Some questions while I self-learn Applied Calculus

[gmax137]

Here is some different letters

@user079622 are you aware that these derivatives you have copied here are "partial derivatives" (that's why they have the curly "d" as in: ${\partial f}$)? The normal nomenclature is:

$\frac {df} {dx}$, the derivative of function f with respect to x

$\frac {\partial f} {\partial x}$, the partial derivative of function f with respect to x

These are not the same, and if you do not know what a partial derivative is, that is OK, you just need to slow down and learn about "regular" derivatives first.

 

[user079622]

@user079622 are you aware that these derivatives you have copied here are "partial derivatives" (that's why they have the curly "d" as in: ${\partial f}$)? The normal nomenclature is:

$\frac {df} {dx}$, the derivative of function f with respect to x

$\frac {\partial f} {\partial x}$, the partial derivative of function f with respect to x

These are not the same, and if you do not know what a partial derivative is, that is OK, you just need to slow down and learn about "regular" derivatives first.

No, I haven't gotten to that part yet, I'm still searching what is the most effective way to learn...

 

[Mark44]

Learning from book is very hard, it is hard to figure out what you need to do with this "hieroglyphics".

What book are you using? As I mentioned before, some of what you call "hieroglyphics" is pretty much nonstandard, such as the notation $f_{'}$ to mean $\frac{\partial f}{\partial y}$ or $f_{''}$ to mean $\frac{\partial^2 f}{\partial y^2}$. As someone else and I already mentioned, these are all partial derivatives. It looks to me like you have skipped too far ahead in this book.

just one example:
In book I cant find how they get du, because book skips steps.

What's the example from the book? The first example I saw in the Youtube video seemed pretty obvious.

 

[user079622]

What book are you using? As I mentioned before, some of what you call "hieroglyphics" is pretty much nonstandard, such as the notation $f_{'}$ to mean $\frac{\partial f}{\partial y}$ or $f_{''}$ to mean $\frac{\partial^2 f}{\partial y^2}$. As someone else and I already mentioned, these are all partial derivatives. It looks to me like you have skipped too far ahead in this book.

What's the example from the book? The first example I saw in the Youtube video seemed pretty obvious.

Book for math I for engineering university on my mother language.

I just read text about u-substitution, in every book they dont show all steps..So for me it easier to watch video where professor solve tasks with u-substitution.

 

[user079622]

@Mark44

Find integral of sin 2x dx
step 1. substitution
2x=t
x=t/2 differentiate........dx= dt/ 2

Why x become dx and t become dt, if derivation of x is 1, (x)´=1 ?

 

[Mark44]

Find integral of sin 2x dx
step 1. substitution
2x=t
x=t/2 differentiate........dx= dt/ 2

The letter u is often used for substitutions like this.
Let u = 2x.
Then du = 2 dx.
So $u = 2x \Rightarrow x = \frac u 2$, and $dx = \frac {du} 2$.

Then the integral $\int \sin(2x)dx$ becomes $\int \sin(u) \frac{du}2 = \frac 1 2 \int \sin(u) du$.
Can you carry out the remainder of this problem?

To answer your question...

Why x become dx and t become dt, if derivation of x is 1, (x)´=1 ?

With substitutions you need to work with differentials, not derivatives.
The differential of x is dx. If u is a function of x, then $du = \frac {du}{dx}\cdot dx$.
Here u = 2x, so $du = 2 \cdot dx$.

 

[user079622]

Then the integral $\int \sin(2x)dx$ becomes $\int \sin(u) \frac{du}2 = \frac 1 2 \int \sin(u) du$.
Can you carry out the remainder of this problem?

Yes
=-cos (u) x 1/2 + C= -cos (2x) /2 +C

With substitutions you need to work with differentials, not derivatives.
The differential of x is dx. If u is a function of x, then $du = \frac {du}{dx}\cdot dx$.
Here u = 2x, so $du = 2 \cdot dx$.

only after watching this video do I understand what it is differential

 

[PeroK]

With substitutions you need to work with differentials, not derivatives.

