#1 BERGKNOFF  #2 BERKOWITZ  #3 BROWN  #4 
#5 ELKHOLY  #6  #7  #8 HEDDY 
#9 KLEYZIT  #10 MCGOWAN  #11  #12 PASHKOVA 
#13 TOZOUR  #14  #15 WILSON  #16 
1.Suppose that F(s,t) is a differentiable function of two
variables and that F(5,3)=A, dF/ds(5,3)=B,
dF/dt(5,3)=C, d^2 F/ds^2(5,3)=D,
d^2 F/ds dt(5,3)=E, and
d^2 F/dt^2(5,3)=F.
If G(x,y)=F(x^2+y^2,x^2y^2), compute d^2 G/dx^2(2,1)
in terms of the information given.
There wasn't much chicanery involved in my solution, it was just a straightforward application of the chain rule twice. I differentiated G with respect to x first, getting dg/dx = (dF/ds)(ds/dx)+(dF/dt)(dt/dx). Then I used the product rule and chain rule to differentiate with respect to x again.
d^2 G/dx^2=2(dF/ds)+(2x)((d^2 F/ds^2)(ds/dx)+(d^2 F/dsdt)(dt/dx))+2(dF/dt)+(2x)((d^2 F/dt^2)(dt/dx)+(d^2 F/dtds)(ds/dx)) d^2 G/dx^2 evaluated at (x=2,y=1) is 2B + 2C + 16D + 32E + 16F (assuming that the last line on my problem is a typo and that should be F rather than a second E). Sent by Jonathan Bergknoff on Sun, 9 Mar 2003 20:40:33 Message from the management The error has been corrected (at least in the version posted on the web). It should indeed have been an F rather than an E. I thank Mr. Bergknoff for his alertness. Of course (sigh!), the F as a constant then conflicts typographically and logically with the F for the function. 
2.
The tangent plane of the graph of a function f(x,y) at point P(xo,yo,zo) is: equation 1: zzo = f_x(x0,y0)(xxo)+f_y(xo,yo)(yyo) where P(x_0,y_0,z_0)= (3,0,9) (we find z_0 by plugging in x_0 and y_0 in f). Here f_x(3,0)=6 and f_y(3,0)=9 which makes equation 1 become: equation 2: (z9) = 6(x3) + (9y) since the equation for a plane is a(xxo)+b(yyo)+c(zzo)=0 where is the normal vector, we rewrite equation 2 to look like this: 6(x3)+9(y)+(1)(z9)=0 The normal vector, , is <6,9,1>. Because we know that this is the normal vector at the point P(3,0,9), we can find parametric equations for the normal line using the equations: x=x_0at y=y_0bt z=z_0ct plugging the known numbers into these 3 equations gives us the paremetric equations: x=36t y=9t z=9t
Sent by Ron Berkowitz on Sun, 9 Mar 2003 18:37:33 
3.The surfaces x^3+5y^2z4z^2=3 and y^47z2x^2=5
intersect in a curve, C. The point p=(3,2,1) is on C. Find
parametric equations for a line tangent to C at p.
Let's call this surface A: x^3+5zy^24z^2=3 and this one B: y^47z2x^2 =5. It is very difficult to find out the exact equation for where A and B intersect, but it is not necessary. Your original intuition could follow that this line tangent to C (lets call it T) is a line that is tangent to both A and B. Therefore, this line is going to reside in the tangent planes to A and B, and go through the point (3,2,1). Unless both tangent planes are equal, there will be 1 distinct line through tangent planes to A and B at (3,2,1). We start by finding each tangent plane. The book says that a tangent plane to a curve S through (X_0,Y_0,Z_0) is ZZ_0=(dz/dx)(XX_0)+(dz/dy)(YY_0).
A: (we must do implicit differentiation of x^3+5zy^24z^2=3)
B: (regular partial differentiation from y^47z2x^2=5) The normal vector for the tangent plane to A at (3,2,1) is
<(27/28),(20/28),1>. Well all lines in the tangent plane are perpendicular to the normal vector, so our tangent line T must be perpendicular to both normal vectors of the planes A and B. The cross product of these normal vectors will be a vector in the direction of T.
