query 114

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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

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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: `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).

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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 #$&*:

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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

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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??????

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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.*@

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

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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 #$&*:

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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.

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Self-critique (if necessary):

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

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Self-critique rating:

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Question: `q003. Use an incremental approximation to estimate f(sqrt(pi) + .01, sqrt(pi) - .01), where f(x,y) = cos(xy)

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Your solution:

f_x = -ysin(xy)

f_y = -xsin(xy)

????I’m confused and going to have to look at the solution?????

confidence rating #$&*:

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Given Solution:

We approximate by first finding f(x, y) at the point (sqrt(pi), sqrt(pi)).

Then we apply the differential with dx = .01 and dy = -.01.

f(sqrt(pi), sqrt(pi)) = cos(sqrt(pi) * sqrt(pi)) = cos(pi) = -1.

df = f_x dx + f_y dy indicates the change in f due to a given change dx in the value of x, and dy in the value of y.

df = -y sin(xy) dx - x sin(xy) * dy.

At the point (sqrt(pi), sqrt(pi) ) we have -y sin(x y) = - pi * sin(sqrt(pi) * sqrt(pi)) = -pi * sin(pi) = -pi * 0 = 0.

In a similar manner we have -x sin(x y) = 0.

That is, both f_x and f_y are zero at this point.

Using our values of f_x and f_y at the original point, along with dx = .01 and dy = -.01 we get

df = 0 * .01 + 0 * (-.01) = 0.

The same procedure would apply to approximate the function at, say, the point (sqrt(pi) + .01, sqrt(pi) / 3 - .02):

At the point (x0, y0) = (sqrt(pi) , sqrt(pi) / 3 ) we have f(x, y) = cos(sqrt(pi) * sqrt(pi / 3) ) = cos ( pi / 3) = 1/2 or .5.

The differential of our function is still df = -y sin(xy) dx - x sin(xy) * dy.

At the point (x0, y0) = (sqrt(pi), sqrt(pi) / 3 ) we have

f_x = -y sin(x y) = - pi / 3 * sin(sqrt(pi) * sqrt(pi) / 3) = -pi * sin(pi / 3) = -pi / 3 sqrt(3) / 2, approximately -0.9.

f_y = In a similar manner we have -x sin(x y) = -pi sin(pi/3) = -pi sqrt(3) / 2 = -2.7.

Using our values of f_x and f_y at our (x0, y0) point, along with dx = .01 and dy = -.02 we get

df = -0.9 * .01 + (-2.7) * (-.02) = -.063.

Our approximation to the value of f at the given point is therefore .5 + (-.063) = .437. That is

f(x, y) = f(sqrt(pi) + .01, sqrt(pi) / 3 -.02) = f(x0, y0) + df = .5 - .063 = .437.

You can assess the accuracy of this approximation by evaluating cos(x y) at the given point.

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Self-critique (if necessary):

I have several questions:

1) Why do we mult x_0 by y_0, I see why it works in this problem but what is the significance of x_0*y_0?????

@& (x_0, y_0) = (sqrt(pi), sqrt(pi) ) gives us the starting point for our approximation. We can easily evaluate cos(x y) at this point.*@

2) df = f_x dx + f_y dy , is this a vector relation or the same as saying df = |x|(change in length of x) +

|y|( change in length of x) because this seems like it would only apply to vector quantities?

@& The function cos(xy) changes because x changes, and it changes because y changes.

The change due to changing x values is f_x * dx, where f_x is evaluated at (x0, y0) and dx is the change in x required to get us to the approximation point.

Similarly the change due to the change in y is f_y * dy.

As you will soon see, f_x `i + f_y `j is the gradient of the function. When multiplied by the displacement vector dx * `i + dy * `j, you do get the change in the value of f. This is closely related to the directional derivative, so your instinct to think of this in terms of vectors is sound. But it's not necessary. The result follows directly from the definition of the partial derivatives, without any consideration of vectors.*@

3) Using this technique allows us to find the change in f but just knowing the change in f is there a

way to find the change in x and y where we can find the new point which f ends????

????I think I really just kind of confused what’s going on during this process????

