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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) =
โซ (^) x
a
f (t) dt.
Then Aโฒ(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 gโฒ(x) = f (x). That is, suppose that g(x) is an antiderivative or f (x). Then โซ (^) b
a
f (x) dx = g(b) โ g(a).
(2) Let A =
[
]
, B =
[
]
, v =
[
]
. We compute ...
(a) 2A โ 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 โ 2 y + 2z = 8
The system has augmented coefficient matrix
๏ฃป. We perform
elementary row operations as follows. ๏ฃฎ
๏ฃฐ
๏ฃป (^) โr^1 โโโrโ^2
๏ฃป (^) โrโ^2 โโrโ^2 โโโ^2 รโrโ^1
r 3 โr 3 โ 2 r 1 โโโโโโโ
๏ฃป (^) โr^2 โโโrโ^3
๏ฃป (^) โrโ^2 โโโโ^1 โ/โ^6 รโrโ^2
r 3 โr 3 +3r 2 โโโโโโโ
This last matrix is in row echelon form. We read off our solution by back substitution: z = โ 3 , y = โ 1 , x = 1 โ 2 y โ z = 1 + 2 + 3 = 6. That is,
x = 6, y = โ 1 , z = โ 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 = 4eโx (^2) โy 2
z = y^2 โ 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 โ 4 x^3 y^7 + 9x^2 y โ 8
fx(x, y) = 30x^4 y^3 โ 12 x^2 y^7 + 18xy, fy(x, y) = 18x^5 y^2 โ 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) = โ 2 x sin(x^2 + y^2 ), fy(x, y) = โ 2 y sin(x^2 + y^2 )
(d) f (x, y) =
x โ y x + y
fx(x, y) =
(x + y)(1) โ (x โ y)(1) (x + y)^2
2 y (x + y)^2
, fy(x, y) =
(x + y)(โ1) โ (x โ y)(1) (x + y)^2
โ 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
