George's explanations of some of the most common  
mistakes people made on the exam.

Problem 1

The mistakes on this problem were mostly small errors.  One common  
error was dividing by the wrong thing to get T and N, so remember  
that |T|=|N|=1.

Problem 2

When it says "find and graph" "find" means write an equation for the  
circle and "graph" means draw it.

Problem 3

There were two big mistakes on this problem that cost people a lot  
of points.  First of all, it gives a function f(x,y) and it asks for  
the gradient.  Since f is a function of two variables, the gradient  
vector should have two terms.

In the problems where you have to find the normal vector to a  
surface defined by z=f(x,y), you define a new function  
g(x,y,z)=f(x,y)-z and take the gradient of g.  Some people did the  
same thing on this problem, even though the problem asked for the  
gradient of f and not the normal vector of the surface.   
Furthermore, since the equation involving x, y, and z expressed z^3  
in terms of x and y, rather than z.  This meant that if you defined  
g(x,y,z)=3x^3+3y^3-z^3, then the gradient of g is significantly  
easier to compute than the gradient of f.

The other error was that some people got that f(x,y) was the cube  
root of 3x^3+3y^3, but then claimed that this was equal to the cube  
root of 3 times x+y.  NEVER DO THIS!

For the record (x+y)^3=x^3+3x^2y+3xy^2+y^3, not x^3+y^3.  The only  
way for those to be equal is if x, y or 3 are equal to zero.

Problem 4

First of all, it says "find the extreme values of f(x,y)".  At the  
point (2,0) the VALUE OF f(x,y) is 32, so your answer for maximum  
value should be 32, not (2,0).  As long as you calculated that  
f(2,0)=32 somewhere on the page you still got full credit even if  
you put the box around 32 instead of (2,0), but if you only  
calculated the points and not the values you did lose points.

Also, when you get the equation 4y=lambda2y this means EITHER y=0 OR  
lambda=2.  Make sure you do both cases.

Problem 5

In part (a), the cross sections are mostly ellipses of the form  
9x^2+16y^2=c-1.  In the case where c=0 the equation has no solutions.

In part (b), it was not asking for the entire three-dimensional  
graph, but the two dimensional graph on the (x,z)-plane (i.e.  
setting y=0).  This should be the parabola z=9x^2+1.

In part (c), there was a little booboo in designing the problem.   
The point in question is the vertex of the elliptic paraboloid, so  
the gradient is zero at that point.  Under normal circumstances, you  
would take the gradient and divide it by its length but, in this  
case, you can't divide by zero, so that doesn't work in this case.  
However, here, the rate of change in any direction is zero, so  
f(x,y) changes equally rapidly in every direction.  Consequently any  
unit vector will be the right direction.  If you got a unit vector  
and got zero as the rate of change, you should get full credit.  If  
you divided the gradient by zero you didn't.

Sorry about that.

Problem 6

This was the problem with the most perfect scores.  The biggest  
thing that caused people to lose points was not showing the work on  
the second derivative test.  If you write the correct answers  
without writing the equations for (f_xx)(f_yy)-(f_xy)^2 and then  
there's no way to show that you didn't just make a lucky guess.   
Furthermore, if you did the second derivative test incorrectly in  
your head there is no way of assigning partial credit.  Other than  
that, and a few scattered arithmetic errors, there weren't many  
problems with this.

Problem 7

People also did really well on this one.

When you use the normal vector in your equations, make sure you plug  
the point in.  Otherwise the equation for your tangent plane will no  
longer be a plane.  Also, remember that, for this problem, you are  
defining an equation for the tangent plane and not a function  
T(x,y). So your answer should be (some expression with x, y and  
z)=(some other expression with x, y and z) and not (The Tangent  
Plane)=(some expression with x, y and z).

For the normal line, remember that the normal line and the normal  
vector aren't the same thing.

--
George Martin Fell Brown