Scribe Post 11/20

For the first half of class we took the first Unit 3 quiz, some of us crossing our fingers that the one question we didn't do on the IW was not of the quiz. Just remember, you can redo the IWs, but you cannot redo a quiz, so try to keep up with the IWs, even with the behemoth of stress from applying to college. If your having trouble with stress, take this gentlemen's word of advice:

      "The greatest weapon against stress is our ability to choose one thought over another."

       - William James

So, if your stressed out about all of the things going on in senior year, take a breath, sit back, and just think about one thing–calculus! Don't let your mind wander to the pending applications to your first choice school, or that environmental science forestry report, just sit there and focus on math, and the stress will just melt away like hot butter on pancakes...It's really that easy!

We then went over the Unit 2 Supercorrection Test; remember if you scored lower than a 70, you can retake it for a score up to 70 points. We went over every question during class, so if you need extra help, or missed class, Mr. O'Brien will be more than happy to help you out. If you are in that situation, keep in mind that since the infamous nordic team is beginning, Mr. O'Brien will not be around after school to assist you when there is practice. Plan to get help during a study hall, or ask another teacher.

Then the new stuff begins. Today was a discussion of all of the functions we've been looking at, that are increasing. Increasing is determined by the derivative, which is always positive when a function is increasing. The concept we explored today was the difference between functions that are increasing more rapidly as the value of x increases, or in other words, the derivative gets larger and larger. Also functions that are increasing more slowly as the value of x is greater, or in other words, the derivative gets smaller as x increases, but never becomes negative. Below is a homemade diagram that I think will help you visualize some of these concepts. 

You'll notice two things: one, each of the graphs is increasing (the derivative is greater than zero); two, there are two new vocabulary words. Concavity is visual for me, but incase you're more of a definition kind of guy, then here are the straight facts. Concave up is when the derivative is increasing, and the second derivative is positive. Concave down is where the derivative is decreasing, and the second derivative is negative. 

If you are wondering what happens if you have a function where the derivative is increasing, but then switches to decreasing, then you are on the right track because that's called the Inflection Point. The inflection point is simply the point or coordinate where function changes concavity, or in other words when the derivative switches from increasing to decreasing, or vice versa. Here is a homemade diagram to enhance your understanding.

You can also think about this inflection point, as where the second derivative changes sign, from negative to positive, or positive to negative. Also, when the second derivative is undefined, or possibly zero. Remember, just because the second derivative is zero, does not guarantee there is an inflection point.  

Lastly, we reviewed the first and second derivative tests. Instead of just showing you a diagram, let's watch a video of a wonderful lady who provides examples to apply these two tests. The video is below.

A quick tip though, whether a critical point is a min or a max can be determined by, the max is when the second derivative is negative, and the min is where the second derivative is positive. It's not what you would like to think, that the max is positive, and the min is negative, but just know that.

Remember to do the homework, it's IW #4: p. 219/7, 13, 21, 23, 39, 45-55 odd, 56-60 all. Good luck, and hope you enjoyed your holiday!

UPDATE: 
So. It's been a long quarter–we've learned a lot. However you can remember to look back on your Unit 3 Finely Crafted Opprotunity Day, whip out those supercorrections, and take a look at problems 20, 17, 7, 10, and 11, and that should help you a lot to prepare for the Midterm exam, or even the AP exam. The relationships between the first derivative and second derivative are not so scary after taking a fourth look (IWs, test day, supercorrections, and make-up test) at those O'B test questions.

Scribe Post 11/16

We began class with the excitement of the follow-up test looming over us. Since Mr. O'B was leaving early, we would save that for the 2nd half of the class. 

We began by looking at this function, and trying to think about what the absolute value signs do to this derivative. To try and understand this equation better, and see how the absolute value sign would affect it's derivative, we split it to make it a piecewise function:

Now, Using our derivative knowledge that (lnx)' = 1/x, we can simplify the equation:

Therefore, 

The absolute value signs don't affect the derivative of ln, however adding absolute value signs to the equation allows us to use negative values for x.

