query 114

#$&*

course Mth 277

3/21 1

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.

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

Question: `q001. Give the standard form equation for the tangent plane to the surface z(x,y) = ln(x^2 + y^2) at the point P_0 = (e,0,2).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

z_x(x, y) = f(g(h))’, f(z) = ln(z), g(h) = x^2 + y^2

f’(z) = 1/z, g’(h) = 2x

f(g(h))’ = 1/(x^2 + y^2)*2x = 2x/(x^2 + y^2), for the P_0 = (e, 0, 2)

z_x(e, 0) = 2e/(e^2 + 0^2) = 2e/e^2

z_y(x, y) = 2y/(x^2 + y^2) z_y(e, 0) = 2*0/(e^2 + 0^2) = 0

Equation of tang line is

z - 2 = (2e/e^2)(x - e) + 0(y - 0)

z - 2 = (2e/e^2)(x - e)

Standard form

e^2z-2e^2 = 2e(x - e)

2ex - 2e^2 = e^2z - 2e^2

2e*x - e^2*z = 0

confidence rating #$&*:

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

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

Given Solution:

z_x = 2 x / (x^2 + y^2).

At (e, 0, 2) we have z_x = 2 * e / (e^2 + 0^2) = 2 / e (very approximately .75)

A unit vector tangent to the plane, in the x z plane, is therefore i + (2/e) k.

z_y = 2 y / (x^2 + y^2).

At (e, 0, 2) we have z_y = 2 * 0 / (e^2 + 0^2) = 0.

The j vector is therefore tangent to the plane, in the yz plane.

The cross product of two tangent vectors is a normal vector.

The cross product of the two tangent vectors obtained above is k + (2 / e) * (-i) = k - (2 / e) * i.

The point (x, y, z) lies on the tangent plane if (x - e) i + (y - 0) j + (z - 2) k is perpendicular to the normal vector.

Setting the dot product of the two vectors equal to zero we get the equation

-(2/e) * (x - e) + (z - 2) = 0, which we simplify to

-(2/e) x + z = 0

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

Self-critique (if necessary):

I didn’t use dot product and just got x and z terms on the same side which gave the correct solution but to better understand what’s going on could you explain what’s going on when you talked about taking dot product and how it relates to this problem??????

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

Self-critique rating:

@& Two vectors are perpendicular when, and only when, their dot product is zero.

For the point (x, y, z) to lie on the plane, the vector

(x - e) i + (y - 0) j + (z - 2) k

must be perpendicular to the normal vector.*@

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

Question: `q002. Find the total differential of f(x,y,z) = 2xzy^3*cos(xy)*sin(z)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

df = [sin(z)(2zy^3*cos(xy) + 2xy^3*-ysin(xy)) dx] + [sin(z)(6xzy^2*cos(xy) + 2xzy^3*-xsin(xy)) dy] +

[cos(xy)(2xy^3*sin(z) + 2xzy^3cos(z)) dz]

confidence rating #$&*:

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

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

Given Solution:

The total differential is f_x ds + f_y dy + f_z dz.

f_x = ( 2 z y^3 * cos(xy) - 2 x z y^4 sin(x y) ) * sin(z)

f_y = (6 x z y^2 cos(x y) - 2 x^2 z y^3 sin(xy)) * sin(z)

f_z = (2xy^3*cos(xy)*sin(z) + 2 xzy^3*cos(xy)*cos(z)

The total differential is therefore

( 2 z y^3 * cos(xy) - 2 x z y^4 sin(x y) ) * sin(z) * f_x + (6 x z y^2 cos(x y) - 2 x^2 z y^3 sin(xy)) * sin(z) * f_y + ( 2xy^3*cos(xy)*sin(z) + 2 xzy^3*cos(xy)*cos(z) ) * dz

The expression would be simplified by factoring 2 z y^3 out of the coefficient of f_x , 2 x z y^2 out of the coefficient of f_y, and 2 x y^3 out of the coefficient of dz.

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

Self-critique (if necessary):

I believe we have the same answer unless I’ve over looked something.

