Mathematics College

Let C be a circle of radius 9 centered at (0,0), traversed counterclockwise. Use this curve to answer the questions below. (a) Let F = y i + x j Find a potential function. f(x,y) = xy Does the Fundamental Theorem of Line Integrals apply to F · dr C ? Does Green's Theorem apply to F · dr C ? (b) Let G = y x2 + y2 i − x x2 + y2 j . Find a potential function. g(x,y) = Where is the potential function not defined? Does the Fundamental Theorem of Line Integrals apply to G · dr C ? Does Green's Theorem apply to G · dr C ? (c) Let H = x x2 + y2 i + y x2 + y2 j . Find a potential function. h(x,y) = Where is the potential function not defined? Does the Fundamental Theorem of Line Integrals apply to H · dr C ? Does Green's Theorem apply to H · dr C ? Submit Answer Save Progress

Answers

Answer 1

a. We're looking for a scalar function [tex]f(x,y)[/tex] such that [tex]\vec F(x,y)=\nabla f(x,y)[/tex]. This means

[tex]\dfrac{\partial f}{\partial x}=y[/tex]

[tex]\dfrac{\partial f}{\partial y}=x[/tex]

Integrate both sides of the first PDE with respect to [tex]x[/tex]:

[tex]\displaystyle\int\frac{\partial f}{\partial x}\,\mathrm dx=\int y\,\mathrm dx\implies f(x,y)=xy+g(y)[/tex]

Differentating both sides with respect to [tex]y[/tex] gives

[tex]\dfrac{\partial f}{\partial y}=x=x+\dfrac{\mathrm dg}{\mathrm dy}\implies g(y)=C[/tex]

so that [tex]\boxed{f(x,y)=xy+C}[/tex]. A potential function exists, so the fundamental theorem does apply.

Green's theorem also applies because [tex]C[/tex] is a simple and smooth curve.

b. Now with (and I'm guessing as to what [tex]\vec G[/tex] is supposed to be)

[tex]\vec G(x,y)=\dfrac y{x^2+y^2}\,\vec\imath-\dfrac x{x^2+y^2}\,\vec\jmath[/tex]

we want to find [tex]g[/tex] such that

[tex]\dfrac{\partial g}{\partial x}=\dfrac y{x^2+y^2}[/tex]

[tex]\dfrac{\partial g}{\partial y}=-\dfrac x{x^2+y^2}[/tex]

Same procedure as in (a): integrating the first PDE wrt [tex]x[/tex] gives

[tex]g(x,y)=\tan^{-1}\dfrac xy+h(y)[/tex]

Differentiating wrt [tex]y[/tex] gives

[tex]-\dfrac x{x^2+y^2}=-\dfrac x{x^2+y^2}+\dfrac{\mathrm dh}{\mathrm dy}\implies h(y)=C[/tex]

so that [tex]\boxed{g(x,y)=\tan^{-1}\dfrac xy+C}[/tex], which is undefined whenever [tex]y=0[/tex], and the fundamental theorem applies, and Green's theorem also applies for the same reason as in (a).

c. Same as (b) with slight changes. Again, I'm assuming the same format for [tex]\vec H[/tex] as I did for [tex]\vec G[/tex], i.e.

[tex]\vec H(x,y)=\dfrac x{x^2+y^2}\,\vec\imath+\dfrac y{x^2+y^2}\,\vec\jmath[/tex]

Now

[tex]\dfrac{\partial h}{\partial x}=\dfrac x{x^2+y^2}\implies h(x,y)=\dfrac12\ln(x^2+y^2)+i(y)[/tex]

[tex]\dfrac{\partial h}{\partial x}=\dfrac y{x^2+y^2}=\dfrac y{x^2+y^2}+\dfrac{\mathrm di}{\mathrm dy}\implies i(y)=C[/tex]

[tex]\implies\boxed{h(x,y)=\dfrac12\ln(x^2+y^2)+C}[/tex]

which is undefined at the point (0, 0). Again, both the fundamental theorem and Green's theorem apply.

Answer 2

Final answer:

The potential functions for the given vector fields F, G and H are f(x,y) = xy, g(x,y) = (1/2)xy² - arctan(x/y), and h(x,y) = (1/2)(x² + y²). Also, Green's theorem and the Fundamental Theorem of Line Integrals apply under certain conditions and constraints on the fields.

