UMTYMP Calculus II
Spring 2005
Exam 1 Review Problems

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

One possibility is f(t) = t, 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) = t.   The color scheme for the vectors is:

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


Problem 3

One possibility 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 4

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

Problem 5

(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 6

(a) One possibility is r(t) = 〈0, t^2 + 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 7

(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 8

(* Cartesian *)Sqrt[π^2 + π^2 + π^2] = 3^(1/2) π

(* Cylindrical *)Sqrt[π^2 + π^2] = 2^(1/2) π

(* Spherical *)π

Problem 9

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 11

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

(b) s (t) = ∫_0^t (3 + 4u^2)^(1/2) du

Created by Mathematica  (January 29, 2005)