Cheesecake on the "x" Axis

Now that we have covered the two major concepts of Calculus (derivatives and integrals), we can move on to combining these two to solve problems.

Think of these two things as your cheesecake ingredients. Like cheesecake is made up of cream cheese and eggs, Calculus is built upon these two concepts.

One of the more applicable situations that these can be used in is the relationships between position, velocity, and acceleration:

 A nice graphical representation

 The order goes like this derivative wise:

The derivative of the position gives you velocity, and the derivative of the velocity gives you acceleration.

And the opposite is true of integrals:

The integral of acceleration gives velocity, and the integral of velocity gives position.

These don't have to exclusively apply to position - velocity - acceleration either. This concept can be universally applied to any function of "something".

If you take the derivative of "something", then you'll get the rate of "something", and if you take the derivative of that then you'll get the change in rate of the rate of "something". 

If this is confusing, applying some Cheesecake to this will clear this up:

At the Cheesecake Factory, they make a lot of cheesecake. As a matter of fact, they make such a great amount of cheesecake that the actual amount of cheesecake cannot be counted. Luckily, the rate at which they make cheesecake per hour can be measured:

Where r(t) is the rate of cheesecake created and t is time in hours. Everyday has 24 hours, which is why time is bounded from 0 hours to 24 hours.

(a) Calculate the amount of cheesecake produced per day.

(b) The rate of cheesecake production starts at one point and then begins falling from there, but at some point in the day cheesecake production begins increasing and increases until the end of the day. Find when in the day this happens.

(c) At what time is a total of 234.85 cheesecakes made? Yes, the reason why there's a fraction of a cheesecake is so that your answer is nice and even.

Solutions:

Will be posted later due to my laziness.

If you crazy people out there did not think this problem is up to AP problem standards, don't worry. The next problem will be BASED upon a REAL AP problem, but, there will definitely be cheesecake.

And the Results are In!

In our first poll, a grand total of twelve people participated.

In the beginning there were five: Cheesecake, Tea, Soup, John Adams, and Strom Thurmond.

The delectable desert was a favorite to win (for obvious reasons), but ironically was barely able to scrape up one vote.

A complimentary beverage and a certain racist politician fought bravely, managing to at one point tie with Soup and John Adams, but eventually lost steam. Tea reportedly returned to its home country of England, while Thurmond reportedly had a heart attack and died at the ripe age of 101.

Then there were two...

Our ever present Minister Plenipotentiary and viscous meal fought it out tooth and nail; spoon and bowl. 


But, in the end there could only be one...

Oh Adams... if you only stopped from passing the Alien and Sedition Act...

Integrals Featuring John Adams

Cheesecake and John Adams... How does one combine these two great tastes?

A historically correct hypothetical situation in which both objects are located in the royal courts of France is the way to go.

 The man... the legend... 

John Adams, as we all know, went to France as a diplomat during the Revolutionary War. As he was there, he experienced French court life which he disliked due to the slowness of their political system. 

One day, as Adams attended a court dinner in which a 18th century equivalent of cheesecake was served. Enthralled with the exquisite food, Adams began eating it at a rate expressed by:

Where c(t) expresses the rate at which cheesecake is being consumed in bites and 't' is in time in minutes. The equation is also bounded by:

Also, at t = 0 minutes, Adams had not taken any bites of cheesecake mostly because it is hard to eat cheesecake before its served.

(a) Write the the indefinite integral for the rate at which cheesecake is being consumed. Explain what this equation represents. (Hint: make sure you find a way to solve for C using the given information)

(b) Find out the rate at which John Adams is eating cheesecake at t = 8 minutes. Indicate units.

(c) Find the amount of cheesecake that John Adams has consumed between t = 3 and t = 14 minutes. Indicate units.

(d) Assuming that Adams finishes eating his cheesecake at t = 20 minutes and each bite that he took was 5 grams worth of cheesecake, determine the mass of the cheesecake. 

Solutions:

The rate that is given for the consumption of cheesecake is that way because people generally eat at a fast rate and then slow down as they get closer to being full. 

Now, Adams likes to savor his food, which is why he took 20 minutes to eat his cheesecake. 

(a) 

By using the "reverse power rule" you get:

Now, after you take the integral of the rate of something you get the amount of something. Since you just took the integral of the rate of cheesecake consumption in "bites", then the integral represents how much cheesecake has been eaten. In order to solve for "C", you must refer to the information given on how much cheesecake has been consumed. In this case, Adams has eaten 0 bites of cheesecake at t = 0 minutes.

Therefore, you can set the integral equal to 0 and set t = 0. From this, you can see that C = 0.

(b) 6 bites / minute

This is asking for the rate of consumption of cheesecake by Adams. That equation is already given, therefore plug and chug fellows... plug and chug...

The units are bites / min because that is the units that are given for the rate at which John Adams eats cheesecake.

(c) 63.25 bites

If you figured out part (a), then more than half of this problem has already been solved. In order to get the amount eaten, you must take a definite integral with the bounds given:

The unit is bites because if you take the integral of a rate then you get the amount.

