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Past Test 2 with Answers - Calculus for Life Sciences II | MATH 2122, Exams of Mathematics

Material Type: Exam; Professor: Benson; Class: Calculus for the Life Scie II; Subject: Mathematics; University: East Carolina University;

Typology: Exams

2011/2012

Uploaded on 05/08/2012

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MATH 2122 SECTION 1, TEST #2 SOLUTIONS
SPRING 2012
(1) State the Fundamental Theorem of Calculus in its two forms.
First Form of the Fundamental Theorem of Calculus:
Let fbe a continuous function and let
A(x) = Zx
a
f(t)dt.
Then A0(x) = f(x). That is, the area function A(x) is an antiderivative for
f(x).
Second Form of the Fundamental Theorem of Calculus:
Let fbe a continuous function and suppose that g0(x) = f(x). That is,
suppose that g(x) is an antiderivative or f(x). Then
Zb
a
f(x)dx =g(b)โˆ’g(a).
(2) Let A=๎˜”1 2
3 4 ๎˜•,B=๎˜”1โˆ’1
โˆ’2 2 ๎˜•,v=๎˜”2
3๎˜•. We compute ...
(a) 2Aโˆ’B=๎˜”2(1) โˆ’1 2(2) โˆ’(โˆ’1)
2(3) โˆ’(โˆ’2) 2(4) โˆ’2๎˜•=๎˜”1 5
8 6 ๎˜•
(b) AB =๎˜”1(1) + 2(โˆ’2) 1(โˆ’1) + 2(2)
3(1) + 4(โˆ’2) 3(โˆ’1) + 4(2) ๎˜•=๎˜”โˆ’3 3
โˆ’5 5 ๎˜•
(c) Av =๎˜”1(2) + 2(3)
3(2) + 4(3) ๎˜•=๎˜”8
18 ๎˜•
(d) (A+B)v=๎˜”2 1
1 6 ๎˜•๎˜” 2
3๎˜•=๎˜”2(2) + 1(3)
1(2) + 6(3) ๎˜•=๎˜”7
20 ๎˜•
1
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pf4
pf5

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MATH 2122 SECTION 1, TEST #2 SOLUTIONS

SPRING 2012

(1) State the Fundamental Theorem of Calculus in its two forms.

First Form of the Fundamental Theorem of Calculus: Let f be a continuous function and let

A(x) =

โˆซ (^) x

a

f (t) dt.

Then Aโ€ฒ(x) = f (x). That is, the area function A(x) is an antiderivative for f (x). Second Form of the Fundamental Theorem of Calculus: Let f be a continuous function and suppose that gโ€ฒ(x) = f (x). That is, suppose that g(x) is an antiderivative or f (x). Then โˆซ (^) b

a

f (x) dx = g(b) โˆ’ g(a).

(2) Let A =

[

]

, B =

[

]

, v =

[

]

. We compute ...

(a) 2A โˆ’ B =

[

]

[

]

(b) AB =

[

]

[

]

(c) Av =

[

]

[

]

(d) (A + B)v =

[

] [

]

[

]

[

]

(3) A =

๏ฃป (^) has square

A^2 = AA =

(4) Solve by Gaussian Elimination: ๏ฃฑ ๏ฃฒ

๏ฃณ

2 x + y + 3z = 2 x + 2y + z = 1 2 x โˆ’ 2 y + 2z = 8

The system has augmented coefficient matrix

๏ฃป. We perform

elementary row operations as follows. ๏ฃฎ

๏ฃฐ

๏ฃป (^) โˆ’r^1 โˆ’โ†”โˆ’rโ†’^2

๏ฃป (^) โˆ’rโˆ’^2 โ†’โˆ’rโˆ’^2 โˆ’โˆ’โˆ’^2 ร—โˆ’rโ†’^1

r 3 โ†’r 3 โˆ’ 2 r 1 โˆ’โˆ’โˆ’โˆ’โˆ’โˆ’โ†’

๏ฃป (^) โˆ’r^2 โˆ’โ†”โˆ’rโ†’^3

๏ฃป (^) โˆ’rโˆ’^2 โ†’โˆ’โˆ’โˆ’^1 โˆ’/โˆ’^6 ร—โˆ’rโ†’^2

r 3 โ†’r 3 +3r 2 โˆ’โˆ’โˆ’โˆ’โˆ’โˆ’โ†’

This last matrix is in row echelon form. We read off our solution by back substitution: z = โˆ’ 3 , y = โˆ’ 1 , x = 1 โˆ’ 2 y โˆ’ z = 1 + 2 + 3 = 6. That is,

x = 6, y = โˆ’ 1 , z = โˆ’ 3.

(6) Match plots (#1)...(#6) to the following functions.

Function Matches plot # Function Matches plot # z = sin(x^2 + y^2 ) #5 z = 2x^2 + y^2 #

z = ex/^3 cos(3y) #6 z = 4eโˆ’x (^2) โˆ’y 2

z = y^2 โˆ’ x^2 #2 z = 1 +

x 2

y 3

(7) Evaluate the partial derivatives fx and fy for each of the following functions.

(a) f (x, y) = 6x^5 y^3 โˆ’ 4 x^3 y^7 + 9x^2 y โˆ’ 8

fx(x, y) = 30x^4 y^3 โˆ’ 12 x^2 y^7 + 18xy, fy(x, y) = 18x^5 y^2 โˆ’ 28 x^3 y^6 + 9x^2

(b) f (x, y) = e^2 x^ sin(3y)

fx(x, y) = 2e^2 x^ sin(3y), fy(x, y) = 3e^2 x^ cos(3y)

(c) f (x, y) = cos(x^2 + y^2 )

fx(x, y) = โˆ’ 2 x sin(x^2 + y^2 ), fy(x, y) = โˆ’ 2 y sin(x^2 + y^2 )

(d) f (x, y) =

x โˆ’ y x + y

fx(x, y) =

(x + y)(1) โˆ’ (x โˆ’ y)(1) (x + y)^2

2 y (x + y)^2

, fy(x, y) =

(x + y)(โˆ’1) โˆ’ (x โˆ’ y)(1) (x + y)^2

โˆ’ 2 x (x + y)^2

(e) f (x, y) = (x^2 + 5xy + y^3 )^2012

fx(x, y) = 2012(2x + 5y)(x^2 + 5xy + y^3 )^2011 , fy(x, y) = 2012(5x + 3y^2 )(x^2 + 5xy + y^3 )^2011

(f) f (x, y) = ln(1 + x^2 y^3 )

fx(x, y) =

2 xy^3 1 + x^2 y^3

, fy(x, y) =

3 x^2 y^2 1 + x^2 y^3