Winter Storm Warning

A winter storm warning has been issued for the region including NYC, beginning tonight and extending to Thursday. It is not clear right now (as I write) how and how badly NYC will be affected. It’s a good idea to monitor the situation – check weather forecasts as the storm approaches tonight.

Also, if you have not already done so, it’s a good idea to sign up for CUNY Alerts. You can do that here.

 

Homework for Wednesday 7 March

For the notes from the class, see this post.

Homework:
• Make sure that your citytech email address is in the User information in WeBWorK. There are still a few students who have not done this. (It was supposed to be done at the start of the semester.)

• Review the use of the various derivative rules, particularly the Product Rule and the Quotient Rule

• Do the WeBWorK on Derivatives-ProductQuotient, and at least start the assignment on Higher Derivatives. Problem 8 in Derivatives-ProductQuotient is extra credit.

• Use the Quotient Rule to show that the derivative of $\cot(x)$ is $-\csc^{2}(x)$, and the derivative of $\sec(x)$ is $\tan(x)\sec(x)$ (remember that $\sec(x) = \frac{1}{\cos(x)$).

• Also do the following problems from the textbook:

 P. 94: 15, 17, 18, 19, 23, 25, 29, 31+32 (tangent line only), 34, 39

• There will be a Quiz on Wednesday: it will be on the Product Rule and/or Quotient Rule (and you also need to know the other rules then, of course.)

 

Don’t forget, if you get stuck on a problem, you can post a question on Piazza. Make sure to give your question a good subject line and tell us the problem itself – we need this information in order to answer your question. And please only put one problem per posted question!

Monday 5 March class

Topics:

• Review of the derivative rules (so far)

• The Product Rule and the Quotient Rule

• Notation for higher derivatives

My slide show (with the corrected derivative for cosine) is here:

MAT1475DerivativeRules

Here is a video which uses the “LO dee HI” formula for the Quotient Rule:

https://www.youtube.com/watch?v=jIX0VvwfEko

Examples I worked:

Derivative of $f(x) = x^{3}e^{x}$ using the Product Rule:

$\displaystyle \frac{\textrm{d}}{\textrm{d}x}\left(x^{3}e^{x}\right) = x^{3}\frac{\textrm{d}}{\textrm{d}x}\left(e^{x}\right) + e^{x}\frac{\textrm{d}}{\textrm{d}x}\left(x^{3}\right)$

$ = x^{3}e^{x} +e^{x}3x^{2} = e^{x}\left(x^{3} + 3x^{2}\right)$

Derivative of a product of polynomials (two ways) – first by using the Product Rule, and then by  multiplying it out and taking the derivative. See Example 51 in section 2.4 in the textbook. (This is not the one I did, but it’s similar.)

Derivative of a rational function using the Quotient Rule.  A similar example is in the video above. (Also see Example 58 in section 2.4 in the textbook, which shows how sometimes you may be able to avoid using the Quotient Rule.)

Derivative of $\tan{x}$ using the Quotient Rule. See Example 55 in section 2.4 in the textbook.

Derivative of $f(x) = x\ln(x)$ using the Product Rule

Derivative of $g(x) = x\ln(x) – x$ For both this and the previous, see Example 53 in section 2.4 in the textbook.

Then we found the first, second, and third derivatives of $f(t) = -16t^{2} + 150$, our position function for the object falling from a height of 150 feet.

 

For the homework assignment please see this separate post.

Homework for Monday 5 February UPDATED

• Make sure that you know and understand the following things (which are in my slideshow linked in this post):

The definition of the derivative at a point

What it means to say that a function is differentiable at a point or on an open interval

The definition of the derivative function

The various notations that are used for the derivative function

• Do the WeBWorK assignment “DerivativeDefinition”. You can wait to do the assignment “DerivativeFunction” until after Wednesday’s class (after Test 1 we will have class). Please note that you should be using the definition of derivative, and not any shortcuts you may know, in order to compute the derivatives of problems 2 and 3 in “DerivativeDefinition”.