Next, we looked into finding anti-derivatives. We looked an example which read: 

ex/ Find the function f(x) whose derivative is f '(x)= x^2, and which passes through (2,4).

After some thought, we came to realize that the anti-derivative of f'(x) was



 Anti-derivatives must have a constant attached to them because the derivative of any constant is just 0. Unless, that is, we have a condition for our equation. In this case, the condition is that the equation must pass through (2,4). To find our constant, we can plug in our given values:

Since C is 4/3, 

Using this new idea, we looked at a couple examples of some anti derivatives:

Lastly, before taking the Follow-up test we learned of a new Theorem:

Mean Value Theorem

If f(x) is a continuous function on [a,b] and a differentiable function on (a,b), then there is at least 1 value of c in [a,b] such that: 

The example that Mr. O'B used to put the MVT into easier terms is the following:

If you travel 50 miles to Augusta and it takes exactly 1 hour, at some point in the trip the car  must have been going exactly 50 mph. Let's think about this graphically:

What the Mean Value Theorem tells us is that as long as f(x) is continuous on the interval [a,b] and differentiable on the interval (a,b), then a tangent line to the point c will have a secant line with the exact same slope at the point c. 

For another explanation on the MVT, and an applet to help understand it's uses: Click Here!

Don't forget to do IW #3, study for the quiz, and have a nice Turkey Day.

Scribe Post 11/14: Extrema

Hey guys!
**To study, try to do the previous problems on the blank test, which are on the iCal
Remember, problems that were multiple choice before probably won’t be in multiple choice form on the test! Some of these will be the type of problem where you just replace values, but others will not, so make sure you actually know how to do everything!

Now, on to the real mathy stuff.

We started by talking about absolute (global) extreme. The definition can be found on page 191 in our textbooks, but here it is anyway!

• Let f be a function with domain D. Then, f(c) is the

• a) Absolute maximum value on D if and only if f(x) is less than or equal to f(c) for all values of x in D
• b) Absolute minimum value on D if and only if f(x) is greater than or equal to f(c) for all values of x in D
• These are the largest or smallest y values!

Example:
If we look at the function f(x)=x2 on various domains, we realize that there are different maximums and minimums depending on the restrictions. When the domain is all reals, there is no max because the function is continuously increasing, but there is a minimum of zero

• On domain [0,2], there is a max of 4 and a minimum of zero, because it is a closed interval
• On domain (0,2], there is no min because it is an open interval on that side and the max is 4, because that side is still closed
• On domain (0,2) there is no min or max, because both sides are open

 This leads us to our next theorem, the Extreme Value Theorem, which states that:

• If f is continuous on a closed interval [a,b], then there is both an absolute (global) max and an absolute (global) min

Local Extreme Values: The definition can be found on page 193 in our textbooks. It is as follows:
Let c be a point of the domain of the function f. Then, f(c) is either

• a) a local maximum value: at c if and only f f(x) is less than or equal to f(c) for all values x in some open interval containing c
• b) a local minimum value of c, if and only if f(x) is greater than or equal to f(c) for all values x in some open interval containing c

• In other words; a function f has a local max or min at an endpoint c if the appropriate inequality holds for all x in some half open domain interval containing c

We also took some time to review an old term: critical points. Our new definition states that they are values of x where f’ is zero or undefined. More specifically, the points where f’ is zero are called stationary points, because it is where the motion briefly stops.  The local extreme are always either found at endpoints or critical points.

•  One of the absolute max or mins has to be one of the local max or mins!!
• The max/min will be either an endpoint or at a point of change, such as a cusp or corner, where the derivative is undefined; or when the derivative is zero, which can be caused by a gradual change.

**Just remember to be careful, because this can actually be a little tricky. For example, f(x)=x³ has a derivative of zero when evaluated at zero, however, it is not a max or a min. It’s a lot like the argument of a square being a rectangle, but a rectangle isn’t a square kind of thing.




After learning all of the definitions, we put them to practice. We were told to find the extreme values of f(x)=x^2/3 on [-2,3] algebraically and confirm graphically.

