# Math 90: Week 4

It was quiz-day!

The class did pretty well on the quiz. I wrote the quiz, and I’m pleased with the skill-level demonstrated. The average was about a 77%, and the median was an 80%. (For stat-witty folk, this means that the lower scores were somehow ‘lower’ than the average scores).

Anyhow, the solutions to the day’s problems and the quiz are below the fold:

We had three questions in recitation.

1. Let’s prove the power rule!
1. Show that $z^n – x^n=$ $(z-n)(z^{n-1} + z^{n-2}x + \ldots + z^2x^{n-3} + zx^{n-2} + x^{n-1})$
2. Compute $\lim_{z \to x} \dfrac{z^n – x^n}{z – x}$
3. If $f(x) = x^n$, what is $f'(x)$?
2. We had a classic problem that’s based on the following question: A man starts to climb a path on a mountain at noon and ends at 8pm. He sleeps overnight on the mountain. At noon the next day, he climbs down the same path, reaching the bottom at 8pm. Show that there is a time of day where the man is at the same spot on the mountain.
3. A continuity question that was taken from the homework.

Let’s look at the solutions to the first two –

#### Question 1

The first part of the first problem gave a lot of heartache, so in recitation I said to assume it and move on. But let’s look at the solution anyway. We want to show that $(z^n – x^n) = (z-x)(z^{n-1} + z^{n-2}x + \ldots + zx^{n-2} + x^{n-1})$. From what I could tell, the primary source of confusion came from the … bit. To understand that, let’s consider a few quick examples.

• $1, 2, 3, \ldots, 7, 8, 9$ is the same as $1, 2, 3, 4, 5, 6, 7, 8, 9$
• $1, 2, 3, \ldots, n-2, n-1, n$ is the same idea: all numbers from $1$ to $n$ in order
• $x^7 + x^6 + \ldots + x^2 + x^1$ is the same as $x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x^1$
• $x^{n-1} + x^{n-2} + \ldots + x^3 + x^2 + x^1$ is the same idea: the sum of the powers of $x$ from the $n-1$st post to the $1$st power

The idea here is to distribute out on the right. So $(z-x)(z^{n-1} + z^{n-2}x + \ldots + zx^{n-2} + x^{n-1}) =$ $(z^n + z^{n-1}x + \ldots + z^2x^{n-2} + zx^{n-1})$ $– (z^{n-1}x + z^{n-2}x^2 + \ldots + zx^{n-1} + x^n)$. For every positive term of the first summand except $z^n$, there is a corresponding negative term in the second, except that $x^n$ is left out. So we are left with $z^n – x^n$.

We can use this to compute the limit. Since $(z^n – x^n) = (z-x)(z^{n-1} + z^{n-2}x + \ldots + zx^{n-2} + x^{n-1})$, we can say that $\dfrac{(z^n – x^n)}{(z-x)} = \dfrac{(z-x)}{z-x}(z^{n-1} + z^{n-2}x + \ldots + zx^{n-2} + x^{n-1})$. Then in the limit, we can just let $z \to x$, and we have $n$ copies of $x^{n-1}$. So the limit is $n x^{n-1}$.

Finally, if we recall that $\lim{h \to 0} \dfrac{f(x + h) – f(x)}{h}$ and $\lim_{z \to x} \dfrac{f(z) – f(x)}{z-x}$ are both definitions (equivalent definitions) of the derivative at the point $x$, then we see that the calculation we’ve just done is, in fact, the derivative.

This is known as the Power Rule and is a nice simplification.

#### Question 2

This question is an Intermediate Value Theorem question. If we let $f(t)$ be the position on the trail on the first day, so that $f(\text{noon}) = 0$, and if we let $g(t)$ be the position on the second day, so that $g(\text{noon}) = \text{top of mountain}$, then we can consider the (not at all immediately obvious) function $h(t) = g(t) – f(t)$, then this is a difference of two continuous functions. So it is continuous. At noon, it’s positive. At 8pm, it’s negative. By the intermediate value theorem, since $0$ is between negative and positive numbers, we know there is a $t_0$ such that $h(t_0) = 0$. This is precisely the statement that the position is the same. So regardless of how the man climbs or descends the mountain, there will always be a time of day when he is at the same place.