• Also do the following problems from the textbook:

p. 56 # 19-27 odd (see this post for some notes)

p. 69 #1, 3, 5, 10, 12, 13-19 odd (part (a) only)

p. 85 #5, 11-19 all, 21, 33-38 all (tangent line only)

There will be a Quiz on Monday. The topic will be using the definition of derivative to compute a derivative.

Don’t forget, if you get stuck on a problem, you can post a question on Piazza. Make sure to give your question a good subject line and tell us the problem itself – we need this information in order to answer your question. And please only put one problem per posted question!

Test 1 Review (UPDATED)

Test 1 is scheduled for the first 50 minutes of class on Wednesday 28 February. Please see the course policies

The review problems distributed in class are also here:

The answers are here

 

Here are some resources that may help with these problems: warning, many videos are on YouTube, so you may have to watch ads and another video may autoplay after this one is finished.

For problem 1, see this video from PatrickJMT  on finding a limit from a graph.

For problem 3a, see this video from PatrickJMT on finding a limit by factoring and canceling.

You may also want to take the time to watch lots of limit examples  part 1

For problem 4, you should also answer this question for the graph in problem 1 (as suggested in the answer sheet). Here is a video from PatrickJMT about continuity and the types of discontinuities.

Don’t forget, if you get stuck on a problem, you can post a question on Piazza. Make sure to give your question a good subject line and tell us the problem itself – we need this information in order to answer your question. And please only put one problem per posted question!

Monday 26 February class

Topics:

• Definition of the derivative of a function at a point

• Differentiability

• Definition of the derivative function

Slideshow about definitions of derivatives:

MAT1475DefinitionOfDerivative-slideshow

Here are notes on two of the examples I worked in class: finding the derivative function of $f(x) = \frac{1}{x+1}$, and finding the derivative function of $f(x) = |x|$. Pay special attention in the second example to the way we need to use one-sided limits to see if the derivative exists at $x=0$. (It does not.)

MAT1475DerivativeDefinitionExamples

 

From now on I will post the homework in a separate post from the class notes. The homework assignment from today’s class is here.

Wednesday 21 February class

Topics:

•More on limits which involve infinity:

How to tell if a limit is $\infty$ or $-\infty$ without using the graph

Here are my handwritten (sloppy) notes from class:

MAT1475limitsInfinity page 1

Limits at infinity for polynomial and rational functions

• Derivatives: introduction to the idea

There are two ways of looking at what we are trying to achieve with the derivative:

One way : geometrically, thinking about the graph of a function f(x), we are trying to find something like a slope for functions that are not linear. Why would we want to do this? Because the slope of a line represents the rate at which y (or f(x)) is changing as x increases. If you think of x as representing time, that may be helpful. If the slope is 1.5, that means that if x increases by 1, the value of the function will increase by 1.5.

A line has the same slope (rate of change) at every point, but we can’t expect that to be true for a function whose graph is not a line. But hopefully we can find something like a slope at a single point on the graph. It makes sense that this should be the slope of the tangent line at that point, if you think about it. (That’s if there is a tangent line!)

Another way: Thinking about the function itself, the rate of change of the function may represent something we are very interested in. For example, if we have a position function as in the example, the rate of change of the function is the velocity. (Also, for a velocity function, the rate of change will give the acceleration!) We would like to be able to define velocity for position functions which are not linear, meaning that the velocity is not the same at every point.

Example: I basically used the same example that is used at the start of Chapter 2. The position function represents the height of an object which is dropped from a height of 150 feet, t seconds after it is dropped. The function is

$f(t) = -16t^{2} + 150$

For any function (not just this one), we define the average rate of change (here it’s the average velocity) on an interval $[a,b]$ to be the slope of the secant line that passes through the two points $\left(a, f(a)\right)$ and $\left(b, f(b)\right)$ as

$\frac{f(b) – f(a)}{b-a}$

In class we computed the average rate of change for our position function on the intervals [2,  3] and [2, 2.5] . On your own you can compute the average rate of change on the intervals [2, 2.1] and [2, 2.01] and more if you like… or you can look at the textbook!

