MAT 331 Homework Exercises. Week 3 (Oct 5, 99).

NOTE: [No Maple] means that the problem does not involve Maple, except as a word processor to write your solution. In this case you can alternatively turn in a printed version, if you are more comfortable with that.

#11 (exp. 10/19)

Fit the points (1.02, - 4.30),(1.00, - 2.12),(0.99, 0.52),(1.03, 2.51),
(1.00, 3.34),(1.02, 5.30) with a line, using the least square method we used in class. You will see that this is not a good fit. Think of a better way to do the fit and use Maple to do it.

#12 (exp. 10/19)

[No Maple] Once we have calculated the line (or any other curve, for that matter) that best fits a sets of points, we can get an idea how good the fit is by plotting the line together with the points. It is much more scientific, however, to have a measure for this. Come up with a function of the data and parameters of a given best-fit problem that is small when the fit is good and large when the fit is bad, no matter how many points are used. Justify your answer.

#13 (exp. 10/19)

In this problem we will estimate the charge of the electron: If an electron of energy E is thrown into a magnetic field B, perpendicular to its velocity, its trajectory will be deflected into a circular trajectory of radius r. The relation between these three quantities is:

 

$${\frac{E^2}{m^2 c^4}} \sqrt{E^2 - m^2 c^4}$$ (1)

where e and m are, respectively, the charge and the mass of the electron, and c is the speed of light. The rest mass of the electron is defined as E₀ = mc², and is about equal to 8.817 10^-14Joules. In our experimental set-up the energy of the emitted electrons is set to be E = 2.511E₀.
Use read to make Maple load and execute the commands in the file electron_data.txt, which is located in the directories Worksheets1 and Worksheets2 of the mat331 account. This defines a list called electron. Each element of the list is a pair of the form [B_i, r_i], and these quantities are expressed in Teslas and meters. Use least square fitting to determine the best value for e. [Hint: Notice that the right hand side of () is just a constant--calculate it once and for all and give it a name. Then () is a very easy equation, which is linear in the unknown parameter e. To verify your solution: e $\approx$ 1.602 10^-19 Coulomb].
Physical constants courtesy of N.I.S.T.

#14 (exp. 10/19)

Prove relation (), knowing the following physical facts: In relativistic dynamics Newton's law is replaced by

$${\frac{d}{dt}} \left(\vphantom{ \frac{v} {\sqrt{1 - \frac{v^2}{c^2}}} }\right. {\frac{v}{\sqrt{1 - \frac{v^2}{c^2}}}} \left.\vphantom{ \frac{v} {\sqrt{1 - \frac{v^2}{c^2}}} }\right)$$ (2)

where F is the force acting on a particle, m its mass and v, a function of time, its velocity. In the case at hand, the force exerted by a magnetic field B on an electron is F = e v B/c. Recall that in a circular motion the acceleration a = dv/dt = v²/r, rbeing the radius of the circle. Since () is expressed in terms of the energy, rather than the velocity, you also need Einstein's formula,

$${\frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}}}$$ (3)

which can be solved in terms of v.

#15 (exp. 10/19)

[No Maple] Following Section 4 of the notes, prove that if we describe the circle of center (a, b) and radius r using the parameters (a, b, k), with k = a² + b² - r², rather than the more natural parameters (a, b, r), then the error function H(a, b, k) = E(a, b,$\sqrt{a^2 + b^2 -k}$) is quadratic in a, b and k. What does this imply about the number of critical points?

#16 (exp. 10/19)

[No Maple] With reference to Problem #15, show that, for r > 0, the transformation (a, b, r) $\mapsto$ (a, b, k) is a change of variables, that is, is one-to-one. This should help you prove that E(a, b, r) has only one ``physical'' critical point, which is a minimum, and is mapped, through the transformation, into the unique critical point of H(a, b, k).