@& In this problem we are extrapolating from the point (sqrt(pi), sqrt(pi) ) to the point (sqrt(pi + .01), sqrt(pi - .01).

The displacement from the 'known point' to the 'estimate point' is .01 in the x direction and -.01 in the y direction. Thus dx = .01 and dy = - .01.

f_x tells us the rate at which the function changes as we move in the x direction, so that f_x * dx is the change due to the x displacement.

Similarly f_y * dy is the change due to the y displacement.

Be sure to ask me to draw this for you.*@

Self-critique rating:

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Question: `q004. Find the equation of all horizontal tangent planes to the surface z = 4 - x^2 - y^2 + 6x.

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Your solution:

Any plane is tang. to z = 4-x^2-y^2+6x, whenever it is perpendicular to this surface.

z-4 = -(x^2+6x) + y^2, completing the square

z-4+9 = (x^2+6x+9) - y^2 = (x+3)^2 + y^2

z+5 = - (x+3)^2 - (y-0)^2

(x+3)^2 + (y-0)^2+z = -5

(x+3)^2/-5 + (y-0)^2/-5 + z/-5 = 1

I can’t seem to be able to find the surface I’m supposed to be looking at, z would have to be equal to -5 right because if not this equation is not possible because the sum of squares law says the sum of 2 squares cannot be neg and allowing z to equal 0 we have (x+3)^2 + (y-0)^2 = -5 or for the same reasons we couldn’t have (x+3)^2/-5 + (y-0)^2/-5 = 1?????

confidence rating #$&*:

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Given Solution:

The tangent plane will be horizontal only if all tangent lines are horizontal. That is, all tangent lines have to have slope zero. Thus all the derivatives need to be zero.

This will be the case, for example, at a point where the x and y partial derivatives are both zero.

For this function z_x = -2 x + 6 and z_y = -2 y. Thus our conditions z_x = 0 and z_y = 0 give us the two equations

-2 x + 6 = 0

-2 y = 0.

Each equation has only one solution. We get x = 3 and y = 0.

Thus the point (3, 0) is a critical point.

We need to check to be sure that our critical point isn't a saddle point.

Our second derivatives z_xx and z_yy are both negative, so (3, 0) is a candidate for a relative maximum. So far so good.

We also need to test that the graph doesn't go off into a saddle point when we move at some nonzero angle to the x and y axes. The test for this is that z_xx * z_yy - z_xy ^ 2 must be positive.

In this case z_xy = 0. We get z_xx * z_yy - z_xy ^ 2 = -2 * -2 + 0^2 = 4, which is > 0, so we don't have a saddle point

We conclude that our point (3, 0) does indeed give us a relative maximum.

The relative maximum therefore occurs at (3, 0, f(3, 0) ) = (3, 0, 13).

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Self-critique (if necessary):

I was approaching the problem incorrectly, so I still am having a hard time understanding what’s going on. The derivatives need to be 0 because the tang. plane is perpendicular to the surface correct? Also Can you describe the surface formed by this equation just so I can see what’s going on, I know its not needed for finding the solution just curious? Thanks.

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Self-critique rating:

@& The surface is very similar to the quadric surface z = 13 - x^2 - y^2, which is parabolic in the y-z plane and in the x-z plane, elliptical in the x-y plane. The surface is therefore an elliptic paraboloid.

Note that for z = 0 this gives us x^2 + y^2 = 13, which is perfectly OK. For z < -13 we would have x^2 + y^2 < 0, so the paraboloid has its vertex at z = -13. At that point, x^2 + y^2 = 0 so x = y = 0, and the vertex is (0, 0, -13).

The given equation does differ a bit from the above, but is still an elliptic paraboloid, found by shifting the above 3 units in the x direction (so for example the vertex will be at (3, 0, -13).

Completing the square on x we get

z = 4 - (x - 3)^2 - y^2 + 9

so

z = 13 - (x-3)^2 - y^2,

verifying the shift of 3 units in the x direction.

This particular surface can easily be analyzed as a quadric surface, so the analysis in terms of critical points is not necessary here. However this method will work for more complicated surfaces and can't be analyzed by simply completing the square.*@

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&#Good responses. See my notes and let me know if you have questions. &#