Test Drive #4 over Chapter 9

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Chapter 9 Unit Test. Name: Section: Use v = 3.2i – 2.5j + 2.7k and w = 4.6i – 3.2j – k to find: 1a) – 7v + 4w. 1b) ||v + w|| – 3||v||. Decompose v into two vectors, one ...
Chapter 10 Unit Test Name:

Section:

Use v = 3.2i – 2.5j + 2.7k and w = 4.6i – 3.2j – k to find: 1a)

– 7v + 4w

1b)

||v + w|| – 3||v||

Decompose v into two vectors, one parallel to w and the other orthogonal to w: 2) v = – 4.6i + 2.1j and w = 1.8i – 5.2j Given v and w, find a) v•w, b) the angle between v and w, and c) v×w: 3) v = – 5.2i – 2.3j + 6k and w = 3.7i + 1.9j – 3.4k Find the following: 4a)

Write the complex number z = 3 – ( 5 )i in polar form:

4b)

Use the result from part a to evaluate [3 – ( 5 )i]11. Write your final answer in rectangular form. €

€ in polar form: Find the following. Leave your answers z = 4.5(cos(234.5˚) + isin(234.5˚)) and w = 1.8(cos(84˚) + i sin(84˚)) 5a)

zw

5b)

z w

Transform the equation from polar to rectangular form: 6)

r=

4 1+sin(θ)

Find all complex roots. Write your answer in polar form with the argument in degrees: € 7) The complex fifth roots of – 4 – 7 i Solve the following: 8) Sailing a boat, Juanita maintains a constant speed of 25 nautical miles per hour bearing N45˚E. € If the current speed is 8 nautical miles per hour due North, find the actual speed and direction of the boat.

Write each complex number in standard form a + bi and graph the number on the complex plane (Be sure to label the axes): 9) 4.6(cos(135.3˚) + isin(135.3˚)) Find the direction angles of each vector and write the vector in the magnitude and direction angle form: 10) v = – 5.6i + 1.7j – 3.3k Given the following polar equation r = 1 – 3cos(θ): 11a) Determine which tests for symmetry the equation satisfies. Show your justification. 11b) Create an appropriate table of values and sketch the graph. Given the following polar equation r2 = 15sin(2θ): 12a) Determine which tests for symmetry the equation satisfies. Show your justification. 12b) Create an appropriate table of values and sketch the graph. Answers: 1a) – 4i + 4.7j – 22.9k 1b) ≈ – 4.820 2) v1 ≈ – 1.141i + 3.297j, v2 ≈ – 3.459i – 1.197j 3a) – 44.01 3b) 172.331˚ 3c) – 3.58i + 4.52j – 1.37k 4a) ≈

14 [cos(– 36.7˚) + isin(– 36.7˚)]

4b) 1,455,072 – 1,390,082.963i 5a) 8.1[cos(318.5˚) + isin(318.5˚)] 5b) 2.5[cos(150.5˚) + isin(150.5˚)] 6) x2 = – 8(y – 2) € 10 7) z0 = 23 [cos(43˚) + isin(43˚)], z1 = 10 23 [cos(115˚) + isin(115˚)], z2 = 10 23 [cos(187˚) + isin(187˚)], z3 = 10 23 [cos(259˚) + isin(259˚)],

€z4 = 10 23 [cos(331˚) + isin(331˚)] € € €

8) The boat's speed was ≈ 31.17 € knots at a bearing of N34.55˚E.

9) z ≈ – 3.270 + 3.236i 5 4 3 2 1 0 -5 -4 -3 -2 -1-1 0

1

2

3

4

5

-2 -3 -4 -5

10) α ≈ 146.46˚, β ≈ 75.34˚, γ ≈ 119.42˚ v = 45.14 [cos(146.46˚)i + cos(75.34˚)j + cos(119.42˚)k] 11a) It is symmetric with respect to the Polar Axis. € 11b) π

12a) It is symmetric with respect to the pole. 12b) π

2

2

3π 4

3π 4

π 4

€ €

π 4

€ €

π

0





π

1 2 3 4 5 6

5π 4

1 2 3 4 5 6

7π 4

5π 4

7π 4

3π 2



3π 2

€ €

0



€ €