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Multivariable Calculus UAF MATH F253X

Dr. Latrice Bowman

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Section Vector Applications

Worksheet Fun-1 Reading

All of the readings for this course come from the OpenStax Calculus Volume 3 textbook, which is available for free online at https://openstax.org/details/books/calculus-volume-3. You can read the textbook online or download a PDF version. The textbook is also available in print from various retailers if you prefer a physical copy.
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For this lesson, you should read sections 2.1 Vectors in the Plane, 2.2 Vectors in Three Dimensions, 2.3 The Dot Product, and 2.4 The Cross Product. You should read through the sections carefully, making sure to understand the definitions, examples, and key ideas. You should also work through the exercises at the end of each section to test your understanding and practice applying the concepts. Some of the exercises at the end of the section are assigned in the WP assignments.
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Worksheet FUN-1 Videos

Vectors in the Plane
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Vectors in Three Dimensions
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The Dot Product
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The Cross Product
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Worksheet FUN-1 Written Practice

INFORMATION ABOUT ALL WRITTEN PRACTICE (WP) ASSIGNMENTS:
  • All of the WP problems come from our online textbook: https://openstax.org/details/books/calculus-volume-3
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  • Complete solutions to all WP assignments can be found in Canvas within the section Module.
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  • Links to the text and PDFs of the text can be found in the Canvas module under the Reading and Information.
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THE WP ASSIGNMENTS ARE DESIGNED TO PREPARE YOU FOR THE QUIZZES AND ASSESSMENTS
  • Work on every problem on every assignment.
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  • If you get stuck or do not understand a solution, ask questions.
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  • For each problem, try it once or twice before looking at solutions or asking for help.
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  • Mark the problems you did correctly on the first try.
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  • Highlight problems where you got help (using solutions, videos, a tutor, etc) -These are the problem types that you will need more practice on to be ready for the assessments.
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  • Understand that just copying the solutions instead of working through the problems will greatly reduce your chances of success in this class.
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WP1
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Vectors in the Plane: Complete problems 3, 7, 13, 15, 25, 45
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Vectors in Three-Dimensions: Complete problems 77, 81, 89, 99, 115
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The Dot Product: Complete problems 125, 131, 145, 171, 177
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The Cross Product: Complete problems 185, 189, 199, 209, 235, 237
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In addition to these problems, upload a link to a solution video: Pick one problem from each of the above sections (the problems should not be any listed above). Create a short video of you presenting the solution to your classmates. You may use any video software (zoom, flip, explain everything, etc). The video does not need to be perfect or fully edited (mistakes are welcome), but it should show your thought process as you go through the solution.
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Worksheet FUN-1 Sample Outcomes

For the quiz on this outcome you would be given 30 minutes to complete 3 problems. Here are samples of the types of problems you will encounter:
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1.

Given two points in space, P(-1,5,-10) and Q(3,2,4) find the vector connecting them.
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Hint.
To find the vector connecting two points, subtract the coordinates of the first point from the coordinates of the second point.
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Answer.
\(\vec{PQ} = \lt 4, -3, 14 \gt\)
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Solution.
To find the vector connecting two points, subtract the coordinates of the first point from the coordinates of the second point. So, \(\vec{PQ} = Q - P = (3,2,4) - (-1,5,-10) = \lt 4, -3, 14 \gt\text{.}\)
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2.

Given two vectors \(a= \lt 2,2 \gt\) and \(b= \lt -1-\sqrt{3},1-\sqrt{3} \gt\)
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(a)
Find the unit vector, \(\vec{u}\text{,}\) in the direction of \(a\text{.}\)
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Hint.
A unit vector has a magnitude of 1. To find it, divide the vector by its magnitude.
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Answer.
\(\vec{u} = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\)
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Solution.
To find the unit vector in the direction of \(a\text{,}\) we first need to find the magnitude of \(a\text{.}\) The magnitude is given by \(|\vec{a}| = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}\text{.}\) Then, we divide the vector \(a\) by its magnitude to get the unit vector: \(\vec{u} = \frac{\vec{a}}{|\vec{a}|} = \frac{\lt 2, 2 \gt}{2\sqrt{2}} = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\text{.}\)
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(b)
Find the angle between vectors \(a\) and \(b\text{.}\)
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Hint.
The angle between two vectors can be found using the dot product formula: \(\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\text{.}\)
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Answer.
\(\theta = \frac{5\pi}{6}\)
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Solution.
To find the angle between vectors \(a\) and \(b\text{,}\) we use the dot product formula. First, we find the dot product: \(\vec{a} \cdot \vec{b} = (2)(-1-\sqrt{3}) + (2)(1-\sqrt{3}) = -2-2\sqrt{3}+2-2\sqrt{3} = -4\sqrt{3}\text{.}\) Next, we find the magnitudes of both vectors: \(|\vec{a}| = \sqrt{8} =2\sqrt{2}\) and \(|\vec{b}| = \sqrt{(-1-\sqrt{3})^2+(1-\sqrt{3})^2} = \sqrt{8} = 2\sqrt{2}\text{.}\) Then, we substitute into the formula: \(\cos \theta = \frac{-4\sqrt{3}}{(2\sqrt{2})(2\sqrt{2})} = \frac{-4\sqrt{3}}{8} = \frac{-\sqrt{3}}{2}\text{.}\) Therefore, \(\theta = \frac{5\pi}{6}\text{.}\)
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(c)
Find the projection of vector \(a\) onto vector \(b\text{.}\)
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Hint.
The projection of vector \(a\) onto vector \(b\) is given by: \(\text{proj}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b}\text{.}\)
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Answer.
\(\text{proj}_{\vec{b}}\vec{a} = \left\lt\frac{3+\sqrt{3}}{2}, \frac{3-\sqrt{3}}{2}\right\gt\)
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Solution.
To find the projection of vector \(a\) onto vector \(b\text{,}\) we use the formula: \(\text{proj}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b}\text{.}\) We already found that \(\vec{a} \cdot \vec{b} = -4\sqrt{3}\) and \(|\vec{b}|^2 = 8\text{.}\) Substituting these values, we get: \(\text{proj}_{\vec{b}}\vec{a} = \frac{-4\sqrt{3}}{8} \vec{b} = -\frac{\sqrt{3}}{2} \vec{b}\text{.}\) Since \(\vec{b} = \lt -1-\sqrt{3}, 1-\sqrt{3} \gt\text{,}\) the projection is: \(\text{proj}_{\vec{b}}\vec{a} = -\frac{\sqrt{3}}{2}(-1-\sqrt{3}, 1-\sqrt{3}) = \left\lt\frac{3+\sqrt{3}}{2}, \frac{3-\sqrt{3}}{2}\right\gt\text{.}\)
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3.

