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 0 answersPost by annac97 on April 23, 2015For the first example, can you just use a -2 to plug into the equation? So you'd do inverse tan(-2/5) instead?

### Vector Functions

• Displacement represented by magnitude and angel from +x-axis
• Vector has x (horizontal) and y (vertical) components
• Solving for vectors:
• Pythagorean theorem

### Vector Functions

Find the magnitude and direction of the vector represented by ( − 1,6)
• Use the the Pythagorean Theorem to find r
• ||r|| = √{x2 + y2}
• = √{( − 1)2 + (6)2}
• = √{37}
• Use the arctangent function on x,y components to find θ
• θ = tan − 1([y/x])
• = tan − 1([6/( − 1)])
(√{37} , − 1.41)
Find the magnitude and direction of the vector represented by ( − 5, − 12)
• Use the the Pythagorean Theorem to find r
• ||r|| = √{x2 + y2}
• = √{( − 5)2 + ( − 12)2}
• = √{25 + 144}
• = √{169}
• = 13
• Use the arctangent function on x,y components to find θ
• θ = tan − 1([y/x])
• = tan − 1([( − 12)/( − 5)])
(13,1.18)
Find the magnitude and direction of the vector represented by (u,√{15} u)
(Leave in terms of u)
• Use the the Pythagorean Theorem to find r
• ||r|| = √{x2 + y2}
• = √{(u)2 + (√{15} u)2}
• = √{u2 + 15u2}
• = √{16u2}
• = 4u
• Use the arctangent function on x,y components to find θ
• θ = tan − 1([(√{15} u)/u])
• = tan − 1(√{15} )
(4u,1.32)
What is the vector ordered pair of a vector with magnitude 8 and direction [(7π)/4]?
• Apply definitions for x and y
• x = rcosθ
• = 8cos([(7π)/4])
• = 4√2
• y = rsinθ
• = 8sin([(7π)/4])
• = − 4√2
(4√2 , − 4√2 )
What is the vector ordered pair of a vector with magnitude 32 and direction [(5π)/6]?
• Apply definitions for x and y
• x = rcosθ
• = 32cos([(5π)/6])
• = − 16√3
• y = rsinθ
• = 32sin([(5π)/6])
• = 16
( − 16√3 ,16)
What is the vector ordered pair of a vector with magnitude 3 and direction [(13π)/8]?
• Apply definitions for x and y
• x = rcosθ
• = 3cos([(13π)/8])
• = 1.15
• y = rsinθ
• = 3sin([(13π)/8])
• = − 2.77
(1.15, - 2.77)
Find the vector coordinates with the given value: v( − 4) = ( − 2,6t2,t − 9)
• Substitute −4 for t
= ( − 2,6( − 4)2, − 4 − 9) = ( − 2,6(24), − 13) = ( − 2,144, − 13)
Find the vector coordinates with the given value: v( − [(2√2 )/2]) = (t − 2,4cos − 1t,t)
• Substitute − [(2√2 )/2] for t
= (( − [(2√2 )/2]) − 2,4cos − 1( − [(2√2 )/2]), − [(2√2 )/2]) = (( − [2/(2√2 )])2,4([(5π)/4]), − [(2√2 )/2]) = ([1/2],5π, − [(2√2 )/2])
If the vector function v(t) is given with t = 2, what 3 functions will have the coordinate values of (4,16, − 28)?
Answers will vary. An example would be
v(t) = (2t,t4, − 30 + t)
Sketch the graph of the vector function created in problem # 9 for t = 3
• Answers will vary. In the case of v(t) = (2t,t4,30 − t)
• v(3) = (2(3),(3)4, − 30 + 3)
• = (6,81, − 27)

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Vector Functions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Vector Functions 0:11
• 2 Parts
• Example 1 1:44
• Example 2 3:52
• Example 3 4:59
• Example 4 6:03
• Example 5 8:00
• Example 6 9:23
• Example 7 10:16