B Some questions while I self-learn Applied Calculus

[user079622]

In free time I start to solve differentials and integrals, I am doing fine, I just follow rules and solve the tasks.
I start solve some applied calculus tasks, but I dont really understand why for exmple second derivative represent acceleration, why first is speed, why I need to derivate Area=axb to find a and b with max area of recntangle if I know diagonal..etc

I know that differential is ratio of delta y / delta x. If I change x how much is change y, or simple tangent angle..
Can someone explain this?

 

[Vanadium 50]

What texts (physics and calculus) are you using?

 

[Ibix]

I dont really understand why for exmple second derivative represent acceleration, why first is speed,

Derivatives of displacement with respect to time, you mean. It's important to say what you are differentiating and with respect to what.

How fast you are going is a measure of the rate of change of your position. Rate of change of a function (position) as you vary something (time) is what derivatives measure. Acceleration is the rate of change of velocity with respect to time - so ditto.

why I need to derivate Area=axb to find a and b with max area of recntangle if I know diagonal..etc

Imagine doing it the hard way - draw all the possible rectangles, measure the area as you vary $a$. You could draw a graph of area as a function of one side length. Note that the maximum area is the top of the curve you draw on the graph. So the slope is changing from positive to negative at that point - it's zero. So to find the maximum area you take the derivative of that graph and set it to zero.

 

[user079622]

What texts (physics and calculus) are you using?

I learn and solve tasks from youtube videos, I also bought book mathematics for engineering university, but I didnt learn from this book yet.

Do you think is better to learn from book or what is best source to learn?

Derivatives of displacement with respect to time, you mean. It's important to say what you are differentiating and with respect to what.

How fast you are going is a measure of the rate of change of your position. Rate of change of a function (position) as you vary something (time) is what derivatives measure. Acceleration is the rate of change of velocity with respect to time - so ditto.

Imagine doing it the hard way - draw all the possible rectangles, measure the area as you vary $a$. You could draw a graph of area as a function of one side length. Note that the maximum area is the top of the curve you draw on the graph. So the slope is changing from positive to negative at that point - it's zero. So to find the maximum area you take the derivative of that graph and set it to zero.

I rember from my university time(long time ago) integral is area under curve, so wondered why I derivate function A=axb for this...
I am struggle to connect what has rate of something change(differential) with area, speed..

 

[Mark44]

I start solve some applied calculus tasks, but I dont really understand why for exmple second derivative represent acceleration, why first is speed, why I need to derivate Area=axb to find a and b with max area of recntangle if I know diagonal..etc

Derivatives are rates of change of some quantity with respect to another. Velocity is the rate of change of position with respect to time (also described as the time rate of change of position). If s(t) represents the position at time t, the velocity v(t) is the time rate of change of position. (Note that physics books often use s to represent position -- I don't know why.) So $v(t) = \frac d {dt}(s(t))$ also written as s'(t).
Acceleration is the time rate of change of velocity, so $a(t) = \frac{d^2}{dt^2}(s(t)) = \frac d{dt} v(t)$, also written as s''(t) or v'(t).

I know that differential is ratio of delta y / delta x.

No, what you're talking about is the derivative, which is different from the differential of something. Also, the derivative is a limit.
If y = f(x), the derivative is $\frac{dy}{dx} = \lim_{h \to 0}\frac{f(x +h) - f(x)}h$.
The differential of y is defined as $dy = f'(x) dx$.

I learn and solve tasks from youtube videos,

Not the best source.

Do you think is better to learn from book or what is best source to learn?

From a reputable textbook.

I rember from my university time(long time ago) integral is area under curve, so wondered why I derivate function A=axb for this...

You don't "derivate" a function -- you differentiate it. Doesn't make sense, but it is what it is.

I don't understand what you mean when you say you "derivate function A = axb for this." Another member has already mentioned what happens when you integrate a function. The most basic example of integration is to find the area under a curve.

 

[SammyS]

I don't understand what you mean when you say you "derivate function A = axb for this." Another member has already mentioned what happens when you integrate a function. The most basic example of integration is to find the area under a curve.

I think OP is referring to maximizing Area - as for a rectangle when given some constraint such as perimeter.

 

[Vanadium 50]

I can say with a high degree of confidence that you will make more progress with a good textbook than a bunch of YouTube videos.

 

[gmax137]

... I rember from my university time(long time ago) integral is area under curve, so wondered why I derivate function A=axb for this...
I am struggle to connect what has rate of something change(differential) with area, speed..

The relationship between derivatives and integrals (that the integral is the anti derivative) is "the fundamental theorem of calculus." Look it up.

 

[user079622]

I don't understand what you mean when you say you "derivate function A = axb for this." Another member has already mentioned what happens when you integrate a function. The most basic example of integration is to find the area under a curve.

I dont know what is difference of differential vs derivation , I mean on sign ´
English is not my first language

I think OP is referring to maximizing Area - as for a rectangle when given some constraint such as perimeter.