The cross product is <9/7,3/4,33/49>. We can use <3/7,1/4,11/49> (just divided by common factor of 3 which does not affect the direction). Sent by Matt Brown on Fri, 7 Mar 2003 04:03:57 Message from the management I might have just taken the gradient of the lefthand sides of each of the equations x^3+5y^2z4z^2=3 and y^47z2x^2=5, remembering that gradient vectors are perpendicular to level "sets" (curves in two dimensions, surfaces in three dimensions, etc.). Then I would "plug in" (3,2,1). That way I have to think and remember less. You can try these Maple commands: with(plots): F:=(x,y,z)>x^3+5*y^2*z4*z^2; G:=(x,y,z)>y^47*z2*x^2; RED:=implicitplot3d(F(x,y,z)=3,x=1..5,y=0..4,z=3..1,axes=normal,color=red,grid=[20,20,20]): BLUE:=implicitplot3d(G(x,y,z)=5,x=1..5,y=0..4,z=3..1,axes=normal,color=blue,grid=[20,20,20]): display3d({RED,BLUE});A view of what is produced is included here, so you perhaps can "see" the two surfaces and the intersection curve. It is remarkable to me that the tangent line can be described so easily. The curve itself is difficult to describe algebraically. 
5.Consider the function f(x,y,z)=x2y+5z if
(x,y,z) is not (0,0,0) and 10 if (x,y,z)=(0,0,0). a) Use the definition of continuity to verify that f is continuous if (x,y,z)=(2,6,7). b) Use the definition of continuity to verify that f is not continuous if (x,y,z)=(0,0,0).
The point that we are asked to find whether the function is continuous
or not lies on the part of the function where f(x,y,z)=x2y+5z. To
verify that the function was continuous at (2,6,7), we need to check
the following statement: Given K>0, there is H>0 so that if
sqrt((x2)2+(y6)2 +(z+7)2) Now we must prove that the function is discontinuous when (x,y,z)=(0,0,0). Here P2=(a,b,c)=(0,0,0) and f(P2)=10. For (x,y,z) near (0,0,0) but not equal to (0,0,0), f(x,y,z)=x2y+5z. This is very small. So let's take K=4. I claim that no H>0 will work. For example, if H=1/(10,000), then (x,y,z)=(1/(20,000),0,0) is close enough to (0,0,0), and f at that (x,y,z) is 1/(20,000). 1/(20,000)10>4. If we try any H>0, then take (x,y,z)=(*,0,0) where * is the minimum of 1/(20,000) and H/2. (H could be smaller than 1/(20,000): we need to consider this possibility). Then f at this (x,y,z)f(0,0,0)>4, so f is not continuous at (0,0,0). (The management helped with this case a bit.) Sent by Emir Elkholy on Tue, 11 Mar 2003 11:51:11 
8. a) What is the sum, V, of the twelve vectors from the
center of a clock to the hours? b) If the 4 o'clock vector is removed, find V for the other eleven vectors. c) If the vectors to 1, 2, 3 are cut in half, find V for the twelve vectors. Source Calculus by Gilbert Strang. In a) the twelve vectors are composed of six pairs of vectors with equal magnitude and opposite direction. The twelve o'clock vector (R12) will add with the six o'clock vector (R6), and likewise around the face R1+R7=0, R2+R8=0, R3+R9=0, R4+R10=0, R5+R11=0. Thus the sum of the vectors, V, will be 0. In b), R4 is removed. Ten of the eleven remaining vectors are still paired with their opposites, however, and add to 0. The only vector that will have an effect on the vector sum V is R10, the vector opposite R4, which is now no longer available to counteract R10. The vector sum, then, will be R10. In c), R1, R2, and R3 are reduced to half their initial magnitude. In this case, we look at the opposite vectors, R7, R8, and R9. Since R1+R7, R2+R8, and R3+R9 all equal zero, (R1/2)+R7 = (R7)/2 and so forth. The contributing vectors are half the magnitude of the original vectors, in the original directions of R(79). The vector sum V will be (R7+R8+R9) divided by two.
Sent by Ryan Heddy on Sat, 8 Mar 2003 15:43:07 Message from the management
Of course if this is a conventional clock, we can imagine the angles
between vectors. Mr. Heddy kindly offered these additions to his
original answer:

9. Suppose S is the surface defined by the equation
z cos(x^3+y^2)+xz^2+2=0. The point (1,1,2) is on S. Assume that
the equation defines x implicitly as a differentiable
function of y and z near (1,1,2). If y is changed from 1 to
1.03 and z is changed from 2 to 1.96, use linear approximation
to find an approximate x so that (x,1.03,1.96) is on S. Note You are not asked to solve a nonlinear equation. You are asked to use a certain approximation strategy to get an answer.