Which is, of course, true of the technique normally used for substitutions. Just this morning, however, I showed how this technique is related to the integration by change of variables, which does not depend on differentials!

https://www.physicsforums.com/threads/question-about-change-of-variables.1060606/#post-7065893

 

[Mark44]

only after watching this video do I understand what it is differential

In every calculus textbook I've ever seen (and there are many), the technique of integration by substitution is almost the first method shown for evaluating integrals. All of the examples show what the substitution is, and from it how to replace the integrand and dx. Somewhere after derivatives are presented, there is usually some discussion of differentials. You didn't mention the title of your calculus textbook or its authors, but I'm starting to think that it isn't a very good source for learning calculus.

 

[Mark44]

I showed how this technique is related to the integration by change of variables, which does not depend on differentials!

From that post, you wrote

The change of variables is really the inverse of the chain rule. Formally, if $f$ and $u$ are suitable functions:
$$\int_a^bf(u(x))u'(x) \ dx = \int_{u(a)}^{u(b)}f(t) \ dt$$

Although differentials aren't explicitly mentioned, the substitution evidently is t = u(x), so dt on the right side corresponds to u'(x) dx on the left side of the equation above,

 

[PeroK]

Although differentials aren't explicitly mentioned, the substitution evidently is t = u(x), so dt on the right side corresponds to u'(x) dx on the left side of the equation above,

That observation motivates the idea of differentials, in terms of something that might makes sense outside the integrand. And, motivates the substitution technique. The proof of the change of variables formula, however, involves only the chain rule and the FTC. It doesn't involve an appeal to dfferentials or infinitesimals. Which is an important point, IMO.

 

[user079622]

I will report back in a few days when I pass the differentials.

 

[bhobba]

In the interim, here is something that might help. Recently, I have thought about how I would teach beginning calculus, i.e., intuitive, non-rigorous calculus. I wrote a few insights articles using the concept of infinitesimal, but now I believe that is not the correct approach to start with. Here are my latest thoughts

Let's break it down step by step. This approach will make learning more manageable and less daunting.

First, here is a preliminary incorrect definition of the derivative that will be improved later.

If f(x) is a function, then Δf/Δx is written as (f(x + Δx - f(x))/Δx.

The derivative of f(x), written as f’(x), is defined as Δf/Δx when Δx is zero. Ok, let's substitute Δx = 0. You get 0/0. Yikes. Calculus is a trick that allows sense to be made of 0/0. Why you would want to do that will be left to when you look at its applications. For now, accept it is very, very useful.

Here is an example of the trick. Let f(x) = x^2. Δf/Δx = ((x + Δx)^2 - x^2)/Δx = (2x*Δx + (Δx)^2)/Δx = 2x + Δx. Now, Δx can be taken as zero without problems, so f’(x) = 2x.

The alert reader will notice that when dividing by Δx, we assume delta x is not zero. But later, it is taken as zero. This is a problem. At this stage, we live with the problem, but as you know, the concept of limits I will show how it circumvents the issue.

Instead of taking Δx as zero, take the limit of both sides, which is legit even though it is never zero, and you get f'(x) = 2x. So the correct definition of f'(x) is limit Δx→0 (f(x + Δx - f(x))/Δx.

Now you can see why the concept of limit has been introduced.

However, at the start, and intuitively, using the incorrect preliminary definition may help.

Thanks
Bill

 

[gmax137]

In the interim, here is something that might help. Recently, I have thought about how I would teach beginning calculus, i.e., intuitive, non-rigorous calculus. I wrote a few insights articles using the concept of infinitesimal, but now I believe that is not the correct approach to start with. Here are my latest thoughts

.
..
...

However, at the start, and intuitively, using the incorrect preliminary definition may help.

I'm pretty sure that's how I was taught calculus in 11th grade, back in the early 1970s.

 

[bhobba]

I'm pretty sure that's how I was taught calculus in 11th grade, back in the early 1970s.

That's how I learned it at 13, in the late 60s, armed with a book from the local library.

For years, I've held the belief that calculus should be introduced sooner in the education system. I'm convinced that a decent student can grasp the basics in US Middle School. There are even standard textbooks and syllabi that support this, such as Shoreman Math and Kuman Math. Kuman math is self-paced. Get this: Some complete it to Calculus level by year 6, but they start at age 3. Most do it by year 8. Shoreman math is a bit too religious for my taste, but it is not uncommon for those using it to complete it by year 8 or 9.

Thanks
Bill