Diagonal is 4cm, I must find sides a and b for rectangular with max area.
A=a x b --> MAX
d^2=a^2 x b^2

I can say with a high degree of confidence that you will make more progress with a good textbook than a bunch of YouTube videos.

But problem in book there is no professor who explain, you must grasp everything alone.
Khan academy?
https://www.khanacademy.org

 

[Vanadium 50]

But problem in book there is no professor who explain, you must grasp everything alone.

And when you ask questions of a recorded video, does it answer you back? And for how long has this been happening?

 

[user079622]

And when you ask questions of a recorded video, does it answer you back? And for how long has this been happening?

No. How long I study calculus? Few days

 

[Mark44]

I dont know what is difference of differential vs derivation , I mean on sign ´

I explained the difference between a derivative and a differential in post #5.

Diagonal is 4cm, I must find sides a and b for rectangular with max area.
A=a x b --> MAX
d^2=a^2 x b^2

One of the most common applications of differentiation is to find the maximum value of some function. To lessen the chance of confusion I'm going to call the sides l and w (for length and width). For this problem, since the diagonal is 4 cm, then $w^2 + l^2 = 16$, so $l^2 = 16 - w^2 \Rightarrow l = \pm \sqrt{16 - w^2}$
We can assume that neither side is negative, so $l = \sqrt{16 - w^2}$

Then we can write the area of the rectangle as a function of side w as $A(w) = w \cdot \sqrt{16 - w^2}$.
A maximum value occurs at a point (w, A(w)) at which

• A'(w) = 0 or
• w is an endpoint of some restricted domain or
• a point at which A(w) is undefined.

But problem in book there is no professor who explain, you must grasp everything alone.
Khan academy?

Khan Academy is better than most, but IMO, not as good as a textbook. As pointed out already, after you have watched a video, if you have a question, you can't ask the video for an answer to your question.

 

[user079622]

I explained the difference between a derivative and a differential in post #5.


One of the most common applications of differentiation is to find the maximum value of some function. To lessen the chance of confusion I'm going to call the sides l and w (for length and width). For this problem, since the diagonal is 4 cm, then $w^2 + l^2 = 16$, so $l^2 = 16 - w^2 \Rightarrow l = \pm \sqrt{16 - w^2}$
We can assume that neither side is negative, so $l = \sqrt{16 - w^2}$

Then we can write the area of the rectangle as a function of side w as $A(w) = w \cdot \sqrt{16 - w^2}$.
A maximum value occurs at a point (w, A(w)) at which

• A'(w) = 0 or
• w is an endpoint of some restricted domain or
• a point at which A(w) is undefined.

Why you use A(w) and A(ω) ? Is this tipfeler?

Khan Academy is better than most, but IMO, not as good as a textbook. As pointed out already, after you have watched a video, if you have a question, you can't ask the video for an answer to your question.

Yes but you cant ask question to a book as well, plus book has lots of proofs that is not maybe necessary.

 

[Mark44]

Why you use A(w) and A(ω) ? Is this tipfeler?

I used A(w), where w represents the width of the rectangle, because I didn't want to have both A and a in the same equation. I didn't use A(ω) (ω is lower-case Greek $\Omega$). Maybe you thought that the w in the LaTeX expression looked like $\omega$.
No, it wasn't tipfeler (or in English, typo).

Yes but you cant ask question to a book as well, plus book has lots of proofs that is not maybe necessary.

Most of the calculus textbooks I've seen (lots) have very few proofs.

 

[gmax137]

Then we can write the area of the rectangle as a function of side w as
$A(w) = w \cdot \sqrt{16 - w^2}$
A maximum value occurs at a point (w, A(w)) at which

• A'(w) = 0 or
• w is an endpoint of some restricted domain or
• a point at which A(w) is undefined.

So, @user079622 , can you differentiate the expression $A(w) = w \cdot \sqrt{16 - w^2}$

to get $\frac {d} {dw} (w \cdot \sqrt{16 - w^2})$

You will need to use the product rule (https://en.wikipedia.org/wiki/Product_rule)

 

[user079622]

Most of the calculus textbooks I've seen (lots) have very few proofs.

And when you ask questions of a recorded video, does it answer you back? And for how long has this been happening?

Do you think that book is even better option from watching class video lecture, what site/channel do you suggest?
For example like this

So, @user079622 , can you differentiate the expression $A(w) = w \cdot \sqrt{16 - w^2}$

to get $\frac {d} {dw} (w \cdot \sqrt{16 - w^2})$

You will need to use the product rule (https://en.wikipedia.org/wiki/Product_rule)

Yes I can do it.

 

[PeroK]

Official lecture series from a reputable university are a good option. Especially if they follow a textbook that you have.

There are also some good unofficial sources, like this guy:

https://thebrightsideofmathematics.com/

The main problem is that these series rarely give the viewer access to problem sheets. I quite like pausing the video and trying to figure out the next bit myself. If you can do that, then you definitely understand the material.

 

[user079622]

Official lecture series from a reputable university are a good option. Especially if they follow a textbook that you have.