• First, find the derivative (you can use the power rule in this case!) 
• Then, decide on the critical points. Include the end points and the value that makes the denominator undefined (-2,0,3). In this case, the derivative can never equal zero! Remember, these are x values!
• It can be helpful to make a chart of these values. Use it to plug the x values back into the original equation. From there, just take the smallest value produced for the minimum and the largest value for the maximum! Tada!
  
  
  What if we were asked to find the extreme values of a more difficult function? Well, we can do it!
  
  
• We find the derivative by using the chain rule, shown above
• Then, identify the critical points, which are x=0,2,-2
• Plug into original function and pick out largest and smallest values
• We then can see that the function has two undefined values. Therefore, there is no max, but there is a minimum of 1/2.
  
  
  
  
  
  
  

Weston 10/26

cos^-1 '= 1/√1-x^2

We started off class today going over the quiz #3. After going over next weeks schedule and things to remember for the TEST on Monday we immediately went into a complicated lesson on derivatives of inverse trig functions.

On question 3 of the quiz it should be remembered that   is similar to just with different variables and they are both asking you to find the derivative. Using the quotient rule ( (u*v'-u'*v)/ u^2 ) we can find the derivative
when which is or and after simplifying equals






Question 5 was just a matter of understanding what velocity, speed and acceleration. Velocity is the rate of change of a function or the derivative which in terms of seconds and meters is m/sec. Speed at a certain time is the absolute value of the velocity of the certain value or abs(m/sec). Acceleration is the second derivative or the rate of change of velocity. Therefore when finding the velocity each time the acceleration is zero means finding the zero of (second derivative) which is .  Then plug it into the velocity equation to get -12 m/sec. The speed at t=2 is the absolute value of velocity which equals 12 m/sec.

Question 7 When using the h approaches 0 form of the derivative remember to use a FUFOO to simplify and use the derivatives rules to check your answer.

The Bonus question uses the idea that we can take the constant multiple out of the parenthesis of an equation to find the derivative more easily.

  taking out the constant multiple and taking the derivative of inside the parenthesis using the product rule to get which simplifies to

  

*Things to memorize before the test on Monday, November 5th:
-Trig Values
-Trig Derivatives
-Product Rule
-Quotient Rule
-Chain Rule
-h approaches zero form
-h approaches c form 

At the end of the unit we should be able to find any derivative!


Inverse Trig Functions

Mr. O'Brien says that the best way to think of inverse functions is through the function machine (what make comes into the function and what comes out). In a regular sinx function the input is a number and the output is a ration. ex- when x=1/2, sinx = √2/2. Yet in a inverse function like has an input of a number and an output of an angle. ex- when x=1/2, = π/6.

You can also compute this: , using Math - Frac on your calculator you will find the answer is . Using radian mode will make your life a lot easier. Also not that negatives reflect to the other side of the function ex-


Graphing Inverse Trig Functions:

Sinx- The inverse of sinx doesn't pass the vertical line test and therefore is not a function so in order to graph the function there needs to be restrictions. Domain:[-1,1] and Range:[-π/2.π/2]

Cosx- The inverse of cosx doesn't pass the vertical line test either therefore it's not a function unless there are restrictions. To make the calculations easier the range starts at zero. Domain:[-1,1] and Range:[o,π]

 

Tanx- The inverse of tanx has asymptotes at y=-π/2 and π/2 rather x=-π/2 and π/2. Yet for it to be a function it also needs restrictions. Domain:[all real #'s] and Range:[-π/2 and π/2]



To prove the derivative of the inverse trig functions we use inverse properties to cancel out the inverse and then we use implicit differentiation (meaning we take the derivative of each side) then using the identities below we simplify and use division to equal out both sides and find the derivative. Since the derivatives of the inverse trig functions are angles we used a triangle to show what angle y equals.

CHea!

Homework for next class is to work on the IW #10 and #11 packet. 

Here is a list of all the derivatives of inverse trig values

*Update- in certain functions it is easier to find the derivative after simplifying or reworking the original function. ex-

in which the derivative is

that is a lot easier than applying the product rule.