An object weighing 5 lbs is located at the point (2,1,5) and is being pulled by a cable in the direction of (0,-3,-1) with a force of \(\sqrt{14}lbs\text{.}\)
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(a)
Find the force of the cable on the object.
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Hint.
The force of the cable on the object is given by the product of the force magnitude and the unit vector in the direction of the cable.
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Answer.
\(\vec{F} = \lt -1, -2, -3 \gt\)
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Solution.
The force of the cable on the object is given by: \(\vec{F}_{cable} = |\vec{F}| \vec{u}\text{,}\) where \(|\vec{F}| = \sqrt{14}\) and \(\vec{u}\) is the unit vector in the direction of the cable. The unit vector is: \(\vec{u} = \frac{\lt 0-2, -3-1, -1-5 \gt}{\sqrt{(0-2)^2 + (-3-1)^2 + (-1-5)^2}} = \frac{\lt -2, -4, -6 \gt}{\sqrt{4 + 16 + 36}} = \frac{\lt -2, -4, -6 \gt}{\sqrt{56}} = \frac{\lt -1, -2, -3 \gt}{\sqrt{14}}\text{.}\) Therefore, \(\vec{F}_{cable} = \sqrt{14} \cdot \frac{\lt -1, -2, -3 \gt}{\sqrt{14}} = \lt -1, -2, -3 \gt\text{.}\)
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(b)
Give the total force acting on the object.
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Hint.
The total force is the sum of all forces acting on the object.
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Answer.
\(\vec{F}_{total} = \lt -1, -2, -8 \gt\)
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Solution.
The total force acting on the object is the sum of all forces. Since we have the force from the cable and the force due to gravity, the total force is the sum of these forces. The force from the cable is: \(\vec{F}_{cable} = \lt -1, -2, -3 \gt\) and the force due to gravity is: \(\vec{F}_{gravity} = \lt 0, 0, -5 \gt\text{.}\) Therefore, the total force is: \(\vec{F}_{total} = \vec{F}_{cable} + \vec{F}_{gravity} = \lt -1, -2, -3 \gt + \lt 0, 0, -5 \gt = \lt -1, -2, -8 \gt\text{.}\)
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4.

(a)
Find the area of the triangle with vertices at (-1,0,0), (2,3,0), and (0,4,6).
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Hint.
The area of a triangle with vertices at \(\vec{a}\text{,}\) \(\vec{b}\text{,}\) and \(\vec{c}\) is given by: \(\text{Area} = \frac{1}{2}|\vec{ab} \times \vec{ac}|\text{.}\)
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Answer.
\(\text{Area} = \frac{27}{2}\) square units
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Solution.
Let \(\vec{a} = (-1,0,0)\text{,}\) \(\vec{b} = (2,3,0)\text{,}\) and \(\vec{c} = (0,4,6)\text{.}\) Then: \(\vec{ab} = \lt 2-(-1), 3-0, 0-0 \gt = \lt 3, 3, 0 \gt\) and \(\vec{ac} = \lt 0-(-1), 4-0, 6-0 \gt = \lt 1, 4, 6 \gt\text{.}\) The cross product is: \(\vec{ab} \times \vec{ac} = \lt 18, -18, 9 \gt\text{.}\) The magnitude of the cross product is: \(|\vec{ab} \times \vec{ac}| = \sqrt{18^2 + (-18)^2 + 9^2} = \sqrt{324 + 324 + 81} = \sqrt{729} = 27\text{.}\) Therefore, the area is: \(\text{Area} = \frac{1}{2} \cdot 27 = \frac{27}{2}\) square units.
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5.

(a)
A snow machine drags a sled along a flat trail. The tether makes an angle of \(25^{\circ}\) with the sled and the tension in the tether is \(1050 N\text{.}\) How much work is done by the machine in pulling the sled 0.7km?
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Hint.
Work is given by the product of the force and the distance moved in the direction of the force.
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Answer.
\(W = 735000 \cos(25^\circ) \approx 666,136\) Nm
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Solution.
The work done by the machine is given by: \(W = F \cdot d \cdot \cos(\theta)\text{,}\) where \(F = 1050 N\text{,}\) \(d = 0.7 km = 700 m\text{,}\) and \(\theta = 25^{\circ}\text{.}\) Therefore, \(W = 1050 \cdot 700 \cdot \cos(25^{\circ}) \approx 735000 \cos(25^{\circ}) \approx 666,136.22\) Nm.
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