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Multivariable Calculus UAF MATH F253X

Dr. Latrice Bowman

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Section Line Integrals and Vector Fields

Worksheet APP-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 6.1 Vector Fields, 6.2 Line Integrals, and 6.3 Conservative Vector Fields. 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 APP-1 Videos

Vector Fields
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Line Integrals
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Conservative Vector Fields
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Worksheet APP-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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WP14
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Vector Fields: Complete problems 5, 8, 23, 25, 32, 33, 34
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Line Integrals: Complete problems 51, 54, 57, 94
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Conservative Vector Fields: Complete problems 102, 113, 116, 119, 126, 143
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Worksheet APP-1 Sample Outcomes

For the quiz on this outcome you would be given 40 minutes to complete 3 problems. Here are samples of the types of problems you will encounter:
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1.

Given the vector field \(F(x,y)=(x+y)\hat{i}+(y-1)\hat{j}\)
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(a)
Sketch the vector field on the integer values of \([-1,1]\times[-1,1]\)
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Hint.
For each point \((x,y)\) in the grid, plot the vector \(F(x,y)\) using the position, direction, and the magnitude
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Answer.
Diagram Exploration Keyboard Controls
Key Action
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X Toggle expert mode
W Extra details if available
Space Repeat speech
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Comma Activate direct magnification
N Deactivate magnification
Z Toggle subtitles
C Cycle contrast settings
T Monochrome colours
L Toggle language (if available)
K Kill current sound
Y Stop sound output
O Start and stop sonification
P Repeat sonification output
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(b)
A particle moves through the vector field. If its position at \(t=2\) is \((4,-6)\text{,}\) estimate the position at \(t=2.02\)
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Hint.
Use the vector field to approximate the velocity at the given point and then use that to estimate the position at the new time.
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Answer.
The estimated position at \(t=2.02\) is \((3.96,-6.1)\text{.}\)
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Solution.
First we evaluate the vector field at the given point to get the velocity vector.
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At \(t=2\text{,}\) the velocity vector is \(F(4,-6) = (-2,-5)\text{.}\)
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The change in time is \(\Delta t = 0.02\text{,}\) so we can estimate the change in position as \(\Delta \vec{r} = \vec{v} \Delta t = (-2,-5)(0.02) = (-0.04,-0.1)\text{.}\)
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Then we use the velocity vector to estimate the position at the new time.
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At \(t=2.02\text{,}\) the estimated position is \((4,-6) + (-0.04,-0.1) = (3.96,-6.1)\text{.}\)
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2.

Given \(f(x,y,z) = e^z-\ln(x^2+y^2)\text{.}\) Find the gradient of \(f\text{.}\)
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Hint.
The gradient of a function is the vector of its partial derivatives.
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Answer.
\\(nabla f = \lt-\frac{2x}{x^2+y^2}, -\frac{2y}{x^2+y^2}, e^z \gt\)
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Solution.
To find the gradient, we compute the partial derivatives:
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\(\frac{\partial f}{\partial x} = -\frac{2x}{x^2+y^2}\)
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\(\frac{\partial f}{\partial y} = -\frac{2y}{x^2+y^2}\)
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\(\frac{\partial f}{\partial z} = e^z\)
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Therefore, \(\nabla f = \lt-\frac{2x}{x^2+y^2}, -\frac{2y}{x^2+y^2}, e^z \gt\)
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3.