Explanation:

The potential function for F is indeed f(x,y) = xy as the partial derivatives of this function are equal to the components of F. The Fundamental Theorem of Line Integrals applies to F • dr C because F is the gradient of the scalar potential f and C is a closed curve. Green's Theorem applies as well since both F = y i + x j and C satisfy the appropriate conditions of continuity and differentiability.

The potential function for G is g(x,y)= (1/2)xy² - arctan(x/y) when y !=0. It is not defined where y=0. The Fundamental Theorem of Line Integrals does not apply because G is not conservative, meaning it doesn't have a potential function with continuous derivatives everywhere. Green's Theorem does apply to G • dr C because both the vector field G and C satisfy the conditions it requires.

The potential function for H is h(x,y)= (1/2)(x² + y²) and it is defined everywhere except at the origin (0,0) where it becomes undefined because of division by zero in its components. The Fundamental Theorem of Line Integrals does apply to H · dr C but only for curves which avoid the origin, because H is not conservative everywhere. Green's Theorem does not apply to H · dr C because H's components are not differentiable at the origin.

Learn more about Potential Functions here:

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Related Questions

what is m angle ABC

Answers

Answer:

3rd option: 60 degrees

Step-by-step explanation:

We can see in the diagram that the angle on C is a supplementary angle, which means that the sum of 135 and internal angle will be equal to 180 degrees.

Let x be the internal angle,

Then

x+135 = 180

x = 180-135

x = 45 degrees

So now we know that two interior angles of the triangle.

Also we know that sum of all internal angles of triangle is 180 degrees.

Using the same postulate:

A+B+C = 180

75 + B + 45 = 180

120+B = 180

B = 180 - 120

B = 60 degrees

So,

third option is the correct answer ..

Answer:

It’s 120 I got it right on the test !

A probability experiment is conducted in which the sample space of the experiment is S={7,8,9,10,11,12,13,14,15,16,17,18}, event F={7,8,9,10,11,12}, and event G={11,12,13,14}. Assume that each outcome is equally likely. List the outcomes in F or G. Find P(F or G) by counting the number of outcomes in F or G. Determine P(F or G) using the general addition rule.

Answers

Answer:

F or G = {7,8,9,10,11,12,13,14}

n(F or G) = 8

n(S) = 12

By counting the no. of outcome

P(F or G) = n(F or G) / n(S)

P(F or G) = 8 /12

P(F or G) = 2/3

By using the general addition rule

P(F or G) = P(F) + P(G) - P(F and G)

= 6/12 + 4/12 - 2/12

= 2/3

A composite figure is divided into two congruent trapezoids, each with a height of 4 cm.
What is the area of this composite figure?

Answers

Answer:

The area of this composite figure is [tex]64\ cm^{2}[/tex]

Step-by-step explanation:

we know that

If the composite figure is divided into two congruent trapezoids, then the area of the composite figure is equal to the area of one trapezoid multiplied by two

The area of the composite figure is

[tex]A=2[\frac{1}{2}(b1+b2)h][/tex]

[tex]A=(b1+b2)h[/tex]

substitute the values

[tex]A=(6+10)4[/tex]

[tex]A=64\ cm^{2}[/tex]

Answer:

64cm

Step-by-step explanation:

I looked at the guy's answer on top of mine and it was correct, go thank him!

An actor invests some money at 7%, and $24000 more than four times the amount at 8%. The total annual interest earned from the investment is $29220. How much
did he invest at each amount? Use the six-step method.

He invested $_____at 7% and ____at 8%.

Answers

Answer:

$70,000 at 7%$304,000 at 8%

Step-by-step explanation:

Given:

Total interest earned on two investments is $29,220.

An amount is invested at 7%.

$24,000 more than 4 times that amount is invested at 8%.

Find:

The amount invested at each rate.

Solution:

Our strategy will be to define a variable representing the amount invested at 7%, use that variable to write an expression for the amount invested at 8%, then write an equation for the total return on the investments.

Let x represent the amount invested at 7%. Then (24,000+4x) will be the amount invested at 8%. The total interest earned will be ...

interest on 7% account + interest on 8% account = total interest

0.07x + 0.08(24000+4x) = 29220

0.39x + 1920 = 29220 . . . . . . . . . . . simplify

0.39x = 27300 . . . . . . . . . . . . . . . . . .subtract 1920

x = 27300/0.39 = 70000 . . . . . . . . . divide by the coefficient of x

24,000 +4x = 24,000 +280,000 = 304,000 . . . . amount invested at 8%

He invested $70,000 at 7% and $304,000 at 8%.

