Class 5 - Review (Thursday, 9/14/2017)

As always, before we talk about mathematics, here is important logistics:

1st Gateway (Computer based Pass/Fail test)

Where? B069 EH
When? Sept 11 ~ Sept 25 (Hours are the same as Math Lab hours.) 

You have two trials a day, and if you don't pass during the above period, then you will loose 1/3 of your letter grade (i.e., B- to C). The contents will be about derivatives, which is prerequisite for this course.

Early pass event. If you pass by 9/15, then I will give you 10% extra credit on your next quiz.

Web Homework

Here are your web homework dues:
  • 6.4 : 9/18 Mon

Review of Quiz 1

We will start by reviewing Quiz 1.

Team Homework

Hints for Team Homework 1

Caution. These hints may contain errors. They are unofficial, so use them with caution. I only wrote them in order for you to get started with problems. For example, I will not promise any score just because you wrote down something identical to the hints.

1 (a). Review what a "right-hand Riemann sum" means.

1 (b). We have $f(x) = x^{4} - 8x^{2} = x^{2}(x^{2} - 8) = x^{2}(x - \sqrt{8})(x + \sqrt{8})$. Thus, we see the graph $y = f(x)$ means $x$-axis when $x = -\sqrt{8}, 0, \sqrt{8}$. At $x = 0$, something special happens. What is it? When $x \rightarrow \infty$, where does $f(x)$ go? When $x \rightarrow -\infty$, where does $f(x)$ go? To see where $f(x)$ increases or decreases, consider the sign of $f'(x) = 4x^{3} - 16x = 4x^{2}(x + 2)(x - 2)$.

1 (c). Write $g(x) = e^{f(x)}$. Then $g'(x) = e^{f(x)}f'(x)$ by chain rule. Try to understand why the following is true: $x$ values that make $g'(x) = 0$ are precisely those that make $f'(x) = 0$. (Hint: $e^{\text{anything}} \neq 0$.) Conclude that the critical points of $g$ are the same as the critical points of $f$.

1 (d). Recall that saying "$F$ is an anti-derivative of $f$" is the same as saying $F' = f$. In this case, you can even explicitly compute what $F(x)$ is for all $x$.

2 (a), (b). You have a full ability to do this on your own.

2 (c). This is merely because the sigma notation introduced here varies over integers. For example, it is difficult to write down the sum of prime numbers (e.g., $2, 3, 5, 7, 11, 13, \dots$) between $1$ and $100.$ 

Optional Reading. However, you can actually use the sigma notation varying over any set of objects that you can add. This may be outside the scope of the course, although I am also not sure whether people teach this at all. For example, the difficulty you had in 2 (c) could be resolved by writing
\[\sum_{\substack{p \text{ prime} \\ 1 < p < 100}}p.\]

3 (a). You need to see whether $f(t)$ is differentiable at $t = 1, 2, 2.5$. Recall that we say that $f(t)$ is differentiable at $t = a$, if $f'(a) = \lim_{t \rightarrow a}f(t)$ exists. Checking this can be broken down in the following three steps:
  1. Check $\lim_{t \rightarrow a-}\frac{f(t) - f(a)}{t - a}$ exists (i.e., slope from the left exists).
  2. Check $\lim_{t \rightarrow a+}\frac{f(t) - f(a)}{t - a}$ exists (i.e., slope from the left exists).
  3. Check whether the answers in the above two steps agree.
If all the three steps above go through, then $f(t)$ is differentiable at $t = a$ and the coinciding answers you get from Step 1 and Step 2 is called the derivative of $f(t)$ at $t = a$, which is also written as $f'(a)$ or $\frac{d}{dt}|_{t = a}f(t)$.

If any of the above three steps fail, then $f(t)$ is not differentiable at $t = a$.

Recall from elementary physics that:
  • velocity is the rate of change of position;
  • acceleration is the rate of change of velocity.
In our terms, we can say
  • velocity is the derivative of position (in time);
  • acceleration is the derivative of velocity (in time).
I won't give hints for answering the last sentence.

3 (b). The difference between the velocities of Car 1 and Car 2 is $f(t) - g(t)$. Read the question again, if this does not make much sense.

3 (c). This is an important question. The answers for two questions are different. The position is given by the integral. The distance traveled is given by the sum of areas. Why are they different? Recall that we defined (definite) integrals as areas, but areas "with sign".

3 (d). The definition of average value of $g(t)$ for $0 \leq t \leq 3$ is
\[\frac{1}{3}\int_{0}^{3}g(t)dt.\]

Review the word: Average Rate of Change.

3 (e) - i. Let $F(t)$ be the position of Car 1 at time $t$ and similarly define $G(t)$ for Car 2. Then $F'(t) = f(t)$ and $G'(t) = g(t)$. Your task is to maximize $|F(t) - G(t)|$. If you are keen enough, from the graph you may realize that $F(t) > G(t)$ from $t = 0$ to $t = 2$. Hence for $0 \leq t \leq 2$, we can just erase absolute value and maximize $F(t) - G(t)$. It is more subtle to check whether we can erase the absolute value for $0 \leq t \leq 3$. Please write out in detail whether you can do that, and if so, also write how you concluded so.

3 (e) - ii. The distance between Car 1 and Car 2 at time $t$ is $|F(t) - G(t)|$, where $F, G$ are given as in the previous part.

3 (e) - iii. No hints given for this part.


4 (a). Try to actually compute what $P(x)$ should be. Abstractly, we can write $P(x) = -2 + \int_{0}^{x}p(t)dt.$ Can you see why? If so, compute the integral by dividing into the following four cases:
  1. $-3 < x < -2$;
  2. $-2 \leq x < 0$;
  3. $0 < x < 2$;
  4. $2 \leq x < 3$.
Although it seems like the above cases do not cover $x = 0$, we already know $P(0) = -2$.

4 (b). Note that $P'(x) = p(x)$, so the given graph should give you "slope" of your graph at every point.

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