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

Self-critique rating:

query 114

#$&*

course Mth 277

3/21 1

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.

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

Question: `q001. Give the standard form equation for the tangent plane to the surface z(x,y) = ln(x^2 + y^2) at the point P_0 = (e,0,2).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

z_x(x, y) = f(g(h))’, f(z) = ln(z), g(h) = x^2 + y^2

f’(z) = 1/z, g’(h) = 2x

f(g(h))’ = 1/(x^2 + y^2)*2x = 2x/(x^2 + y^2), for the P_0 = (e, 0, 2)

z_x(e, 0) = 2e/(e^2 + 0^2) = 2e/e^2

z_y(x, y) = 2y/(x^2 + y^2) z_y(e, 0) = 2*0/(e^2 + 0^2) = 0

Equation of tang line is

z - 2 = (2e/e^2)(x - e) + 0(y - 0)

z - 2 = (2e/e^2)(x - e)

Standard form

e^2z-2e^2 = 2e(x - e)

2ex - 2e^2 = e^2z - 2e^2

2e*x - e^2*z = 0

confidence rating #$&*:

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

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

Given Solution:

z_x = 2 x / (x^2 + y^2).

At (e, 0, 2) we have z_x = 2 * e / (e^2 + 0^2) = 2 / e (very approximately .75)

A unit vector tangent to the plane, in the x z plane, is therefore i + (2/e) k.

z_y = 2 y / (x^2 + y^2).

At (e, 0, 2) we have z_y = 2 * 0 / (e^2 + 0^2) = 0.

The j vector is therefore tangent to the plane, in the yz plane.

The cross product of two tangent vectors is a normal vector.

The cross product of the two tangent vectors obtained above is k + (2 / e) * (-i) = k - (2 / e) * i.

The point (x, y, z) lies on the tangent plane if (x - e) i + (y - 0) j + (z - 2) k is perpendicular to the normal vector.

Setting the dot product of the two vectors equal to zero we get the equation

-(2/e) * (x - e) + (z - 2) = 0, which we simplify to

-(2/e) x + z = 0

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

Self-critique (if necessary):

I didn’t use dot product and just got x and z terms on the same side which gave the correct solution but to better understand what’s going on could you explain what’s going on when you talked about taking dot product and how it relates to this problem??????

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

Self-critique rating:

@& Two vectors are perpendicular when, and only when, their dot product is zero.

For the point (x, y, z) to lie on the plane, the vector

(x - e) i + (y - 0) j + (z - 2) k

must be perpendicular to the normal vector.*@

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

Question: `q002. Find the total differential of f(x,y,z) = 2xzy^3*cos(xy)*sin(z)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

df = [sin(z)(2zy^3*cos(xy) + 2xy^3*-ysin(xy)) dx] + [sin(z)(6xzy^2*cos(xy) + 2xzy^3*-xsin(xy)) dy] +

[cos(xy)(2xy^3*sin(z) + 2xzy^3cos(z)) dz]

confidence rating #$&*:

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

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

Given Solution:

The total differential is f_x ds + f_y dy + f_z dz.

f_x = ( 2 z y^3 * cos(xy) - 2 x z y^4 sin(x y) ) * sin(z)

f_y = (6 x z y^2 cos(x y) - 2 x^2 z y^3 sin(xy)) * sin(z)

f_z = (2xy^3*cos(xy)*sin(z) + 2 xzy^3*cos(xy)*cos(z)

The total differential is therefore

( 2 z y^3 * cos(xy) - 2 x z y^4 sin(x y) ) * sin(z) * f_x + (6 x z y^2 cos(x y) - 2 x^2 z y^3 sin(xy)) * sin(z) * f_y + ( 2xy^3*cos(xy)*sin(z) + 2 xzy^3*cos(xy)*cos(z) ) * dz

The expression would be simplified by factoring 2 z y^3 out of the coefficient of f_x , 2 x z y^2 out of the coefficient of f_y, and 2 x y^3 out of the coefficient of dz.

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

Self-critique (if necessary):

I believe we have the same answer unless I’ve over looked something.

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

Self-critique rating: