UMTYMP Calculus III
Fall 2006
Exam 1 Review Problems
Hints/Solutions

Note: this page has hints and solutions for many, but not all, of the problems on the review sheet.  If you're stuck on a problem not listed here, let me know!

Problem 1

(a) If you find the velocity vector correctly, you'll find that it has length k R.  Hence the speed is 1 whenever k=1/R.
(b) All values.
(c) All values.
(d) See (a)
(e) All values.
(f) No values.
(j) The curvature is 1/R; this shouldn't depend on t or k.  It's a geometric property of the curve, and we've already shown in class that a circle has constant curvature of one divided by its radius.

The osculating circle at any point is the circle itself.  (!)

Problem 2

If the speed is constant, then the acceleration vector is always perpendicular to the tangent vector.  In the review session we described a number of ways to see this.  My favorite is to take the formula =v'+κ, where v is the length of the velocity (or tangent) vector, i.e. speed.  If v is constant, then v'=0, and the claim in the first sentence follows.

Problem 3

One possibility is f(t)=t2, although the dot product of the acceleration and velocity vectors is still zero at the one point t=0.  Can you find any other examples where this doesn't happen?

Here's a picture for t=π/4 with f(t)=t2.   The color scheme for the vectors is:

black: position vector
red: velocity vector
green: acceleration vector
blue: unit tangent vector.

Problem 4

Remeber that I mentioned in class to replace "z-axis" with "x-axis" in the first sentance.  One possible answer is given by

r(t) = < t/2, -2 cos(t)+6, 2sin(t) + 4 >

A graph of this curve is shown below for the range 0≤t≤2π.

Problem 5

From smallest curvature to largest: B, A, then C.  If you're intersted in the osculating circles, then the osculating circle will have the largest radius at B, then A, and the smallest at C.

Problem 6

I think everybody assumed this, but it's important to say that this closed curve doesn't intersect itself at all.  In that case, the answers are:

(a) All three could be different.

(b) There is no way to know in which direction any one of these vectors point, but at least two of them must be equal.  (There are only two possible unit tangent vectors, and we have three.)

(c) The curvatures must all be the same; curvature is a geometric property of the point A on the curve and does not depend on the parametric functions.

Problem 7

(a) One possibility is r(t)=<0,+t+2,0> with 0≤t≤1.

(b) One possibility is r(t)=<0, 3-cos(π t), sin(π t) > with 0≤t≤1.

Problem 8

(a) In polar coordinates, from -π/2 to π/2 this give the line x=3.  If you consider angles π/2 to 3π/2 then you also get the line x=-3.   In cylindrical coordinates you would get vertical planes at x=±3 instead.

(b) This gives a right circular cylinder of radius 1whose central axis is parallel to the z-axis, but shifted over to where x=1/2.

(c) In pictures:

Here's a cross section so you can get a better idea of what it looks like:

Problem 9

Problem 10

Although it often helps to begin thinking in spherical (or cylindrical) coordinates, parametrizations should always be given in rectangular coordinates unless we tell you otherwise.

In spherical coordinates, you want θ to range from 0 to 30π in the time it takes φ to go from 0 to π.  This can be accomplished with the following function, whose graph is also shown below.

r(t)=< 10 sin(t)cos(30t), 10 sin(t) sin(30t), 10 cos(t)> with 0≤t≤π.

Problem 10

Problem 12

(a) Ok, this was a bit harder than intended.  The answer is 15+ln(4).

(b) s(t)= du

Problem 15

Lighter shades indicate higher values of z.

Problem 16

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.

Problem 17

In[27]:=

Out[27]=

Lighter shades indicate higher values of z.

Problem 18

Go to http://www.math.umn.edu/~rogness/test/rollerCoaster.html to see the animation of this curve which we viewed in class.  It might help you solve this problem. 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 19

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 20

 Created by Mathematica  (October 9, 2006)