Query 114

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course Mth 277

10/30 7

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:

i've gotten the partial derivatives of f_y and f_x to be 2/e and 0

when putting into the std equation,

I get 2x/e + z = 0.

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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'm not sure why you took the cross product to get k-(2/e)i and sub that in for the i value. this is why your answer is negative, and myn is positive,

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Self-critique rating:

@& (`i + (2/e) `k X `j =

`i X `j + (2/e) (`k X `j) =

`k + (2 / e) * (-`i) =

`k - (2/e) `i.

Check the right-hand rule. I'm pretty sure I've got it straight, but always question and check out my solutions.*@

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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:

f_x = 2zy^3sin(z)(cos(xy) -xysin(xy))

f_y = 2xzsin(z)(-xy^3sin(xy) + 3y^2cos(xy))

f_z = 2xy^3cis(xy)(sin(z) + zcos(z))

df total is all those added together

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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):

my answers are not simplified quite as far as you went. but my partial differinciation is neat and orderly and most importantly correct.

but OK

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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:

delta x is .01

delta y is -.01

do partial derivation of the f(x,y) to get a total deravation of

-ysin(xy)delta x + -xsin(xy)delta y

fill in the change in x and y to get that it equals 0.

you can see here that both f_x and f_y are both zero, to give

total change in f at that point is zero

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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):

OK.

just looked up the equation and steps for that and not bad

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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:

tangent plane is slope of zero.

z_x =

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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):

????????????????????????

ok when doing the partial derivatives for z why would it not be -2x - y^2 +6 = 0. likewise for the partial derivative for y values,

@& y is constant in taking an x derivative. So therefore is y^2. The derivative of a constant is zero.*@

I understand that you have to get a solution for x and y to get a point, but it seems like you just kick

the other values to the side to get that point. also, when you get the point, how are you supposed to know if its a critical number? if it was a critical number, i thought that it was done, plain as that, but why check for ""saddle"" points for inflection

and such. if it was a ""saddle"" point, and had a ""hole"", would that not be the critical point? how would you solve for it to find the real point? this question seems it could get very complicated and a lot of past calculus is getting involved.

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Self-critique rating:"

@& Good question.

If the x axis points from the back to the front of the horse, and the y axis is horizontal and perpendicular to the x axis, then the bottom of the seat is a critical point in both directions. However the second derivative is positive in the x direction, negative in the y direction, giving you the saddle point.

It's also possible for a graph to have a minimum at the origin for both x and y directions, but to fall off as you move at, say 45 degrees to either axis. That constitutes a saddle point at the origin.

In that case the xx and yy derivatives would be positive, but the xy derivative would be negative.*@

Self-critique (if necessary):

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Self-critique (if necessary):

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#*&!

@& Good questions about the saddle point. We'll be talking about that soon, but for now check my notes.

Good questions and good work overall. Again, see my notes and let me know if you have additional questions.

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