There are also some good unofficial sources, like this guy:

https://thebrightsideofmathematics.com/

The main problem is that these series rarely give the viewer access to problem sheets. I quite like pausing the video and trying to figure out the next bit myself. If you can do that, then you definitely understand the material.

Something like this?

 

[WWGD]

I learn and solve tasks from youtube videos, I also bought book mathematics for engineering university, but I didnt learn from this book yet.

Do you think is better to learn from book or what is best source to learn?


I rember from my university time(long time ago) integral is area under curve, so wondered why I derivate function A=axb for this...
I am struggle to connect what has rate of something change(differential) with area, speed..

You'll have to figure out and tweak over time the learning method(s) that work best for you.

 

[user079622]

I am lost in this letters notation, such a shame that didnt use only one universal and the most simplest, like prime..
https://en.wikipedia.org/wiki/Notation_for_differentiation

 

[Hill]

I am lost in this letters notation, such a shame that didnt use only one universal and the most simplest, like prime..
https://en.wikipedia.org/wiki/Notation_for_differentiation

It's a language. Don't synonyms make a language richer?

 

[user079622]

It's a language. Don't synonyms make a language richer?

Why call it df/dx if I can write simple and less confusing f´(x) ?

 

[Hill]

Why call it df/dx if I can write simple and less confusing f´(x) ?

Many reasons. For example, sometimes you need to make a manipulation such as to go from $\frac {df} {dx} = g(x)$ to $df = g(x) dx$.

 

[user079622]

Many reasons. For example, sometimes you need to make a manipulation such as to go from $\frac {df} {dx} = g(x)$ to $df = g(x) dx$.

What this letters "d" mean?

 

[PeroK]

Why call it df/dx if I can write simple and less confusing f´(x) ?
What this letters "d" mean?

The notation $\frac{dy}{dx}$ is so common that there is nothing to be gained by complaining about it.

 

[Hill]

What this letters "d" mean?

Infinitesimal change.

 

[user079622]

The notation $\frac{dy}{dx}$ is so common that there is nothing to be gained by complaining about it.

Here is some different letters

 

[user079622]

Derivatives of displacement with respect to time, you mean. It's important to say what you are differentiating and with respect to what.

How fast you are going is a measure of the rate of change of your position. Rate of change of a function (position) as you vary something (time) is what derivatives measure. Acceleration is the rate of change of velocity with respect to time - so ditto.

Imagine doing it the hard way - draw all the possible rectangles, measure the area as you vary $a$. You could draw a graph of area as a function of one side length. Note that the maximum area is the top of the curve you draw on the graph. So the slope is changing from positive to negative at that point - it's zero. So to find the maximum area you take the derivative of that graph and set it to zero.

I will stop watching youtube and try learn from book.

When I learn from university book, does it make sense to write notes in my notebook something that is already written in book or this is just waste of time?

What learning strategy do you suggest?

​

(for sure I will solve tasks in notebook..)

 

[SammyS]

No. How long I study calculus? Few days

This is rather enlightening to those offering advice to you.

What is your mathematical background?

What math have you studied previously?

What mathematical skills do you have?

 

[WWGD]

I will stop watching youtube and try learn from book.

When I learn from university book, does it make sense to write notes in my notebook something that is already written in book or this is just waste of time?

What learning strategy do you suggest?


(for sure I will solve tasks in notebook..)

Sorry, but this , learning style and methods, is something you need to develop by/for yourself and tweak it as you go.

 

[Mark44]

Here is some different letters

Some of the notation in the image you posted I've never seen before.
For example, I've never seen $f_{'}$ to represent $\frac{\partial f}{\partial y}$, $f_{'}^{'}$ to represent the mixed second partial derivative, or $f_{''}$ to represent $\frac{\partial^2 f}{\partial y^2}$.
I don't believe any of these notations are used in any modern calculus textbooks, with "modern" defined as being published in the last 125 years.

 

[user079622]

This is rather enlightening to those offering advice to you.

What is your mathematical background?

What math have you studied previously?

What mathematical skills do you have?

I had calculus at university 30years ago, all these years I wasnt in touch with this..

 

[user079622]

Some of the notation in the image you posted I've never seen before.
For example, I've never seen $f_{'}$ to represent $\frac{\partial f}{\partial y}$, $f_{'}^{'}$ to represent the mixed second partial derivative, or $f_{''}$ to represent $\frac{\partial^2 f}{\partial y^2}$.
I don't believe any of these notations are used in any modern calculus textbooks, with "modern" defined as being published in the last 125 years.

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

I find it easier to learn when person solve the task and explain step by step what is he doing and why.

 

[PeroK]

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

I find it easier to learn when person solve the task and explain step by step what is he doing and why.

Presumably this is why people part with considerable sums of money to be taught at university.

 

[user079622]

Presumably this is why people part with considerable sums of money to be taught at university.

Yes there is good reason why professors exists, otherwise everyone would study alone.

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

Instead just watch one video and find how he get du.. everything is clear now

 

[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