This is another example of the idea that the great difficulty of this sort of math isn’t the calculation and arithmetic aspects, but having the knowledge to see when it can be applied.

#### Question 3

There were a few homework questions that were a lot like this question, and this was perhaps the most boring of the questions. If there is anything unclear, let me know and I’ll expand this problem.

So now, perhaps the part that you were waiting for – let’s go over the quiz!

1. Find with justification the following limit: $\lim_{x \to \infty} \dfrac{2\sqrt{x} – x^{-1}}{3x – 7}$
2. At $t$ seconds after liftoff, the height of a rocket is $2t^2$ feet. By explicitly using the definition of a derivative, determine how fast the rocket is climbing $5$ seconds after liftoff. In addition, no credit will be given for using only the power rule.

#### Solutions to the Quiz

There were many ways to approach the first problem. A basic result on horizontal asymptotes of rational functions states that when dealing with ratios of polynomials, if the degree on the bottom is bigger than the degree on the top, then there is a horizontal asymptote at $y = 0$. This means that both the limit as $x \to \infty$ and as $x \to – \infty$ is $0$.

The most common solution was to divide the top and the bottom by $x$. This would give $\dfrac{\frac{2 \sqrt x}{x}- x^{-2}}{3 – \frac{7}{x}}$. Now, in looking at the limit as $x \to \infty$, we get something that looks like $\dfrac{0 – 0}{3 – 0}$, which we can safely evaluate as $0$.

These were the two acceptable solution routes. Let’s talk about some of the errors.

One error came up that worried me, because it’s not a skill that we’re going to spend any classtime on. $\dfrac{2 \sqrt x – x^{-1}}{3x – 7} \neq \dfrac{2 \sqrt x}{3x – 7 – x}$. That is not how that works. The thing that people must have been thinking of is that $x^{-1} = \frac{1}{x}$, so that if you had something like $\dfrac{y^2 x^{-2}}{z^2}$, then you could rewrite it like $\dfrac{y^2}{x^2z^2}$. But all this means is that you could rewrite $\dfrac{2 \sqrt x – x^{-1}}{3x – 7}$ as $\dfrac{2 \sqrt x – \frac{1}{x}}{3x – 7}$. And that is that.

Other than that, there were a few algebra errors here and there, and that happens sometimes. For grading, my standard rubric was: a point for the right answer, up to 2 points for valid justification, and up to 2 points for correct algebraic manipulations towards the solution.

The second problem had a lot less variability:

This is a derivative problem. The derivative of $f(x)$ at the point $c$ is $\lim_{h \to 0} \dfrac{f(x+h) – f(x)}{h}$. We can plug in $t = 5$ either before or after. I will plug it in after here. $\lim_{h \to 0} \dfrac{2(t+h)^2 – 2t^2}{h} = \lim_{h \to 0} \dfrac{2t^2 + 4th + 2h^2 – 2t^2}{h} =$ $\lim_{h \to 0} \dfrac{(4t + 2th)h}{h} = 4t$.

So at $t = 5$, we have that the speed of the rocket is $4(5) = 20$ feet per second.

Common errors were not remembering that the limit was as $h \to 0$ or taking $h \to 5$. This surprised me, as we evaluated 2 derivatives in recitation prior to the quiz, and we talked explicitly about the definition of the derivative. The only systematic error that worried me was thinking that $f(t + h) = 2t^2 + h$ This indicates a certain lack of fluency with the idea of a function – if this describes you, I highly recommend coming and talking to either Tom or I during our office hours. This is a very important concept to have nailed down.

My standard grading rubric was as follows: 1 point for the right answer, 1 point for knowing and explicitly using the definition of a derivative (including the proper limit), 1 point for properly plugging in the right things into the function (i.e. $f(t + h) = 2(t + h)^2$ and so on), 1 point for doing the necessary algebraic manipulations to evaluate the limit, and 1 point for evaluating the limit.

I’d like to remind you that it’s much easier for me to grade your work if you write your work cleanly and by clearly indicating your logic.

I’ll see you in recitation next week. Little bonus – Tom may be observing recitation for some or all of it too. Yay!

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