In this problem x is written implicitly as a function x=f(y,z) in
the
form F(x,y,z)=0 with z*cos(x^3+y^2)+x*z^2+2=0. By Taylor's Theorem:
Returning to the linear approximation:
f(1,2)+f_y(1,2)*h+f_z(1,2)*k, where h=.03 and k=.04. Hence x+w=1.03=1.03 where (1.03,1.03,1.96) is the approximate point on S. Sent by Max Kleyzit on Sun, 09 Mar 2003 12:15:41 Message from the management The Maple command fsolve allows approximation of roots to equations. The reported root is 1.030533773: the ``higher order'' effects are not felt until the 4th digit! 
10. Find the area of the triangle whose vertices are the points
A=(1,2,1) and B=(2,1,0) and C=(0,0,4) in R^3. Also
find the coordinates of a point D which together with A and B
and C will form the vertices of a parallelogram. The area is one half of the square root of two hundred thirty The points that would make the triangle a parallelogram are (1,1,3),(3,3,3)and(3,3,5). The area of the triangle is half of the length of the cross product of two vectors that define the triangle. To find the vectors, subtract the x, y, and z coordinates of a two of the points. Make sure one is a reference point, EXAMPLE use AB, and AC to get vectors, calculate their cross product, use the Pythagorean Theorem to find the length of the vector and divide by 2. We divide by 2 because the length of the cross product gives the area of a parallelogram, and that parallelogram has twice the area of the triangle.
To find another point that would make the triangle a
parallelogram, we use the parallelogram method of adding vectors.
Sent by William McGowan on Thu, 6 Mar 2003 10:45:24 Message from the management For those who may not believe that there can be more than one answer to the second part of the problem, here is a possible picture of the situation, drawn in the plane containing the points A and B and C. Only one possible solution (one of the D's) is requested in the problem. 
12. Find the point (x,y) in the plane for which the sum
of the squares of the distances from the three points (a_i,b_i),
i=1,2,3, is a minimum. Give a physical interpretation of your
answer. Source An MIT calculus exam.
The distance between each point (a_i, b_i) and (x,y) is given by:
sqrt( (xa_i)^2 + (yb_i)^2) ).
So the sum of all three distances squared is:
f(x,y)=(xa_1)^2+(yb_1)^2+(xa_2)^2+(yb_2)^2+(xa_3)^2+(yb_3)^2. What we found to be the point (x,y) is the centroid of a triangle with vertices (a_1,b_1), (a_2,b_2), (a_3,b_3). The centroid can be found by constructing 3 medians (connecting a midpoint of a side to the opposite vertex). The centroid is known to be the center of gravity of a triangle i.e. you can balance the triangle on that point. Sent by Anastasia Pashkova on Tue, 11 Mar 2003 13:00:15 
13. Find and classify all the critical points of these
functions. a) K(x,y)=3x^2+6xy+12y^3+12x24y. Source An MIT calculus exam. b) J(x,y)=8xyx^4y^4.
K(x,y)=3x^2+6xy+12y^3+12x24y
so dK/dx=6x+6y+12 and
dK/dy=6x+36y^224.
We can use the same method in this part as we did in Part a to find
the critical points. Sent by Elizabeth Tozour on Thu, 13 Mar 2003 00:50:40 
15. Sketch z=x^2+y in R^3. The normal line to the
surface at a point is the line perpendicular to the tangent plane to
the surface at that point. Find all normal lines to z=x^2+y which
pass through (0,0,3).
I will rewrite the equation as an implicit function of three
variables, F(x,y,z)=x^2+yz=0. The normal vector N to a tangent
plane at P=(a,b,c) equals grad
F(P)=
Since this line must pass through the point (0,0,3), we can set x=0,
y=0, and z=3. We solve the first equation assuming that a is not 0;
1+2t=0, t=1/2. Therefore, b=t=1/2, and c=3+t=5/2. P is on
the surface of F, so we can solve the function for a. a^2+bc=0,
a^2=cb=2, a =+/sqrt(2). Plugging these values of a, b, and c back
into the equations for the normal line, we get
x=+/sqrt(2)(1+2*t)
We must also solve if a=0, in which case x=0 always, and b=c3
(solve y and z for t and set them equal). We then plug
a, b, and c back into the function equation again and find
0^2+(3c)c=0 or 3=2c, c=3/2. Therefore, b=3c=3/2.
Plug these values into the parametric representation of the
normal line. Sent by Neil Wilson on Sun, 9 Mar 2003 22:55:56 Message from the management I think this is a difficult problem. To the right is an approximate sketch of the situation. To the left is the output of a Maple command Mr. Wilson suggested: plot3d(x^2+y,x=5..5,y=5..5,axes=normal); 