Given \(F(x,y,z) = \lt y+z, x+z, x+y \gt\)
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(a)
Determine if \(F\) is a conservative vector field.
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Hint.
\(F(x,y,z)=M\hat{i}+N\hat{j}+P\hat{k}\) is conservative if \(\frac{\partial P}{\partial y} = \frac{\partial N}{\partial z}\text{,}\) \(\frac{\partial P}{\partial x} = \frac{\partial M}{\partial z}\text{,}\) and \(\frac{\partial N}{\partial x} = \frac{\partial M}{\partial y}\text{.}\)
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Answer.
This vector field is conservative.
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Solution.
To see if this is a conservative vector field, we need to find the partial derivatives:
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\(\frac{\partial P}{\partial y} = 1\)
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\(\frac{\partial N}{\partial z} = 1\)
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\(\frac{\partial P}{\partial x} = 1\)
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\(\frac{\partial M}{\partial z} = 1\)
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\(\frac{\partial N}{\partial x} = 1\)
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\(\frac{\partial M}{\partial y} = 1\)
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Since \(M_y=N_x, M_z=P_x,\) and \(N_z=P_y\text{,}\) the vector field is conservative.
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(b)
Find a potential function for \(F\text{.}\)
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Hint.
A potential function \(f\) satisfies \(\nabla f = F\text{.}\)
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Answer.
The potential function is \(f(x,y,z) = xy + xz + yz + C\text{.}\)
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Solution.
We need to find a function \(f\) such that \(\frac{\partial f}{\partial x} = M\text{,}\) \(\frac{\partial f}{\partial y} = N\text{,}\) and \(\frac{\partial f}{\partial z} = P\text{.}\)
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Integrating the first equation with respect to \(x\text{:}\)
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\(f(x,y,z) = xy + xz + g(y,z)\)
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where \(g(y,z)\) is a function of \(y\) and \(z\text{.}\)
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Taking the partial derivative with respect to \(y\text{:}\)
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\(\frac{\partial f}{\partial y} = x + \frac{\partial g}{\partial y} = N = x + z\)
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This gives us:
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\(\frac{\partial g}{\partial y} = z\)
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Integrating with respect to \(y\text{:}\)
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\(g(y,z) = yz + h(z)\)
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where \(h(z)\) is a function of \(z\text{.}\)
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So far, we have:
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\(f(x,y,z) = xy + xz + yz + h(z)\)
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Taking the partial derivative with respect to \(z\) :
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\(\frac{\partial f}{\partial z} = x + y + \frac{\partial h}{\partial z} = P = x + y \)
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This gives us:
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\(\frac{\partial h}{\partial z} = 0\)
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Integrating with respect to \(z\text{:}\)
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\(h(z) = C\)
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Therefore, the potential function is:
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\(f(x,y,z) = xy + xz + yz + C\)
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4.

Find the line integral of \(f(x,y,z)=x+y+z\) over the straight line segment from \((1,2,3)\) to \((0,-1,1)\text{.}\)
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Hint.
Use a parameterization of the line segment to rewrite the integral.
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Answer.
\(\int_C f(x,y,z) dS=3\sqrt{14}\)
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Solution.
Let \(\vec r(t) = (1-t)(1,2,3) + t(0,-1,1) = (1-t, 2-3t, 3-2t)\) for \(0 \leq t \leq 1\text{.}\)
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Then \(d\vec r = (-1, -3, -2) dt\text{.}\)
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The line integral becomes:
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\(\int_0^1 f(\vec r(t)) \|\frac{d\vec r}{dt}\| dt\)
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where \(f(\vec r(t)) = (1-t) + (2-3t) + (3-2t) = 6-6t\text{.}\)
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The magnitude of the derivative is:
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\(\|\frac{d\vec r}{dt}\| = \sqrt{(-1)^2 + (-3)^2 + (-2)^2} = \sqrt{14}\text{.}\)
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Therefore:
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\(\int_C f(x,y,z) dS = \int_0^1 (6-6t) \sqrt{14} dt\text{.}\)
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\(\int_C f(x,y,z) dS = = \sqrt{14} \int_0^1 (6-6t) dt\)
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\(\int_C f(x,y,z) dS = \sqrt{14} [6t - 3t^2]_0^1\)
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\(\int_C f(x,y,z) dS = \sqrt{14} (6 - 3)\)
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\(\int_C f(x,y,z) dS = 3\sqrt{14}\text{.}\)
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