Query132

#$&*

course Mth 277

205pm 10/27/2012

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.

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Question: Evaluate Int[ (2x - 3y) ds, C] where C is defined by x = sin t, y = cos t, 0 <= t <= pi. ( Int[ f(x,y) ds, C] is the line integral of f(x,y) over C).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

r(t) =

r'(t) =

ds = sqrt(cos^2t + sin^2t) = 1

Integral of 2x - 3y ds from 0 to pi

= Integral 2sint - 3cost dt from 0 to pi

= -2cost - 3sint at pi and 0

= 4

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

The distance increment is based on the 'velocity' function x ' i + y ' j = cos(t) i - sin(t) j, which implies displacement increment `ds = (x ' i + y ' j) dt = cos(t) i - sin(t) j and therefore distance increment

`ds = | `ds | = sqrt( x ' ^ 2 + y ' ^ 2 ) dt = sqrt( cos^2(t) + sin^2(t) ) dt = sqrt( 1 ) dt = dt.

The function (2 x - 3 y) is expressed in terms of t by (2 sin(t) - 3 cos(t)), so our integral is

int ( (2 sin(t) - 3 cos(t) ) dt, t from 0 to pi).

Our antiderivative is -2 cos(t) - 3 sin(t), which between t = 0 and t = pi changes from -2 to 2, a change of 4.

Thus our line integral has value 4.

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Self-critique (if necessary):OK

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Question: Evaluate Int[-y dx + 3y dy, C] where C is defined by y^2 = x from the point (1,1) to the point (9,3).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y^2 = x

2y = dx

integral of -ydx + 3ydy

= integral from 1 to 3 of -y*2y + 3ydy

= -2/3*y^3 + 3/2*y^2 from 1 to 3

= -2/3*27 + 3/2*9 - [-2/3 + 3/2]

= -16/3

16/3 is the magnitude, a negative sign just shows you the direction of the change

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

First verify that (1, 1) and (9, 3) are on the curve. This is easily verified for the first point, since 1^2 = 1, and for the second, since 3^2 = 9.

This curve is part of the parabola x = y^2, which has vertex at the origin and opens to the right. The curve C is part of the 'upper half' of this parabola.

If x = y^2, then dx = d(y^2) = 2 y dy.

On this curve y changes from 1 to 3 (and x = y^2 changes from 1^2 = 1 to 3^2 = 9).

Expressing our integrand -y dx + 3 y dy in terms of y we get

-y ( 2 y dy) + 3 y dy = (-2 y^2 + 3 y) dy.

Our integral is therefore

integral((-2 y^2 + 3 y) dy, y from 1 to 3).

Our antiderivative is -2/3 y^2 + 3/2 y^2.

Between y = 1 and y = 3 this antiderivative changes from -2/3 + 3/2 to -6 + 27/2, a change of ...

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Self-critique (if necessary): OK

*********************************************

Question: Evaluate Int[(x^2 - y^2)dx + x dy, C] where C is the circular path given by x = 2 cos(theta), y = 2sin(theta), 0 <= theta <= 2pi.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

x= 2cos(theta)

dx = -2sin(theta)d(theta)

y = 2sin(theta)

dy = 2cos(theta)d(theta)

Integral from 0 to 2pi of (x^2 - y^2)dx + xdy

= int from 0 to 2pi of [4cos^2(theta) - 4sin^2(theta)] * [-2sin(theta)d(theta)] + 2cos(theta)*2cos(theta)*d(theta)

= Int of 0 to 2pi -8cos^2(theta)sin(theta)d(theta) + 8sin^3(theta)d(theta) + 4cos^2(theta)d(theta)

u = cos(theta)

du= -sin(theta)d(theta)

= 8/3cos^3(theta) at 2pi and 0 + 8*Int from 0 to 2pi of [1-cos^2(theta)]*sin(theta)d(theta) + 4*Int from 0 to 2pi of {[1+cos(2theta)]/2*d(theta)

8/3cos^3(theta) at 2pi and 0 is equal to 0

=8*Int 0 to 2pi of sin(theta) - cos^2(theta)*sin(theta)*d(theta) + Int 0 to 2pi of [2 + 2cos(2theta)*d(theta)]

u = cos(theta)

du = -sin(theta)

= 8cos(theta) + (8/3)cos^3(theta) at 2pi and 0 + Int 0 to 2pi of [2 + 2cos(2theta)*d(theta)]

u = 2(theta)

du = 2d(theta)

(1/2)du = d(theta)

= 8cos(theta) + (8/3)cos^3(theta) + 2(theta) + sin(2theta) at 2pi and 0

= -8cos(2pi) + 8/3 + 4pi + 0 + 8 - 8/3 + 0 + 0 = 4pi

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

Our parameter here is theta, so our derivatives will be with respect to theta. Thus ' will indicate derivative with respect to theta.

On this path we have

dx = x ' dTheta = -2 sin(theta) dTheta

and

dy = y ' dTheta = 2 cos(theta) dTheta

Thus

(x^2 - y^2) dx = ( 4 cos^2(theta) - 4 sin^2(theta) ) * 2 sin(theta) dTheta = 4 (cos^2(theta) sin(theta) - sin^3(theta)) dTheta

and

x dy = 2 cos(theta) * 2 cos(theta) dTheta.= 4 cos^2(theta) dTheta

Thus our integral becomes

integral( 4 (cos^2(theta) sin(theta) - sin^3(theta)) dTheta + 4 cos^2(theta) dTheta , theta from 0 to 2 pi)

= integral( 4 (cos^2(theta) sin(theta) - sin^3(theta) + 4 cos^2(theta) ) dTheta , theta from 0 to 2 pi).

