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4/30/16, 7:15 AM Print Test Page 1 of 7 https://www.casa.uh.edu/CourseWare2008/Root/Pages/CW/Users/Student/Grades/PrintTest.htm PRINTABLE VERSION Quiz 24 You scored 0 out of 100 Question 1 You did not answer the question. Express the curve by an equation in and given . a) b) c) d) e) Question 2 You did not answer the question. Express the curve by an equation in and given . a) b) c) d) e) Question 3 You did not answer the question. x y x(t)= and y(t)=2 +4 t 2 t 4 x =4 + 5, y 0 y 2 x =4 + 3, y 0 y 2 y =2 + 3, x 0 x 2 y = + 4, x 0 x 2 y =2 + 4, x 0 x 2 x y x(t) = cos(t) and y(t) = 5 sin(t) 25 = 25 x 2 y 2 25 + = 25 x 2 y 2 25 + =5 x 2 y 2 + 25 = 25 x 2 y 2 25 =5 x 2 y 2 x(t)= (t) + 2 and y(t) = 5 + tan(t) 2

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Page 1: PRINTABLE VERSION - UHbekki/1432sp16/notes/1432_quiz... · 2016-04-30 · PRINTABLE VERSION Quiz 24 You scored 0 out of 100 Question 1 You did not answer the question. ... 16x2 +

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PRINTABLE VERSIONQuiz 24

You scored 0 out of 100Question 1

You did not answer the question.Express the curve by an equation in and given .

a)

b)

c)

d)

e)

Question 2

You did not answer the question.Express the curve by an equation in and given .

a)

b)

c)

d)

e)

Question 3

You did not answer the question.

x y x(t) = and y(t) = 2 + 4t 2 t 4

x = 4 + 5, y ≥ 0y2

x = 4 + 3, y ≥ 0y2

y = 2 + 3, x ≥ 0x2

y = + 4, x ≥ 0x2

y = 2 + 4, x ≥ 0x2

x y x(t) = cos(t) and y(t) = 5 sin(t)

− 25 = 25x2 y2

25 + = 25x2 y2

25 + = 5x2 y2

+ 25 = 25x2 y2

25 − = 5x2 y2

x(t) = (t) + 2 and y(t) = 5 + tan(t)2

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Express the curve by an equation in and given .

a)

b)

c)

d)

e)

Question 4

You did not answer the question.Express the curve by an equation in and given .

a)

b)

c)

d)

e)

Question 5

You did not answer the question.Express the curve by an equation in and given .

a)

b)

c)

d)

x y x(t) = (t) + 2 and y(t) = 5 + tan(t)sec2

x = + 2(y − 5)2

y = + 3(x − 5)2

x = + 3(y − 5)2

y = + 2(x − 5)2

x = + 1(y + 5)2

x y x(t) = sin(t) and y(t) = 9 + (t)cos2

− + y = 10, −9 ≤ x ≤ 9x2

+ x = 10, 0 ≤ x ≤ 1y2

+ y = 9, −1 ≤ x ≤ 1x2

+ y = 10, −1 ≤ x ≤ 1x2

− + x = 9, −9 ≤ x ≤ 9y2

x y x(t) = and y(t) = 5 −et e3 t

+ y = 5, x > 0x3

− + y = 6, 0 ≤ x ≤ 5x4

+ x = 5, x > 0y3

− + y = 5, x ≥ 1x3

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e)

Question 6

You did not answer the question.

Express the curve by an equation in and given and identify the correct

sketch of the curve.

a) ;

− + x = 5, 0 ≤ x ≤ 1y3

x y ( , ), t ∈ (0, 1]1t

2t 2

y = 2 x2

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b) ;

c) ;

y = 2 x−2

y = 2 x−1

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d) ;

e) ;

y = 2 x2

y = −2 x2

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Question 7

You did not answer the question.

Give a parameterization for the ellipse that begins at the point (3,0) and traversesonce in a counterclockwise manner.

a)

b)

c)

d)

e)

Question 8

You did not answer the question.Find a parametrization for the line segment from .

a)

b)

c)

d)

e)

Question 9

You did not answer the question.Find a parametrization for the line segment from .

a)

b)

16 + 9 = 144x2 y2

x(t) = 4 cos(t), y(t) = 3 sin(t), t ∈ [0, 2π]

x(t) = 16 cos(t), y(t) = 9 sin(t), t ∈ [0, 2π]

x(t) = 9 sin(t), y(t) = 16 cos(t), t ∈ [0, 2π]

x(t) = 3 cos(t), y(t) = 4 sin(t), t ∈ [0, 2π]

x(t) = 4 sin(t), y(t) = 3 cos(t), t ∈ [0, 2π]

x = x(t), y = y(t), t ∈ [0, 1] (−8, −2) to (5, 8)

x(t) = 13 t − 8, y(t) = 10 t − 2

x(t) = 13 t − 8, y(t) = −2 + 11 t

x(t) = −8 − 14 t, y(t) = 10 t − 2

x(t) = −10 t − 8, y(t) = −13 t − 2

x(t) = −13 t − 8, y(t) = −10 t − 2

x = x(t), y = y(t), t ∈ [0, 1] (−5, −3) to (−7, −3)

x(t) = −5, y(t) = 2 t − 3

x(t) = −2 t − 5, y(t) = −3

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c)

d)

e)

Question 10

You did not answer the question.Find a parametrization for from .

a)

b)

c)

d)

e)

x(t) = 2 t − 5, y(t) = 0

x(t) = −5 + t, y(t) = −3

x(t) = −2 t − 5, y(t) = −3 + t

x = x(t), y = y(t) f (x) = − 8 − 9x8 x2 (9, −10) to (10, −8)

x(t) = − 8 − 9, y(t) = t, t ∈ [9, 10]t 8 t 2

x(t) = − 8 , y(t) = −9, t ∈ [9, 10]t 8 t 2

x(t) = , y(t) = − 8 t − 9, t ∈ [81, 100]t 2 t 4

x(t) = t, y(t) = − 8 − 9, t ∈ [9, 10]t 8 t 2

x(t) = − 8 − 9, y(t) = t, t ∈ [−10, −8]t 8 t 2