 

That's averaging about 5.75 bites a minute. Or almost one bite every ten seconds. Adams like his cheesecake.

(d) 500 grams

 

This requires another definite integral, but this time from t = 0 to t = 20 because you want to know how many bites of cheesecake Adams has taken over the whole period of time he has been eating:

Then you have to take the number of bites and multiply by 5 grams in order to get the total mass of the cheesecake.

Questions? Corrections? Email me at Jonathan352@gmail.com or comment!

Wait, Exacly How Much Cheesecake / Tea Did I Eat / Drink?

When eating this:

The more variety the better

 Or drinking this:

This is indeed a chest of tea

I often forget exactly how much of each of these items I consume. Cheesecake is too heavenly to bother with thinking about the amount of the decadent treat that you are eating. Tea is... well TEA. The only thing that people drink more of is WATER and its the national drink of ENGLAND. Civilized peoples drink this and civilized peoples certainly don't think about how much they drink.

Now that we have the random facts and foodstuffs out of the way, on to the Calculus.

So... What does an amnesic experience with food consumption have to do with Calculus?

INTEGRALS 

Otherwise known as the "antiderivative"

Yes, this would be the other half of Calculus, and like everything else in Mathworld such as addition / subtraction, multiplication / subtraction, and e / ln, the integral is indeed the opposite of a derivative. 

The integral is defined as the area under the curve, which looks something like this:

As indicated, when the area of the curve is above the x axis, then the integral is positive and when it is below the x axis, then it it negative.

The use of an integral is (you guessed it) opposite of the use of a derivative. Instead of finding a rate of change from a function of how much of "something" there is at a point in time, you can use an integral to find exactly how much of "something" there is during a certain period of time.

Now, yet again there are certain shortcuts that you can use whilst doing integrals.

One of the most used and most common is:

  REVERSE POWER RULE!

Oh those mathematicians... Such creative names...

As you (hopefully) have guessed by now, since the integral is the opposite of a derivative then you can apply the opposite of the power rule in order to do integrals.

This means that you add one to the exponent of each term and then divide the term by the new exponent. This can be represented by:

There are actually two types of integrals. The one shown above is called an indefinite integral. This means that instead of getting a numerical answer, we will get an expression. This is similar to taking the derivative of a function without establishing what x equals first. 

NOTE: The "dx" means with respect to 'x'. If the variable you wanted to integrate was 'y', then you would write "dy" instead.

Then there's the problem of the big "C" in the middle of the page. It stands for "constant" and indicates that when you take an integral of something you are missing some information.

Remember that when you take a derivative of a function that has a constant of it, you lose the constant because the power rule causes the constant to be multiplied by 0:

Essentially, if you just had 2 as a constant, then it would look something like the above. Since x^0 = 1, then 2x^0 = 2. And if you apply the power rule when taking the derivative then it would equal 0.

And so, it is apparent here that some information is lost here. The all important "+2" disappears into oblivion.

Which is why when you take the integral:

You have to add the "+C" there to indicate that there may be a constant missing.

Then we move on to the second type of integral... If you had to guess what it is called based on the fact that the first type of integral is indefinite integral... Yes, it is indeed a definite integral.

A definite integral varies from an indefinite integral in that a definite integral will get you a definite answer...

The point of an integral is to get the area under the curve, or change from a rate of change of "something" into how much of "something" over an area or period of time.

In order to do this, "bounds" have to be set. These bounds indicate the period in which you want to know how much of "something" there is.

An example of how the process works looks like this:

The bottom generally contains the lower bound and the top contains the higher bound of the area you want to integrate. If this is not the case for a problem, then your answer will have to be multiplied by "-1".

First, you take the integral of your function and then you use a vertical bar to indicate what your bounds are. Then you plus in your upper bound to the integral and subtract the lower bound plugged into the integral.

Questions? Email me at Jonathan352@gmail.com or comment.

An integral practice problem will be posted soon in order to clear some things up. Oh, and this time it will have cheesecake and a certain historical figure...

Best Thing to Calculus Since Cheese in Cheesecake

Wolframalpha is pretty much the best thing to Calculus since tea in England, tomato in soup, cheese in cheesecake, or teh interwebz to trolls.

Chesecake: Now With 100% More Calculus, Derivatives, and Tea

By now, most of you are probably thinking "Yummy... cheesecake..." and a smaller percentage are thinking "Okay... so where is the Calculus?"

Right here...

This leads us to one of the two major Calculus concepts:

DERIVATIVES

This is one of the words that you should ingrain into your mind as much as 'cheesecake'. That's right, this words is JUST AS IMPORTANT AS CHEESECAKE.

 = 

That's right... Cheesecake + Tiramisu = Calculus

 What a derivative can tell you, if you know how much of 'something' there is as time passes, is how the amount of 'something' is changing at any point in time. This is otherwise known as the instantaneous rate of change, or slope. 

Derivatives can be represented on a graph with tangent lines, which are indicated in red on the above graph. 