Check

The answer must satisfy ...

7% interest + 8% interest = 29,220

0.07×70,000 +0.08×304,000 = 4,900 +24,320 = 29,220 . . . . as required

_____

Comment on 6-step method

We have tried to hit the highlights. Your steps appear to be ...

Identify the given information (Given)Identify the question you are asked to answer (Find)Identify the useless information in the problem statement (is none)Decide on a strategy. Make a model or drawing. (model equation shown)Solve and show work (Solution)Explain why the answer makes sense (Check)

a.
solve
[tex]\frac{1}{n} \pi = \theta - \frac{1}{2}sin(2 \theta)[/tex] for [tex] \theta[/tex] in terms of "n"

(derivation of equation below)

b. Based on your answer in
part a, if [tex] \theta = arccos(1 - \frac{a}{r} ) = {cos}^{ - 1} (1 - \frac{a}{r} )[/tex] or [tex] a = r-2cos( \theta)[/tex]

find "a" as a function of
r & n. (find f(r,n)=a).

alternately, if a+b=r, we can write [tex] \theta = arccos( \frac{b}{r} ) = {cos}^{ - 1} (\frac{b}{r} )[/tex]
then solve for "a" in terms of r and n

show all work and reasoning.
Solve analytically if possible

Answers

Answer:

a) There is no algebraic method for finding θ in terms of n

b) should be a = r(1 -cos(θ))

Step-by-step explanation:

Algebraic methods have been developed for solving trig functions and polynomial functions individually, but not in combination. In general, the solution is easily found numerically, but not analytically.

You would be looking for the numerical solution to ...

f(n, θ) = 0

where f(n, θ) can be ...

f(n, θ) = θ - (1/2)sin(2θ) - π/n

___

The attached shows Newton's method iterative solutions for n = 3 through 6:

for n = 3, θ ≈ 1.3026628373

for n = 4, θ ≈ 1.15494073001

...

a ball is shot straight upward. with it's height, in feet, after t seconds given by the function f(t)=-16t^2+192t. Find the average velocity of the ball from t=1 to t=6

Answers

ANSWER

[tex]80 {ms}^{ - 1} [/tex]

EXPLANATION

The average velocity of the ball is the rateof displacement over the total time.

The height of the ball, in feet, after t seconds is given by the function:

[tex]f(t)=-16t^2+192t[/tex]

At time t=1, the height of the ball is

[tex]f(1)=-16(1)^2+192(1)[/tex]

[tex]f(1)=-16+192 = 176ft[/tex]

At time t=6, the height of the ball is

[tex]f(6)=-16(6)^2+192(6)[/tex]

[tex]f(6)=-16(36)+192(6)[/tex]

[tex]f(6)=-576+1152 = 576[/tex]

The average velocity

[tex] = \frac{f(6) - f(1)}{6 - 1} [/tex]

[tex]= \frac{576- 176}{6 - 1} [/tex]

[tex]= \frac{400}{5} [/tex]

[tex] = 80 {ms}^{ - 1} [/tex]

you begin playing a new game called hooville. You are King of Hooville, a city of owls that is located in the treetops near Fords of Beruna. In order to know how much food to produce each year, you must predict the population of Hooville. History shows that the population growth rate of Hooville is 3.5%. The current population of owls is 80,000. Using the monetary growth formula that you used in the Uncle Harold problem, write a new function for the population of hooville. (let n=1.) PLEASE HELP. I HAVE NO IDEA WHAT IM DOING!!

Answers

Answer:

Part 1) [tex]y=80,000(1.035)^{x}[/tex]

Part 2) The table in the attached figure

Part 3) The graph in the attached figure

Step-by-step explanation:

Part 1) Find the population function

In this problem we have a exponential function of the form

[tex]y=a(b)^{x}[/tex]

where

y ----> is the population

x ----> the time in years

a is the initial value (a=80,000 people)

b is the base (b=100%+3.5%=103.5%=1.035)

substitute

[tex]y=80,000(1.035)^{x}[/tex]

Part 2) Construct the table

For x=0 years