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Self-critique (if necessary): OK"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

Query132

#$&*

course Mth 277

205pm 10/27/2012

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.

*********************************************

Question: Evaluate Int[ (2x - 3y) ds, C] where C is defined by x = sin t, y = cos t, 0 <= t <= pi. ( Int[ f(x,y) ds, C] is the line integral of f(x,y) over C).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

r(t) =

r'(t) =

ds = sqrt(cos^2t + sin^2t) = 1

Integral of 2x - 3y ds from 0 to pi

= Integral 2sint - 3cost dt from 0 to pi

= -2cost - 3sint at pi and 0

= 4

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`ds = | `ds | = sqrt( x ' ^ 2 + y ' ^ 2 ) dt = sqrt( cos^2(t) + sin^2(t) ) dt = sqrt( 1 ) dt = dt.

The function (2 x - 3 y) is expressed in terms of t by (2 sin(t) - 3 cos(t)), so our integral is

int ( (2 sin(t) - 3 cos(t) ) dt, t from 0 to pi).

Our antiderivative is -2 cos(t) - 3 sin(t), which between t = 0 and t = pi changes from -2 to 2, a change of 4.

Thus our line integral has value 4.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):OK

*********************************************

Question: Evaluate Int[-y dx + 3y dy, C] where C is defined by y^2 = x from the point (1,1) to the point (9,3).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y^2 = x

2y = dx

integral of -ydx + 3ydy

= integral from 1 to 3 of -y*2y + 3ydy

= -2/3*y^3 + 3/2*y^2 from 1 to 3

= -2/3*27 + 3/2*9 - [-2/3 + 3/2]

= -16/3

16/3 is the magnitude, a negative sign just shows you the direction of the change

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

First verify that (1, 1) and (9, 3) are on the curve. This is easily verified for the first point, since 1^2 = 1, and for the second, since 3^2 = 9.

This curve is part of the parabola x = y^2, which has vertex at the origin and opens to the right. The curve C is part of the 'upper half' of this parabola.

If x = y^2, then dx = d(y^2) = 2 y dy.

On this curve y changes from 1 to 3 (and x = y^2 changes from 1^2 = 1 to 3^2 = 9).

Expressing our integrand -y dx + 3 y dy in terms of y we get

-y ( 2 y dy) + 3 y dy = (-2 y^2 + 3 y) dy.

Our integral is therefore

integral((-2 y^2 + 3 y) dy, y from 1 to 3).

Our antiderivative is -2/3 y^2 + 3/2 y^2.

Between y = 1 and y = 3 this antiderivative changes from -2/3 + 3/2 to -6 + 27/2, a change of ...

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

*********************************************

Question: Evaluate Int[(x^2 - y^2)dx + x dy, C] where C is the circular path given by x = 2 cos(theta), y = 2sin(theta), 0 <= theta <= 2pi.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

x= 2cos(theta)

dx = -2sin(theta)d(theta)

y = 2sin(theta)

dy = 2cos(theta)d(theta)

Integral from 0 to 2pi of (x^2 - y^2)dx + xdy

= int from 0 to 2pi of [4cos^2(theta) - 4sin^2(theta)] * [-2sin(theta)d(theta)] + 2cos(theta)*2cos(theta)*d(theta)

= Int of 0 to 2pi -8cos^2(theta)sin(theta)d(theta) + 8sin^3(theta)d(theta) + 4cos^2(theta)d(theta)

u = cos(theta)

du= -sin(theta)d(theta)

= 8/3cos^3(theta) at 2pi and 0 + 8*Int from 0 to 2pi of [1-cos^2(theta)]*sin(theta)d(theta) + 4*Int from 0 to 2pi of {[1+cos(2theta)]/2*d(theta)

8/3cos^3(theta) at 2pi and 0 is equal to 0

=8*Int 0 to 2pi of sin(theta) - cos^2(theta)*sin(theta)*d(theta) + Int 0 to 2pi of [2 + 2cos(2theta)*d(theta)]

u = cos(theta)

du = -sin(theta)

= 8cos(theta) + (8/3)cos^3(theta) at 2pi and 0 + Int 0 to 2pi of [2 + 2cos(2theta)*d(theta)]

u = 2(theta)

du = 2d(theta)

(1/2)du = d(theta)

= 8cos(theta) + (8/3)cos^3(theta) + 2(theta) + sin(2theta) at 2pi and 0

= -8cos(2pi) + 8/3 + 4pi + 0 + 8 - 8/3 + 0 + 0 = 4pi

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

Our parameter here is theta, so our derivatives will be with respect to theta. Thus ' will indicate derivative with respect to theta.

On this path we have

dx = x ' dTheta = -2 sin(theta) dTheta

and

dy = y ' dTheta = 2 cos(theta) dTheta

Thus

(x^2 - y^2) dx = ( 4 cos^2(theta) - 4 sin^2(theta) ) * 2 sin(theta) dTheta = 4 (cos^2(theta) sin(theta) - sin^3(theta)) dTheta

and

x dy = 2 cos(theta) * 2 cos(theta) dTheta.= 4 cos^2(theta) dTheta

Thus our integral becomes

integral( 4 (cos^2(theta) sin(theta) - sin^3(theta)) dTheta + 4 cos^2(theta) dTheta , theta from 0 to 2 pi)

= integral( 4 (cos^2(theta) sin(theta) - sin^3(theta) + 4 cos^2(theta) ) dTheta , theta from 0 to 2 pi).

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Self-critique (if necessary): OK"

Self-critique (if necessary):

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Self-critique rating:

#*&!

Query132

Query132

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