx = (x^4 + 1)^5 20
+ C
(c)
sin^5 x cos xdx
Solution:
sin^5 x cos xdx = sin^6 x 6
+ C
(d)
xeโx 2 dx
Solution:
xeโx
2 dx =
โ 2 xeโx
2 dx =
2 e
โx^2 + C
(e)
2
x
x
Solution:
2
x
x + 4)dx^ =
2
(xโ^3 /^2 + 4)dx = xโ^1 /^2 โ 1 / 2 + 4x^ |
5 2 =^ โ^2 x โ 1 / (^2) + 4x | 5 2 =^ โ2(5)
(โ2(2)โ^1 /^2 + 4(2)) =
11.(5 marks) Find the area of the region bounded by the graphs y = x^2 + 1 and y = x^3 between x = โ 1 and x = 1.
Solution:
โ 1
x^2 + 1 โ x^3 dx = x^3 3 +^ x^ โ^
x^4 4 |
1 โ 1 =
3 + 1^ โ^
4 โ^ (
(โ1)^3
3 โ^1 โ^
(โ1)^4
4 ) = 2.^67
12.(5 marks) Use the disk method to find the volume of the solid generated when the region enclosed by the curves y =
x, y = 0 and x = 4 is revolved the xโaxis.
Solution: V = ฯ
0
x)^2 dx = ฯ
0
xdx = ฯ x
2 2 |^40 = ฯ(^4
2 2
2 2 ) = 8ฯ
