#1 Ellway  #2 Graziadio  #3 Grubin  #4 Gurkovich  #5 Hort 
#6  #7  #8 Kravtsov  #9 Lunemann  #10 Odoi 
#11 Pergament  #12  #13 Ryslik  #14 Sullivan  #15 
#16 Tokayer  #17a Vander Valk  #17b  #18 Walsh  #19 Zhu 
List of textbook problems students have asked me to review on Monday,
October 14: 14.5:5153, 14.3:85, 14.4:41,42.
A draft of formulas requested by students on the first
exam [Postscript
 PDF]
(first version
posted 10/14/2002; revised 10/15/2002)
3. Find an equation of the plane containing the points
(1,2,0) and (0,2,1) and
parallel to the line x=1+t, y=1+t, z=2t. The parametric equations x=1+t, y=1+t, z=2t can be converted into a vector v. Taking the coefficients of t in each parametric equation yields v =i+j+2k. The plane should be parallel to this vector. Using the point (1,2,0), define the vector r as i+2j. Then, define the vector s as r + tv. This vector represents a line parallel to vector v that lies on the plane. s equals (1+t)i+(2+t)j+(2t)k. Next, plug in t=1 (this value is arbitrary) to s to obtain the point (2,3,2). This point is on vector s which is on the plane, so this point must be on the plane. Now, with (1,2,0) and (0,2,1) given and (2,3,2) obtained, it is easy to find the equation of the plane. Find the vector from (1,2,0) to (0,2,1), which equals i+k. Also find the vector from (1,2,0) to (2,3,2), which equals i+k+2k. Take the cross product of these two vectors to get the vector normal to the plane. This vector is equal to i+3jk. Using this vector and the point (1,2,0), the equation of the plane is found to be (x1)+3(y2)z=0. Check this equation by plugging in the three points above. Also, dot the normal vector with vector v. The result equals zero, so the two are indeed perpendicular. Sent by Jeremy Grubin on Mon, 14 Oct 2002 01:53:17 
5. For the surface defined by x^5+y^5+z^3+xyz=0,
find an equation of the tangent plane at (1,1,1), and the value
of z_x at (1,1,1).
Suppose f(x,y,z)=x^5+y^5+z^3+xyz=0.
Equation of the tangent line: use the grad of the function: grad f =
(5x^4+yz)i + (5y^4+xz)j + (3z^2+xy)k = <5x^4+yz, 5y^4+xz, 3z^2+xy>
Evaluated at (1,1,1)= 4i + 4j + 4k = <4,4,4> Sent by Greg Hort on Sat, 12 Oct 2002 16:26:50 
10. Suppose that w=f(x,y), where f is a function satisfying
f(1,2)=3, f_x(1,2)=1, f_y(1,2)=2, f_{xx}(1,2)=3,
f_{xy}(1,2)=2, and f_{yy}(1,2)=0. Suppose further that
x=u+v1 and y=3uv1. Find
w_u{u=1,v=1} and
w_vu{u=1,v=1}.
By applying the chain rule to w_u, we get
w_u=w_x(x_u)+w_y(y_u)
=w_x*1+w_y*3v
=f_x(x,y)*1+f_y(x,y)*3v. Call this equation (#).
Using w_u in (#) from the previous part and applying the product
rule to
the part of w_y*3v, we rewrite:
w_vu=(w_u)_v
=(w_x*1+w_y*3v)_v
=(w_x)_v+(w_y)_v*3v+w_y*3. Sent by Atsuko Odoi on Sun, 13 Oct 2002 05:40:14 
13. Find & classify as well as you can all critical points of f(x,y) =
x^22yx^2+2y^2.
