Antiderivatives and when all starts to fit together

As I near the end marker of this journey of learning calculus I meet anti-derivatives. And I am glad I did. Somehow this concept helped me put into perspective all the concepts that have come before it and makes me feel I am back on track again. Just in case you were wondering where I lost my way, it was somewhere into L'hopital's Rule.

First things first: What is an Anti-derivative?

Well an anti-derivative answers the question of: what is this formula a derivative of? Or from what original formula could we have gotten this derivative. For example, if we have x² (the squaring function), it's derivative is 2x. Therefore, x² is an anti-derivative of 2x. (For formal definition go here)

Notice that I wrote that x² is an anti-derivative of 2x. This is important because there can be many anti-derivatives for a given function. If we consider this formula for an anti-derivative: ^xn+1/_n+1+C, (which looks a lot prettier in pictures, see bellow) there is a constant C that is introduced. The way I understand it is that there is only so much information a anti-derivative can give you. In order to recover a specific formula from it's derivative you need to know where that functions "started".

In the anti derivative formula, if x=0 the all we get is C. For example, imagine that x is time, then x=0 is time 0 or your starting point. At that starting point then f(0)=C.

To complete the argument above, let now imagine this scenario. I know that -2x+5 is a derivative of a function I am interested in knowing. Using the ati-derivative formula I get that the function I am interested in is:

But what is C? I have no clue with the information I was given. Let's say I know C is a whole number between 1 and 4. A graph can show me what to expect the graph to be.

But until I know what that constant actually is, I will not know the original formula.

In the picture opposite I have 4 possible graphs, each passing through the y (vertical axis) at 1,2,3 and 4. All 4 graphs are exact copies of the formula I am looking for, but they "start" at different points C.

For a great explanation about how C is relates to anti-differntiation in terms of position and velocity. Check out this video from my Mooculus course. I liked this video not only because Dr. Fowler seemed to have had had too much coffee, but because his explanation incorporates the steps to solve an anti-differentiation equation that has "physical" applications.

Anti-differentiation is an important bridge in my road to understand calculus. At least that is the promise that was made by my professor when he introduced the topic. Whether that is the case or not, I find it fascinating that having information about a function, I can derive other functions that are related, and give me additional information about the original one.

Let me know what you think in the comments.

 

Words of Encouragement

 

My understanding of the chain rule and 5 videos in case I blew it

A few weeks ago I was introduced to the chain rule in my Coursera/Mooculus course. And I found it to be one of those concepts that is straight forward to understand but hard to our into practice.

The chain rule deals with composition of functions. In other words, it deals with the derivative of functions within functions.

The best example I can think of to explain this is,the following: Imagine you are a salesperson whose job is to call customers, set up an appointment for a consult, give them a sales presentation and close a sale.

That sale depends on how many presentations are given, which are dependent of how many appointments were made, which are in turn dependent on how many calls were made. If you had a formula that described this process and you wanted to know how do a change in calls affect sales you might need to use the chain rule.

Let's imagine that such a formula exists. We will use the following variables for it: Sale (S), Presentation (P), Appointments (A), and Calls will be our (x). The formula is S(x)= P(A(x)), in order to know how changes in x affect S(x) we will need to differentiate the function with using the chain rule:

S'(x) = P'(A(x))(A'(x)) which is the same as saying we are taking the derivative of the outside function evaluated at the inside option and we multiply that with the derivative if the inside option.

Let's imagine than in the example above these formulas can be substituted for S(x)= √(3/4x). How will a change in x affect S(x)?

S'(x)= [1/2(3/4x)^-1/2](3/4) and simplifying we get:

S'(x)=	3
	8√(3/4x)

 

Let's imagine the sales person makes 100 calls a day according to the original formula he will get approximately 8.66 sales. What if the sales person makes 50 extra calls by what amount would sales change?

S'(150)=	3
	8√(3/4(150))

 

According to the derivative sales would change .03535 times multiplied by the 50 extra calls which would yield approximately 1.76 extra sales. Now that's the way I understood it. But, since I might be wrong, here are five videos explaining the chain rule.

Professor Jim Fowler produced this amazing video explaining the concept. It is 10 minutes long but for someone strugling to understand what the chain rule is and how it works, its worth the time.

 

Here is the chain rule introduction by Khan Academy.

