Learning some humility from a technical interview

I had a brutal interview today for a tutoring job. Writing it up as a lesson to myself and others about under-preparation and overconfidence going into an interview with a test component.

It started nicely enough –resume questions from the president, who is an econ major, and some jovial joking with a Caltech engineering major who had worked at JPL. I bedazzled with my complicated one-sentence statement of my previous research and my more understandable explanation of what “non-redundant aperture masking with adaptive optics” really meant. I displayed a comfort level with the subjects I’d be expected to tutor (biology, chemistry, physics, calculus, pre-calculus) and even tied in my behavioral econ knowledge to indicate how I can relate to individuals from different disciplines and career aspirations.

I was anticipating a diagnostic test, which I had prepared for by taking a sample Physics B AP test. I did reasonably well, missing a couple questions concerning induction and the lensmaker equation (apparently I forgot the sign convention for the focal length, where it is positive if the lens is convex and negative if it is concave).

Just as I’m about to be asked some qualitative physics questions from a high school physics textbook, another tutor arrives. He is apparently a Caltech senior physics major, currently applying to grad schools. His engineering colleague, seeing the physicist arrive, decides to have him ask me some questions.



Question 1: (Part 1) You have a frictionless cart traveling to the right with some velocity. It has an open top, and it is raining vertically. What happens with the speed of the cart with time?


(Part 2): Assume that the rain has stopped. The cart is now moving at some velocity to the right, but has sprung a slight leak at the bottom. What happens to the velocity of the cart?

Ok, easy enough. For part 1, the cart is filling with water, but since the rain is falling vertically, it has no horizontal linear momentum. It is making the cart more massive, which by the conservation of linear momentum, means that the cart’s speed should be decreasing.

Numerically,




On the second part, I think, "Ah trickier, but I see it!" Because the cart and the water are traveling together at the same velocity, the loss of the water doesn’t affect the overall speed of the cart. The cart is losing mass, but since the lost mass also has the same velocity, the cart continues at the same speed.

He informs me that I’m right.

At this point I breathe a little easier. I think, ok, I can do this. Caltech isn’t going to intimidate me!

Then he states the next question.

Question 2 : Imagine a scale with negligible response time. On top of this scale is an hourglass with the sand currently held in the top half by a plug. Draw the weight read by the scale as a function of time after the plug has been removed.

Oh, crap. It looks tricky. But a solution jumps out at me – no, it can’t be, it’s too easy. But I give it a try.

"No, that’s not right."

Hm…. I think, starting to sweat. Maybe there is a response. And without thinking about it too much, I think – aha! The sand is in freefall for a period of time. So I modify my answer, and with some relief, offer it for inspection.

I’m told that no, that’s not right either. Here he starts helping me. He states that if you think of force as a change in momentum over time, and you integrate over the entire time you should get zero.





Ok, I say, fumbling with the whiteboard markers in my hand, briefly wondering whether inhaling them would put me in a more relaxed, happy state. I think to myself, “Well, toward the end, the particles have to be mostly settled, and only a bit is in freefall. So it may well be heavier then." I then, somewhat halfheartedly, draw this:

“You’re halfway there.”

FUCK!

He points out that the sand takes time to fall, and that there will be an intermediate point where roughly equal amounts of sand will be hitting the bottom as will be leaving the top. He puts me out of my misery and draws a final solution.

At this point I’m kind of shaken, but assuage my ego with the knowledge that I was at least able to comprehend the steps.

Problem 3: Imagine two rectangular pieces of glass, separated by about one inch and sealed at the edges. It is partially filled with water. The image is of a very wide, very thin fish tank. It is placed on a flywheel rotating at constant angular velocity. Derive the shape of the water’s final steady-state shape, assuming there is enough water to not have it bottom out, but not so much that it reaches the top of the tank.

What the fuck.


Here’s the thing. I know the answer – but didn’t know how to derive it. I was kind of sure it was a parabola, y = ax^2, because I remember doing this before, possibly in theoretical mechanics. It was definitely a paraboloid in the full 3-d case, since I knew that, in astronomy, there has been research into developing a rotating mercury mirror. Such a rotating pool would form a perfect paraboloid, perfect for focusing light at a single focal point.

But I wasn’t completely sure about the 2-d case. And I had no idea how to derive it.

I basically mumbled and stared at the board for a while. I eventually broke down and said that I think it was a parabola, but I wasn’t sure how to derive it.

“Ok. Imagine a tiny surfer at a point on the curve. What are the forces acting on it.”

“Well… there’s a centripedal force, and gravity…”

“And what supplies the centripedal force?”