Eq. 1 f_x = 2x4yx = 2x(14y) and
Eq. 2 f_y = 2x^2 +4y 2(x^22y).
 f_xx f_xy     f_yx f_yy at the critical point. Assuming that the second partial derivatives of f(x,y) are continuous (which they are) and evaluating at the critical points (a,b) found above, If D>0 and f_xx (a,b) (the second partial derivative with respect to x and to x again at the point (a,b) ) >0 then f(a,b) is a local minimum. If D>0 and f_xx (a,b) < 0 then f(a,b) is a local minimum. If D<0, then f(a,b) neither a maximum or a minimum and is therefore a saddle point. Eq. 3 f_xx = 24y Eq. 4 f_yy = 4 Eq. 5 f_xy = 4x Eq. 5 f_yx = 4x Therefore the determinant (since f_xy is equal to f_yx) is f_xx*f_yy[f_xy]^2. Therefore D = D(x,y) = 816y16x^2. At the critical point (0,0) D = 8 and f_xx = 2. Thus since D>0 and f_xx(0,0)>0, the point (0,0) is a local minimum. At the critical point (sqrt(1/2),1/4), D = 4. Since D<0, the point is automatically a saddle point. Sent by Greg Ryslik on Thu, 10 Oct 2002 20:38:55 
16. Use the epsilondelta definition to verify that
g(x,y)=x^2y^2 is continuous at (1,2).
Definition of continuity:
A function f is continuous at x0 if, given any eps>0, there is a
delta>0 so that
if xx0
sqrt((x+1)^2+(y2)^2)<delta: Since both sqrt((x+1)^2) and
sqrt((y2)^2) are greater than or equal to zero, if the restriction is
satisfied, then each of the parts satisfies it. So then
x+1<delta and y2<delta are true also (because
(sqrt((x+1)^2)=x+1 and sqrt((y2)^2)=y2). We can look for delta
restrictions on x+1 and y2 separately. Sent by Jason Tokayer on Sun, 13 Oct 2002 10:29:19 Special added bonus! Here is the substance of several email messages I wrote to a student explaining another example. Maybe what's written here will be helpful to some people.
let's work through a 1 variable example. the algebra will be
irritating, but, what the heck. i believe that the limit of x^3 as
x> 2 is 8. so i should be able to make x^38 small by making
x2 small. but how could i verify that? (yes, i could graph things,
etc., but this is really preparation for a more than 1 variable
example, so i would prefer to stick with algebraic techniques.)

18. Suppose G(u,v) is a differentiable function over two variables
and g(x,y) = G(x/y,y/x). Prove xg_x(x,y) + yg_y(x,y) = 0.
For easy understandability, I will use u = x/y, and v = y/x.
Thus, g(x,y) = G(u,v). g_x(x,y) = D_1 G(u,v)*(du/dx) + D_2 G(u,v)*(dv/dx) = D_1 G(u,v)*(1/y) + D_2 G(u,v)*(y/x^2) and g_y(x,y) = D_1 G(u,v)*(du/dy) + D_2 G(u,v)*(dv/dy) = D_1 G(u,v)*(x/y^2) + D_2 G (u,v)*(1/x) so xg_x(x,y) + yg_y(x,y) = x*[D_1 G(u,v)*(1/y) + D_2 G(u,v)*(y/x^2)] + y*[D_1 G(u,v)*(x/y^2) + D_2 G(u,v)*(1/x)] = (x/y)D_1 G(u,v)  (y/x)D_2 G(u, v)  (x/y)D_1 G(u,v) + (y/x)D_2 G(u,v) = 0.Sent by Joseph Walsh on Sat, 12 Oct 2002 15:04:34 
19. a) Find parametric equations for the line through
the points (3,2,7) and (1,1,2). b) At what point does this line intersect the plane x+y+z=22?
a. Let vectors A = (3, 2, 7) and B = (1, 1, 2). Then the position
vector (x, y, z) for a point on the line AB is given by A + t(A  B).
So x = 3  4t, y = 2  t, z = 7  5t. Sent by Siwei Zhu on Sun, 13 Oct 2002 19:17:03