Notice that the right-hand endpoints of these intervals are getting closer and closer to 2. The idea is that we want to look at what happens to these slopes (average velocities) as the second point approaches (2, 86). This will mean taking a limit.

Looking at the results from those average velocities, it appears that they do approach a limit and the limit is -64 (or very close to that).

 

Now we will rewrite the formula for the average velocity a bit. Think of taking the interval [2, 2+h] where h is some (small) number, possibly negative. So we would think of our intervals as [2, 2+1], [2, 2+0.5], [2, 2+0.1], etc. Then the slope of the secant line (the average velocity) would be written as

$\frac{f(2+h) – f(2)}{(2+h)-2} = \frac{f(2+h) – f(2)}{h}$

And we will define the instantaneous velocity at t=2 to be the limit of this as h goes to 0:

$\displaystyle \lim_{h\rightarrow 0} \frac{f(2+h) – f(2)}{h}$

Let’s compute that limit. We already know that $f(2) = 86$. To find $f(2+h)$:

\begin{align*}

f(2+h) = -16(2+h)^{2} + 150 & = -16(4 + 4h + h^{2}) + 150\\

& = -64 -64h -16h^{2} + 150\\

& = 86 -64h – 16h^{2}

\end{align*}

So $f(2+h) – f(2) = 86 -64h – 16h^{2} -86 = -64h – 16h^{2}$

\begin{align*}

\displaystyle \lim_{h\rightarrow 0} \frac{f(2+h) – f(2)}{h} & = \lim_{h\rightarrow 0} \frac{-64h – 16h^{2}}{h}\\

& =\lim_{h\rightarrow 0} \frac{h(-64 – 16h)}{h}\\

& = \lim_{h\rightarrow 0}(-64 – 16h)\\

& = -64 – 16(0) = -64

\end{align*}

So the slope of the tangent line is -64, as we predicted.

Homework for Wednesday 21 February

• Review the definitions of continuous function and the three types of discontinuity (notes soon)

• Do the WeBWorK “Limits-Continuity2”

• Also do the following problems from the textbook:

p. 44 #1-21 odd

p. 55 #9, 10

 

Don’t forget, if you get stuck on a problem, you can post a question on Piazza. Make sure to give your question a good subject line and tell us the problem itself – we need this information in order to answer your question. And please only put one problem per posted question!

Homework for Tuesday 20 February

Please note, there are no classes on Monday the 19th because of Presidents Day. On Tuesday the 20th, we follow a Monday schedule, so we will meet.

 

I hope to post notes soon from last Wednesday’s class, but in the meanwhile you should make sure that you have done the problems from the textbook which were assigned the previous class .

 

Also do these problems from the textbook:

p. 28 #25-31 odd

p. 35 #1-21 odd

And review the definition of a continuous function which is given on p. 37

 

There will not be a Quiz on Tuesday and there is no new WeBWorK yet.

 

Don’t forget, if you get stuck on a problem, you can post a question on Piazza. Make sure to give your question a good subject line and tell us the problem itself – we need this information in order to answer your question. And please only put one problem per posted question!

 

Math club meeting Thursday (tomorrow)

This looks interesting! Open to all.

 

Title: “Let’s Play Sudoku”

Speaker: Brad Isaacson (NYCCT)

Date/Room: Thursday Feb. 8, 2018, 12:50-2:00pm, Namm N720

Abstract:

Sudoku is a very popular number puzzle.  In this talk, we will discuss some of the strategies for playing Sudoku, including naked and hidden pairs/triples, unique rectangles, and alternating inference chains.  May there never again be a Sudoku puzzle that you are unable to solve.

Pizza and refreshments will be served at 12:45pm.