Homework25 Solutions S14

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Math 21a Homework 25 Solutions Spring, 2014

Due Monday, April 14th (MWF) or Tuesday, April 15th (TTh)

A small warning: A common mistake at this point of the semester is to say “Whew! Now that midterm two is done, I cantake a little break and catch up in a week or so!” Don’t make this mistake – the material on arc length and surface area andscalar line integrals and surface area, as well as the next few sections are a warm-up for everything we do from this pointforward. The material at the end of the course (which some students find the most challenging) depends heavily on yourknowing and being comfortable with the first few sections of Chapter 13 (and also on parametrization and double and triple

integration). Read Sections 13.1 and 13.2, and skim through Section 13.3 before your next class.

The following 4 questions cover the material on vector fields in Section 13.1.

1. Sketch the vector field F.

(a) (Stewart 13.1 #6 ) F(x, y) = yi − x j

x2 + y2

Solution: All the vectors F(x, y) are unit vectors; you can tell simply by taking the length (squared):

|F(x, y)|2 =

y x2 + y2

2

+

−x x2 + y2

2

= y2 + x2

x2 + y2 = 1.

These unit vectors are tangent to circles that are centered at the origin with radius

x2

+ y2

. (The vector fieldis undefined.)

(b) (Stewart 13.1 #10 ) F(x,y,z ) = i − j

Solution: All vectors in this field have length√

2 and point in the same direction, parallel to the xy-plane.



(c) (Stewart 13.1 #26 ) ∇f , where f (x, y) =

x2 + y2

Solution: The gradient of f (x, y) =

x2 + y2 is

∇f (x, y) = 1

2(x2 + y2)−1/2(2x)i +

1

2(x2 + y2)−1/2(2y) j =

x

x2 + y2i +

y

x2 + y2 j.

As in part (a), this vector field is undefined at the origin, but elsewhere has length 1. Every vector in this vectorfield points directly away from the origin (perpendicular to the level curve f = c, which is a circle of radius c):

2. (a) (Stewart 13.1 #11-14 ) Match the vector fields F with the plots labeled I–IV that are drawn in the textbook onpage 911. Give reasons for your choices. Please don’t use Mathematica to solve this problem, as you will beexpected to be able to do this by hand on the exam.

(i) (Stewart 13.1 #11 ) F(x, y) = y, xSolution: F(x, y) = y, x corresponds to Graph II. In the first quadrant all the vectors have positive x- andy-components, in the second quadrant all vectors have positive x-components and negative y -components, inthe third quadrant all vectors have negative x and y-components, and in the fourth quadrant all vectors havenegative x-components and positive y-components. In addition, the vectors get shorter as we approach theorigin.

(ii) (Stewart 13.1 #12 ) F(x, y) = 1, sinySolution: F(x, y) = 1, sin y corresponds to Graph IV since the x-component of each vector is constant,the vectors are independent of x (vectors along horizontal lines are identical), and the vector field appears torepeat the same pattern vertically.

(iii) (Stewart 13.1 #13 ) F(x, y) = x − 2, x + 1Solution: F(x, y) = x − 2, x + 1 corresponds to Graph I since the vectors are independent of y (vectorsalong vertical lines are identical) and, as we move to the right, both the x- and y -components get larger.

(iv) (Stewart 13.1 #14 ) F(x, y) = y, 1/xSolution: F(x, y) = y, 1/x corresponds to Graph III. All the vectors in the first quadrant have positive x-and y-components, in the second quadrant all vectors have positive x-components and negative y-components,

in the third quadrant all vectors have negative x- and y - components, and in the fourth quadrant all vectorshave negative x-components and positive y -components. This is like Exercise 11.

(b) (Stewart 13.1 #15-18 ) Match the vector fields F with the plots labeled I–IV that are drawn in the textbook onpage 912. Give reasons for your choices. Please don’t use Mathematica to solve this problem, as you will beexpected to be able to do this by hand on the exam.

(i) (Stewart 13.1 #15 ) F(x,y,z ) = i + 2 j + 3k

Solution: F(x,y,z ) = i + 2 j + 3k corresponds to Graph IV, since all vectors have identical length anddirection.