Now, there are two ways you can do a derivative of a function. One way is to plug your function into this:

Warning: May lead to brain implosion

Note: f '(x) is denotes that it is the derivative of the function f (x)

If you remember what the definition of slope is, then the definition of a derivative may look slightly familiar:

 Slope equation

The numerator part of the derivative definition gives you two 'y' points or your 'rise' and the denominator 'h' is the difference between the 'x' points giving you your 'run'. So, therefore you have your rise over run or the definition of slope. Remember that slope also means the rate at which something is happening.

No need to be afraid of the derivative though! There is a shortcut to do your derivatives so you don't have to have six lines of work in order to derive x² (as indicated at the beginning of this post). 

One of the most common saving graces is called the:

POWER RULE 

Black... er... POWER RULE!
  

 There are many other shortcuts, but this one is by far the most used and easiest to perform. What you can do is multiply the coefficient by the exponent and then lower the exponent's power by one.

This can be mathematically represented by:

f '(x) = ax^(a-1)

So, instead of taking up six lines and slowly corroding your brain with the definition of a derivative in order to derive x², you can use the power rule to say it is 2x.

Some other examples are:

f(x) = x³

f’(x) = 3x²

f(x) = 2x⁴

f’(x) = 8x³

^{Since you can take the derivative of these functions, you can now plug in any 'x' value into the derivative function to find the slope at that point in the graph, which is also the change in rate at that point.}

NOTE: The power rule is only applicable with isolated terms with one variable. For example, you cannot use the power rule on (x+y)^2 or x*y.

^{If you are still confused on exactly what a derivative is, this example problem may help:}

^{Boston Tea Party}

 ^{342 chests of tea on a boat... 342 chests... take one down, throw it around... 341 chests of tea...}

^{On the night of December 16th, 1773, a group of men dressed as Mohawk Native Americans headed towards the Boston Harbor and dumped all the tea off three ships. The amount of tea dumped off the ships can be modeled as:}

 ^{f(t) = .01 * t^2 where 0 min < t < 185 min}

^{Where 't' is the time in minutes and f(t) is the number of crates dumped off of the ships.}

^{(a) Take the derivative of f(t) and explain what f '(t) means in context of the problem.}

^{(b) Find how fast tea is being dumped at t = 100 min. Be sure to indicate units.}

<sup>(c) Find how fast tea is being dumped after 144 crates have already been dumped. Be sure to indicate units.

Solutions:

For some reason, most teachers don't explain what the purpose of the problem was or how the problem was created. This is similar to asking someone to make a cheesecake without the knowledge of what the ingredients are. You may have the instructions on the process of making a cheesecake (just as you may know how to do a derivative) but not knowing the ingredients (or the makeup of the Calculus problem) can make making a cheesecake (or doing a Calculus problem) much more difficult than it should be.

The reason why the bounds for the problem are 0 min < t < 185 min is because you can't have negative time and if you plug in 185 minutes for t into f(t) then you get around 342 chests of tea, which was the number of chests of tea that the Sons of Liberty dumped on that day.

(a) f '(t) = .02t</sup>

<sup>This was asked in order to test your knowledge of taking the derivative of a function. For this, you could have used the definition of a derivative (the hard way), or used the power rule (the easy way).

For the power rule, you multiply the coefficient, which in this case is .01, by the exponent, which in this case is 2, and then subtract one from the exponent. Thus, you get f '(t) = .02t

The meaning of the derivative is the rate at which tea is being dumped at certain points in time. Since the units are crates of tea and time in minutes, then the unit that represents this derivative is crates / minute.

(b) 2 crates / minute

This was asked in order to see whether you realized that "how fast" means the rate at which tea is being dumped. Since the derivative of a function that indicates an amount (in this case, the total amount of tea that is dumped at a certain period of time) is the rate that the action is taking place (in this case, how fast they are dumping the tea at a certain period in time).

To solve this, all you have to do is plug in the 100 min. into your derivative. Therefore, at 100 minutes, they are dumping tea at around 2 crates per minute. As you can see, the units include 'per minute' which tells you the rate of an action.

 f '(100) = .02 (100) = 2 crates / minute

(c) 2.4 crates / minute

This problem tested your ability to understand the relationship between the derivative and the original function. Yet again, you needed to understand that "how fast" means the rate at which tea is being dumped, but you also needed to understand that the information it gave you was not a time, but the amount of crates that had been dumped.

This is a two step problem. You had to first find out the time at which there had been 144 crates of tea dumped and then plugged in that time into your derivative in order to find out the rate at which tea was being dumped at that time.

144 = .01 * t^2 
t =  120 minutes

This is done in order to find out the time at which 144 crates of tea had been dumped.

f '(120) = .02 (120) = 2.4 crates / minute

Then you can plug in 120 minutes into your derivative to find the rate at which tea is being dumped per minute when 144 crates were dumped.

If you have any questions or complaints over any of this post, feel free to e-mail me at Jonathan352@gmail.com or post in the comments.

Also, saying that the example problem should have been about cheesecake instead of tea is a silly complaint. It is common knowledge that tea is on the same level as cheesecake. Just look it up in the Fact Book of Facts.</sup>