# Sample Answers to Midterm Test

## Q1

(a) Uniform circular motion is a moving point (x(t), y(t)). The coordinates
of the point varies with time according to the functions

x(t) = R cos(kt+d), y(t) = R sin(kt+d).

where R, k, and d are parameters and t is time.

(b) In polar coordinates we use the inverse transform of

x = r cos q, y = sin q

to obtain

r(t) = R, q(t) = kt + d
where r(t) is radial length and q(t) is angle.

(c) We transform the values to polar coordinates to get

t x y q r
0.0 1.0 0.0 0.0 1.0
1.0 0.71 0.7 0.7783 0.997
2.0 0 1.0 1.5708 1.01

using the formula q = arctan(y/x) and r = sqrt(x^2+y^2).
We fit into the result of (b). For r(t), it is a constant, we use the
average R = (1.0+0.997+1.01)/3 = 1.002. For q(t), t versus q is a straight
line fit (with three data points), q(t) = kt + d, which we use normal equation method (equivalent
to least square), thus (Matlab notation)
A = [ 0.0 1; Q = [0.0;
1.0 1; 0.7783;
2.0 1] 1.5708]

such that A*[k; d] = Q. The solution is A^T A [k; d] = A^T Q.
This gives k = 0.7854 and d = -0.002366, therefore the fitting results
for q(t) is
q(t) = 0.7854 t -0.002366.

## Q2

The result looks similar to tutorial 2.6, except that the bifurcation
diagram is given by epsilon = -5x^4 (instead of quadratic).

## Q3

The dimensions of the relevant quantities are

force, [F] = MLT^(-2)

radius, [r] = L

mass, [m] = M

velocity, [v] = L/T

By balancing the dimensions,
[F] = M LT^(-2) = M^x v^y r^z = M^x L^y T^(-y) L^z
we obtain the equations

1 = x, 1 = y + z, -2 = -y.

So the solution is x = 1, y = 2, z = -1. And

F = const m v^2/r.

## Mark distribution

Two of the highest are 100 (full marks), and third is 97, followed
by 81 and 80. The lowest is 20, and 2nd lowest is 25, and up by 30.
The rest are in 70s, 60s, 50s, and 40s.