slope and intercept, Calculus 3 Practice Test help

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grroheqr

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Mathematics

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Hi! I have a calc exam coming up, and I did this practice test, however we weren't provided an answer sheet to check what we did correctly. So if you could solve each problem and show your answer with work I would greatly appreciate it.

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16 Chapter 17 Exam Name: 1. Evaluate the given line integral. You may use any method you deem e↵ective. Z (a) (8 points) y dx x dy where C is the line segment from (2, 4) to (4, 7). C (b) (8 points) Z x2 y ds where C is the top half of the circle x2 + y 2 = 4. C (c) (14 points) Z e y dx xe y dy where C is any simple curve from (1, 0) to (0, 2). C 1 (d) (14 points) I C x2 + y 2 = 25. F̂ · d r̂ where F̂ = ex sin(y) î + ex cos(y) ĵ where C is the positive orientation of the circle 2. (12 points) Prove div(curl( F̂)) = 0 for any vector field F̂. 2 3. (8 points) Determine if the vector field F̂ = (ex + y 2 exy ) î + (1 + xy)exy ĵ is conservative. If it is, find a function f (x, y) such that rf = F̂. 4. Let F̂ = xy 3 î + 3 2 2 2x y + y ĵ. (a) (8 points) Determine if F̂ is conservative, and if so find a function f (x, y) such that rf = F̂. Z (b) (6 points) Use your answer from part a) to evaluate F̂ · d r̂ where C is the curve defined by r̂(t) = C 3 t h3 cos( ⇡t 2 ), 2 i, 0  t  2. (c) (4 points) Use your answer from part a) to quickly evaluate I C 3 F̂ · d r̂ over any simple closed curve C. 5. (2 points each) Match each function f (x, y) with the plot of the gradient field rf . A. f (x, y) = x2 y2 5 2.5 -10 -7.5 -5 -2.5 0 2.5 5 7.5 10 2.5 5 7.5 10 2.5 5 7.5 10 -2.5 -5 I. B. f (x, y) = (x y)2 5 2.5 -10 -7.5 -5 -2.5 0 -2.5 -5 II. 5 C. f (x, y) = 3 x +y 3 3 2.5 -10 -7.5 -5 -2.5 0 -2.5 -5 III. 6. (12 points) Find the divergence and curl of the vector field F̂ = ln(xyz) î + ln(xy) ĵ + ln(yz) k̂. Determine if F̂ is conservative, incompressible, both or neither. 4
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Explanation & Answer

Hi Please check the attached file for details, let me know if you have any question, thank you. James,

1

a)
Slope of the line:
m = (7-4)/(4-2) = 3/2
y-4 = 3/2(x-2)
y = 3/2x+1
4

3

3

 ydx − xdy =  ( 2 x + 1)dx − xd ( 2 x + 1)

C

2

4

3
3
=  ( x + 1)dx − xdx
2
2
2
4

=  dx
2

=2
b)

y = 4 − x2
−x
y' =
4 − x2
1

2
2
2
 x yds =  x y 1 + y' dx
−1

C
1

=  x2 4 − x2 1+
−1

x2
dx
4 − x2

1

=  2 x 2 dx
−1

=

2 3 1
x
3 −1

=

4
3

c)
Since d ( xe− y ) = e − y dx − xe− y dy
The integral depends only on the endpoints and is independent of the path C, and

−y
−y
 e dx − xe dy =
C

( 0, 2 )



d ( xe− y )

(1, 0 )

= xe− y

(0,2)
(1,0)

= 0 −1
= −1
d)

ˆj
iˆ


  Fˆ ...