Chain rule introduction:

 

I liked this example from That Tutor Guy

 

Another straightforward example from justmathtutoring.com

 

This one is from the IntegralCalc channel in YouTube

 

Reference for derivatives and limits

The hardest things  for me to find while studying are good reference sheets. Some people might object to their use and see them as crutches one uses instead of trying to understand the subjects studied. I see them as facilitators. I love to look at reference sheets for the big picture. There I can look for pattern and similarities between the concept studies. And instead of just relying on them, they make me want to explore more and go further.

 I include a Slideshare presentation with a reference sheet on derivatives and limits in the hopes that other feel as I do and can find it helpful. Let me know what you think in the comments. If you cannot see the slide, follow the link at the bottom of the page.

Calculus Derivatives Limits from surya_putra_2009

Words of Encouragment

 

On derivatives and rote learning

The derivative of a function gives us important information about the function being examined. That is a very cool idea. It is basically metadata about how the original function will behave. The derivative of a function, using a very basic explanation which is what I can manage with my knowledge, describes whether a function is increasing, decreasing, and/or changing direction, in other words, it gives you the rate of change.

The derivative also gives us the slope of the tangent line for any point along a function that is differentiable. In other words, if we know the derivative and a point along the function, we can calculate the tangent line for that point.

It bet there a lot of other things the derivative tells us, but these two are the ones that fascinated me the most.

The more I think about it the more it sinks in that calculus is about change. Or in the words of professor Jim Fowler from Mooculus, about how "wiggling the inputs affect the outputs".

Let's do an example: Imagine we have a toy rocket which we shoot up into the sky and watch it fall to the ground. A formula that describes the trajectory of the rocket is: y=-x^2+4x where x=time in seconds and y is feet. In one second it's 3 feet high, in 2 seconds it's 4 feet high. It's average climb from zero to 1second is 3 feet per second and from 1 to 2 seconds it's 1 feet per second. But what how fast was the rocket climbing at exactly 1 second?

The derivative can help us with that. Using the derivative rules:

The derivative of -x^2+4x is -2x+4. And once we know the derivative, and understand that the derivative is the slope of the tangent line at a given point, and remember that the derivative at a point gives us the instantaneous rate of change of a function, we can do some algebra to know how fast the rocket was climbing at exactly 1 second:

-2(1)+4= instantaneous speed

-2+4= 2 feet per second

So at 1 second the rocket is climbing 2 feet per second.

What happens at 2 seconds?

-2(2)+4=0

At 2 seconds the rocket is moving at 0 feet per second, therefore it's standing still.

And at 3 seconds?

-2(3)+4=-2

Interesting, at 3 seconds the rocket is not climbing but declining at 2 feet per second.

We know the rocket hits the ground at 4 seconds. How fast was it going?

-2(4)+4=-8

The rocket hit the ground declining at 8 feet per second.

When we look at the graph we can see these numbers make sense. The parabola starts steeply (the rocket is climbing fast but slowing down), then levels off to a point where there is no more climb (the rocket stands still), and then the parabola declines steeply, (the rocket falls accelerating until it reaches the ground). That was really cool. Now if we look at the graph formula of the derivative by itself, it would have told some key elements of the original function.

-2x+4 is a line. And this line is positive from time 0 to 2 seconds, at 2 seconds it crosses the y axis and is negative until 4 seconds. Looking at that graph and using the first derivative test, we can conclude that from time 0 to 2 seconds the original graph -x^2+4x is increasing in the interval, that at 2 seconds there is a local extreme value where the graph changes direction, and we can also see that from 2 to 4 second the graph is decreasing in the interval. What this means is the if I had the function of the derivative but not the original function I could make some intelligent guesses as to how the original function would behave. Now that to me is amazing.

Like I mentioned before, I have been taking a calculus course for about two months now. Therefore the basics of the derivative is something that has long passed. If fact, it took me a couple of hours to find an explanation of it that I could use in order to post it. This problem got me thinking about the depth of my learning. I know I already covered some of these insights in earlier posts about learning for testing and grades, versus learning to understand.

You see, I was feeling quite confident going week to week in my calculus course, at least at the beginning. That was because I was doing the minimum study required to pass the quizzes. And then, when the concepts started to stack up, I found myself lost. I fear this is what is wrong with the way we educate in Puerto Rico and the United States. If we are teaching to tests, then the students will study to pass those tests and might not explore what the knowledge in itself means. That might mean that the best product we could be creating is excellent test takers instead of excellent thinkers.