“Uh… I suppose that would be… the normal force from the water…”

“Ok.”

Ok, so I got the point that I needed to make a force diagram. But I struggled to define a coordinate system. First I tried keeping it in the x-y frame. Then the inquisitor – er, questioner – suggested doing it in the surfer’s frame, making use of a centripedal force.

“You mean centrifugal?” I corrected. (I can be damn petty when I’m cornered.)

“That’s right,” he said with a genuine smile. “I don’t remember what it’s called.”

I almost said something that maybe I should be tutoring English instead, but was too shaken to attempt anything clever. Good thing too, because it would’ve come out bitter as hell.

So I try, and fail, and eventually give up exasperated, saying apologetically, “I’m sorry – I’m just drawing a blank.”

Eventually, the principal interviewer (an econ major) may have taken pity on me, told the physics major to start tutoring someone, and handed over the interview to the engineer, who proceeded to ask me some qualitative chem-- Questions that were very easy.

But I was still shaken badly. It took a while to recall how a buffer solution worked, but I was able to do that. I then explained how the boiling points of water, methane, and ethanol compared, and why (high, low, and medium, because the molecules exhibit hydrogen bonding, Van der Waals, and dipole-dipole interactions, respectively). The engineer, who was nothing but supportive, made a comment about how methane had a boiling point of about 4 Kelvin.

In my insecure, addled state, I immediately and tersely corrected him, stating that Titan had liquid methane on the surface, and had a surface temperature well above 4 K.

“Oh, well, I guess that’s a good application of astro knowledge to a chemistry problem!”

Poor guy. I hope he genuinely was as chipper and well-adjusted as he seemed, and not being polite for my benefit. In any case, he definitely knew how to positively reinforce an insecure student.

I shook hands, tried to smile, and concluded the interview.

I mentioned this over dinner with a PhD and a grad student. Both struggled as much as I did, though I suspect that they would’ve fared better than I did at the whiteboard.

As a matter of pride, I decided I needed to go home and derive a solution to the final question. It follows below.

But before I go into that, let me just say that I learned something about myself today.

I learned about the power and folly of my pride, about the importance of preparation, and about how important it is to be able to sit down and just learn from your mistakes. By happy chance, I read part of Craig Ferguson’s memoir, American on Purpose afterward, which has this precious quote:

He will know from and early age that failure is not disgrace. It's just a pitch that you missed, and you'd better get ready for the next one. The next one might be the shot heard round the world. My son and I are Americans, we prepare for glory by failing until we don't.

Here’s to failure, and persistence.


****Solution to problem 3****


Starting from the idea of a surfer, I placed the test particle at some position (x, y(x)) on the curve. I then created a force diagram in the rest frame (found that easier than the person’s frame).

The sum of all forces should equal the centripedal force,




Breaking it into components, along the y-direction,




But since the forces in the y-direction have to equal zero (otherwise the surfer would be moving up and down):



So setting the y-components equal to zero and rearranging terms, we get



Let’s call this Equation 1.

Let’s look in the x-direction. There’s only one force that has an x-component: the normal force



The x-components must equal the centripedal force,


where the centripedal force is



The centripedal force has to be precisely this to keep the water in a stable circular “orbit” of radius r around the axis of rotation (in this case, the y-axis).

Also, note that the radius of the orbit is simply equal to x (r = x), and that the unit direction vector indicates that it always points toward the y-axis.

Thus, in the x-direction, canceling negative signs, we get



Let’s call this Equation 2.

To be perfectly clear, note that the normal force and centripedal force have the same sign in the x-direction (otherwise we’d be in trouble, since it’s only one force that can generate the centripedal force). This is true for both x < 0 and x >0.

Now, I’m going to divide equation 1 by equation 2. In case you’re not familiar with dividing equations, I’m allowed to do this because I’m dividing by the same thing on both sides. For example, If x = y, x/5 = y/5. In this case, the divisor isn’t a number like 5, it’s a function. And I’m dividing the other side by something that looks different, but is equal to the divisor on the other side.




Or



Let’s think about the angle. Theta is the angle that a tangent to the curve makes with the x-axis. The tangent of theta, then is simply the height of the surfer divided by the distance from the y-axis, which is x. The tangent of theta, then is equal to the change in y divided by the change in x at that point:*





Replacing this for tan(theta), we get




I then separate y and x terms:



and integrate both sides




where C is an integration constant that is equal to the height of the water at x = 0.

Note that y is quadratic in x. This means that the function is a parabola.

* I wasn't comfortable with the previous argument about tan theta, so I cheated and looked up a solution. I'm glad I did - of course the tangent should be definied in terms of a derivative. Credit goes to this post.