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course Mth 277
November 11 around 5:45pm. I attempted them all and I put questions on what I don't understand, thanks!
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-z0=fx(x0,y0)(x-x0)+fy(x0,y0)(y-y0)
fx=(2x)/(x^2+y^2)=fx(e,0)=2e/e^2
fy=(2y)/(y^2+x^2)=fy=(e,0)=2(0)/e^2=0
z-2=(2e/e^2)(x-e)+0(y-0)
That will simplify to the correct standard form.
However be sure you understand the geometry of the situation, so you don't have to rely on a formula (you can check the Query document for a better-formatted version of this 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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Question: `q002. Find the total differential of f(x,y,z) = 2xzy^3*cos(xy)*sin(z)
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Your solution:
I know it would be:
df=f_x(dx)+f_y(dy)+f_z(dz)
after simplifying:
f_x= 2zy^3*cos(xy)*sin(z)-2xzy^4*sin(z)*sin(xy)
f_y=6xzy^2*cos(xy)*sin(z)-2zx^2y^3*sin(z)*sin(xy)
f_z=2xy^3*cos(xy)*sin(z)+2xzy^3*cos(z)*cos(xy)
df=[2zy^3*cos(xy)*sin(z)-2xzy^4*sin(z)*sin(xy)]dx+[6xzy^2*cos(xy)*sin(z)-2zx^2y^3*sin(z)*sin(xy)]dy+[2xy^3*cos(xy)*sin(z)+2xzy^3*cos(z)*cos(xy)]dz
Would you leave it as this LONG equation or would you simplify it?
My expression is
( 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,
which appears to agree with your solution.
The expression would then 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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Question: `q003. Use an incremental approximation to estimate f(sqrt(pi) + .01, sqrt(pi) - .01), where f(x,y) = cos(xy) f(sqrt(pi) + .01, sqrt(pi) - .01).
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Your solution:
f(x0,y0)=(sqrt(pi)+0.01, sqrt(pi)-0.01)
f_x=[-y*sin(xy)]
f_y=[-x*sin(xy)]
[since 0.01 is close to 0, we can just use sqrt (pi) to plug in first, then you can adjust the answer with the +0.01 and the -0.01. the sqrt (pi) is approx. 1.77]
f_x=[-1.77*sin(1.77*1.77)]
=-0.015+0.01=-0.005
f_y=[-1.77*sin(1.77*1.77)]
=-0.015-0.01=-0.025
You would multiply f_x by .01 and f_y by -.01. You wouldn't add them.
In the process we are multiplying the rate at which the function changes in each direction, by the change in the corresponding dependent variable. That is, you multiply f_x by dx and f_y by dy.
Did I do this right?
You want to use (x0, y0) = (sqrt(pi), sqrt(pi)). Apply the differential with dx = .01 and dy = -.01 to find the change in f(x, y) between (x0, y0) and your given point.
Your results are good to only two significant figures, so everything you have calculated is within roundoff error of 0.
You generally want to avoid the use of decimal approximations to find f(x0, y0).
cos(sqrt(pi) * sqrt(pi) ) = cos(pi) = -1.
Also note that
sin(sqrt(pi) * sqrt(pi) ) = sin(pi) = 0.
Using 1.77 instead of sqrt(pi) throws both results off.
See the solution in the newly posted query document. The formatting would not come through well in this posting.
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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:
Goal: z-z0=f_x(x0,y0)(x-x0)+fy(x0,y0)(y-y0)
f_x=-2x+6
f_y=-2y
z-z0=(-2x+6)(x-x0)+(-2y)(y-y0)
The examples in the book had points to plug in and then put into the equation of the tangent plane. Since this question doesn’t give me a “point (x,y,z)”, did I do the equation right? Or do I need to distribute the f_x and f_y to the (x-x0) and (y-y0)?
For this problem you need to check out critical points, do a second-derivative test and test for saddle points.
Formatting is much better on the revised original document, which now contains solutions. So if you can't follow this, you can check that document.
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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Given Solution:
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Good start. Check my note.
The Query document for 11.4 now includes solutions.
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