UMTYMP Calculus II
Spring 2005
Exam 2 Review Problems
Hints/Solutions

You can find answers to many of the problems from the textbook in the back of the book.

Problem 1

In[3]:=

ContourPlot[(x - y)^2, {x, -1, 1}, {y, -1, 1}] ;

[Graphics:HTMLFiles/index_2.gif]

Lighter shades indicate higher values of z.

Problem 2

Here's a graph of the level set g(x, y, z) = 0; it's actually the graph of the function in the previous problem.  Other level surfaces of the function g are vertically shifted versions of this surface.

In[11]:=

ContourPlot3D[z - (x - y)^2, {x, -2, 2}, {y, -2, 2}, {z, 0, 2}, AxesTrue, AxesLabel {"X", "Y", "Z"}] ;

[Graphics:HTMLFiles/index_6.gif]

Problem 3

Here's an animation of the curve which might help you solve the problem.  Here red represents the velocity vector, and green represents the acceleration vector.  For good measure we've included the osculating circle, which sits in the osculating plane.  Notice how the circle is always "parallel" to the velocity and acceleration vectors, since they span the osculating plane.

Problem 4

(This is a pictorial problem and was done in class.)

Problem 5

The limit of f(x, y) is 0 in the first problem.  In the second, the limit of g(x, y) does not exist, as explained in class.

Problem 6

The acceleration is defined everywhere except possible when t = 2, or t = 2 + π/2.   Check for yourself to see what happens.  The acceleration is in the direction of motion along the portion of the curve which is a line, and orthogonal to the direction of motion in the second piece of the curve.  The acceleration is always horizontal when it's defined.

Problem 7

[Graphics:HTMLFiles/index_11.gif]

Problem 8

[Graphics:HTMLFiles/index_12.gif]


Created by Mathematica  (March 1, 2005)