(ii) (Stewart 13.1 #16 ) F(x,y,z ) = i + 2 j + z k

Solution: F(x,y,z ) = i + 2 j + z k corresponds to Graph I, since the horizontal vector components remainconstant, but the vectors above the xy-plane point generally upward while the vectors below the xy-planepoint generally downward.

(iii) (Stewart 13.1 #17

) F

(x,y,z ) = xi

+ y j

+ 3k

Solution: F(x,y,z ) = xi + y j + 3k corresponds to Graph III; the projection of each vector onto the xy-plane is xi + y j, which points away from the origin, and the vectors point generally upward because theirx-components are all 3.

(iv) (Stewart 13.1 #18 ) F(x,y,z ) = xi + y j + z k

Solution: F(x,y,z ) = xi + y j + z k corresponds to Graph II; each vector F(x,y,z ) has the same length anddirection as the position vector of the point (x,y,z ), and therefore the vectors all point directly away fromthe origin.

3. Mathematica will plot vector fields for you if you ask nicely. In this problem you’ll learn how to use Mathematica’sDocumentation Center to find the command to plot vector fields. You’ll use the command you find to solve (Stewart

13.1 #28 ).

(a) Find the gradient ∇f of f (x, y) = sin(x + y).

Solution: The gradient of f (x, y) = sin(x + y) is ∇f = cos(x + y)i + cos(x + y) j.

(b) Under the “Help” menu in Mathematica, click on “Documentation Center” to open the built-in documentation.Using the search bar, find out which command will plot vector fields, and what the syntax for that command lookslike.

Hint: You might search for “vector field plot”, “plot vector” or “vectorplot1”, and look at the examples to seehow the command is used.

Solution: VectorPlot[{vx, vy}, {x, xmin, xmax}, {y, ymin, ymax}] generates a vector plot of the vectorfield vx, vy as a function of x and y .

(c) Use the command you found below to plot the gradient vector field of f .

gradientfield = CommandYouFound[ ..., ..., ...]

Solution: The command

VectorPlot[{Cos[x + y], Cos[x + y]}, {x, -4, 4}, {y, -4, 4}, Axes->True, AxesLabel->{x,y}]

gives you the following plot:

(We’ve used some options to plot and label the axes; these extras are totally optional.)

1This is the actual name of the command you want – but you still need to look at the documentation to find out (1) how it’s capitalized!and more importantly (2) what the right inputs are.



(d) Also plot the level curves of f on the same plot using the ContourPlot command we learned in Homework 01:

contours = ContourPlot[ ..., ..., ..., ContourShading->None];

Show[contours, gradientfield]

Solution:

(e) Explain how the contours and the gradient field are related to each other.

Solution: The above graph shows that the gradient vectors are perpendicular to the level curves. Also, thegradient vectors point in the direction in which f is increasing and are longer where the level curves are closertogether.

(f) Use the CommandYouFound to check your sketches from question 1. You don’t need to turn in these plots.

Solution: See the solutions for Problem 1.

4. (Stewart 13.1 #36 ) Before starting this problem, read through (Stewart 13.1 #35 ).

(a) Sketch the vector field F(x, y) = i + x j and then sketch some flow lines. What shape do these flow lines appear tohave?

Solution:

The flow lines appear to be parabolas.

(b) If the parametric equations of the flow lines are x = x(t) and y = y(t), what differential equations (with respectto t) do these functions satisfy? Deduce that dy

dx = x.

Solution: If x = x(t) and y = y(t) are parametric equations of a flow line, then the velocity vector of the flowline at the point (x, y) is x(t)i + y(t) j. Since the velocity vectors coincide with the vectors in the vector field, wehave x (t)i + y(t) j = i + x j, so dx

dt = 1 and dy

dt = x. Thus

dy

dx =

dy/dt

dx/dt =

x

1 = x.



(c) If a particle starts at the origin in the velocity field given by F, find an equation of the path it follows.

Solution: From part (b), dydx

= x. Integrating, we have

y = 1

2x2 + C.

Since the particle starts at the origin, we know (0, 0) is on the curve, so we can deduce that C = 0. Thus the path

the particle follows is y = 1

2x2.

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Homework25 Solutions S14