# Tutor profile: Jordan M.

## Questions

### Subject: Set Theory

Consider the set of three elements, $$X = \{a,b,c\}$$. Recall the definitions of topology, openness, and closedness. a) Give a non-trivial example of a topology on the set $$X$$ such that the subset $$\{a,b\}$$ is closed. b) Give a non-trivial example of a topology such that $$\{a,b\}$$ is neither open nor closed. c) Give a non-trivial example of a topology such that $$\{a,b\}$$ is both open and closed. By "non-trivial," I mean neither the trivial nor the discrete topology.

These are the simplest posible answers: a) $$T = \{X, \emptyset, \{c\}\}$$ b) $$T = \{X, \emptyset, \{a\}\}$$ c) $$T = \{X, \emptyset, \{a,b\},\{c\}\}$$

### Subject: Calculus

Let $$f(x) = 3x^{3} + 36x^{2} + 108x$$. Find all local maxima, local minima, and inflection points of $$f(x)$$

Local maxima and minima are found where the derivative of the function is zero. So we differentiate, $$f'(x) = 9x^{2} + 72 x + 108$$ Set this function equal to zero, $$0 = 9x^{2} + 72 x + 108$$ Solve for $$x$$ using the quadratic formula, $$x = -2, -6$$ To determine whether each of these is a max or a min, we must take the derivative of the function again, $$f''(x) = 18x + 72$$ At a local maximum, this function will be negative. At a local minimum, it will be positive. So, plugging in our $$x$$ values, $$f''(-2) = -36 + 72 = 36 > 0$$ $$f''(-6) = -108+72 = -36 < 0$$ So we have a local max at $$x = -6$$ and a local min at $$x = -2$$. Inflection points are found where the second derivative equals zero, $$0 = f''(x) = 18x + 72$$ Solving for $$x$$, $$x = -4$$ So we have an inflection point at $$x = -4$$.

### Subject: Physics

A solid sphere of radius $$R$$ and mass $$M$$ is fixed to a vertical pole atop a table so that it is free to revolve about the $$z$$ axis. Wound about its equator is a mass-less string which is attached to a falling block of mass $$m$$ by a mass-less, friction-less pulley. Suppose everything in the system starts at rest. Find the velocity of the block as a function of time.

The block is being acted upon by two forces: its downward gravitational force, $$mg$$, and the upward tension in the rope, $$T$$. But because the box is attached to the equator of the sphere by the string, it can only move at the same speed as the surface of the sphere. In other words, if the angular velocity of the sphere is given by $$\omega$$, the velocity of the surface is given by $$R\omega$$, and this must always equal the velocity of the box, $$v = R\omega$$. The mass-less rope transfers the tension force $$T$$ to the sphere, applying a toque of magnitude $$\tau = TR$$ at its equator, since torque is force times radius times the sin of the angle, which here is 90 degrees. Torque is related to angular acceleration by the equation, $$\tau = I\alpha$$, where $$I$$ is the moment of inertia of the object subjected to the torque. The moment of inertia of a solid sphere is $$\frac{2}{5}MR^{2}$$. Plugging this and the torque into the above equation gives us, $$TR = \frac{2}{5}MR^{2}\alpha$$ Simplify, $$T = \frac{2}{5}MR\alpha$$ Since our system starts at rest, the sphere's angular velocity $$\omega$$ is related to angular acceleration $$\alpha$$ by $$t\alpha = \omega$$, or $$\\alpha = \omega / t$$, where $$t$$ is time elapsed. Let's plug this in, $$T = \frac{2}{5}MR\alpha = \frac{2}{5}MR\frac{\omega}{t}$$ Notice that we now have $$R\omega$$ on the right hand side, which is equal to $$v$$. Let's substitute that in instead, $$T = \frac{2}{5}Mv \frac{1}{t}$$ Now, in this equation we have two unknowns, $$T$$ and $$v$$, so in order to find them, we need at least one more equation relating the them. Thankfully, we can also examine the forces on the box itself. We have two forces on the box, $$mg$$ down and $$T$$ up. Since we are moving downwards, let's make that the positive direction. So the net force on the box is, $$F = mg - T$$ By Newton's 2nd Law, $$F = ma$$ $$ma = mg - T$$ Since our block starts at rest, $$v = at$$, or equivalently $$a = v/t$$, $$m\frac{v}{t} = mg - T$$ We can replace the $T$ in this equation with the expression we found for $$T$$ earlier, $$m\frac{v}{t} = mg - \frac{2}{5}Mv \frac{1}{t}$$ Now we simply solve for $$v$$, and we are done, $$v = \frac{m}{m + \frac{2}{5}M}gt$$ Notice how this results relates to the velocity of the box in free-fall, $$v_{free-fall} = gt$$ If the mass of the sphere, $$M$$, is very small compared to $$m$$, the ratio $$\frac{m}{m + \frac{2}{5}M}$$ gets very close to $$1$$, indicating that the box is pretty much in free-fall and the sphere provides little resistance to the box's motion. On the other hand, if $$M$$ is very large, the ration decreases, slowing the box down. This indicates that a heavy sphere keeps the box from moving quickly, which makes intuitive sense!

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