Calculus! For beginners, with a ‘no theorems’ approach!

So what really is calculus?

My preferred (but imperfect) definition:

Calculus is the study of how things change.

In this basic course we’ll concentrate on:

• Derivatives
• Differentiation
• Integrals
• Basic differential equations.

So let’s get stuck in!

The two most important functions you already knew:



Start with the left side:

$$y=x$$

Too stupid for words really, isn’t it? For every value of x it returns that value as y.

Now notice something interesting. If I say:

$$\Delta x=x_2-x_1$$
and:
$$\Delta y=y_2-y_1$$

Then always:

$$\Delta y= \Delta x$$

Always! So that:

$$\frac{\Delta y}{ \Delta x}=1$$

Always! So what is this ratio that is always 1 for y=x? This ratio is the slope of the line. In most textbooks the term gradient is now used instead of slope, so I’ll use it here too.

Now, it can be shown that in this particular case this gradient is the first derivative of y to x. Mathematically, if we write:

$$y=f(x)$$

Where f(x) means function of x, then:

$$[f(x)]’=y’=x’=\frac{\Delta y}{ \Delta x}=1$$

The notation:

$$[f(x)]’$$

... means the derivative of the function f(x) to x.

Another notation but meaning exactly the same is:

$$f’(x)=y’=x’=1$$

It’s an enormous step forward, even though it doesn’t look like much:

The derivative of x to x equals 1.

Now look at the function on the right hand side:

$$f(x)=y=a$$

Where a is any number. So this one’s even dumber: no matter the value of x, it always returns a!

Again there’s something interesting about it: no matter the value of Δx, always is Δy=0.

And that means our gradient:
$$\frac{\Delta y}{ \Delta x}=0$$

And by extension:
$$[f(x)]’=f’(x)=y’=a’=\frac{\Delta y}{ \Delta x}=0$$

The derivative of any number to x equals 0.

It’s important to note that obtaining the gradient of a function (and thus the derivative of f(x) to x) can only be obtained as done above for these simple functions. For other, more complicated functions another way has to be used and we’ll soon get to that. For now I want to run with our two main conclusions:

$$x’=1, a’=0$$

Quote: Originally posted by Metacelsus

No theorems? But theorems are so much fun! (Actually though, they are pretty interesting.) One of my favorites: https://en.wikipedia.org/wiki/Mean_value_theorem

They are fun, Metacelsus but let's just say that this thread is for the theorem-averse, of which there are many.

Quote: Originally posted by Volanschemia

Kudos for undertaking this blogfast, I'm studying Mathematical Methods at school at the moment and what you wrote is making sense so far so I must be doing something right!

Do you plan on teaching beyond what is taught in AS level maths?

Great idea for a thread, and well explained.

Just one thing to add, many (Most?) people use "d" in place of "Δ", so the derivative of x is written as dy/dx.

Quote: Originally posted by Maker

1. Do you plan on teaching beyond what is taught in AS level maths? 2. Great idea for a thread, and well explained. 3. Just one thing to add, many (Most?) people use "d" in place of "Δ", so the derivative of x is written as dy/dx.

1. In a later instalment, I'd be very interested to learn how to integrate or differentiate combined functions, like f(g(x)).
3. Ah yes, I see what you mean, just me trying to rush ahead .

Quote: Originally posted by Maker

1. In a later instalment, I'd be very interested to learn how to integrate or differentiate combined functions, like f(g(x)). 3. Ah yes, I see what you mean, just me trying to rush ahead .

No worries.

Quote: Originally posted by aga

Superb stuff bloggers ! Keep this up and you'll definitely get a Sainthood. For those of us who either didn't get more than a basic education, or went to uni and simply didn't try hard at the time, this is of enormous help. [Edited on 8-3-2016 by aga]

Thanks. Sainthood? Nah. Put me out with the rest of the garbage, when I'm done...

Next instalment in about 30 mins.

Stay tuned!

[Edited on 8-3-2016 by blogfast25]

Derivation of polynomial functions:

Quote: Originally posted by aga

Whoah there hoss ! $$[f(x)g(x)]'$$ In the lingo, does that mean the first derivative of a function of x multiplied by another function of x ? Edit: $$f(x)=a, g(x)=x, u(x)=ax$$ What does the comma operator do ? [Edited on 8-3-2016 by aga]

$$[f(x)g(x)]'= [f(x) \times g(x)]'$$


$$f(x)=a\:\text{and}\:\ g(x)=x,\text{then}\:\ u(x)=ax$$

Is that better for you? The comma is just a comma here (punctuation mark).

In algebra we rarely use the X sign for multiplication anymore, instead (e.g.):

$$a \times x=ax$$

Also sometimes:

$$a \times x=a.x=ax$$

The 'dot' as a symbol of a product.

[Edited on 8-3-2016 by blogfast25]

Quote: Originally posted by woelen

An interesting addition may be: (a*x^n)' = n*a*x^(n-1) which is valid for ANY constant power n. Also for non-integer powers and negative powers. E.g. 1/x can be written as x^(-1). The derivative of this is (-1) * x^(-2) = -1/(x*x). E.g. √x can be written as x^½. The derivative of this is ½*x^(½ - 1) = ½*x^(-½)=1/(2√x)

Exercises: derivatives of polynomials

Return to the slope, I mean: gradient!

Consider a general, smooth and continuous function f(x):



If we define intervals between points 1 and 2 as:

$$\Delta x=x_2-x_1\:\text{and}\:\Delta y=y_2-y_1$$

Then, as in the case of y=x, is:

$$\frac{\Delta y}{\Delta x}$$

... the gradient of f(x)? No. Actually, it's the gradient of the cord between point 1 and 2. It's kind of an approximate, average gradient of f(x) between the points 1 and 2 but nowhere near accurate enough for our purpose.

Now if we move the point 2 along the curve of f(x), towards point 1, then the intervals:

$$\Delta x\:\text{and}\:\Delta y$$

... both decrease and the cord between 1 and 2 starts to resemble the line marked tangent. The tangent line is the line that is parallel to the curve f(x) in the point 1.

The gradient of the tangent line in the point 1 is in fact the gradient of the curve of f(x) in the point 1 and by extension of what we saw in the first instalment, that gradient is also the derivative of f(x) in point 1:

$$f'(x_1)$$

I promised as little theory as possible but a minimum is hard to avoid.

It can be shown (but I won’t do that here) that:
$$f'(x_1)=\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$$
What does this mean? It means that we evaluate the expression for ever and ever smaller values of Δx, so Δx actually tends to zero. “lim” stands for in the limit for Δx going to zero.

More explicitly written:
$$f'(x_1)=\lim_{\Delta x \to 0} \frac{f(x_1+\Delta x)-f(x_1)}{\Delta x}$$
And for the general case of x:
$$f'(x)=\lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$$
It’s also written as:
$$\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}=\frac{dy}{dx}=y'(x)$$
Where dy and dx are so-called differentials.

Let's just try it with a simple Example: let:

$$f(x)=x^2$$

$$y'(x)=\lim_{\Delta x \to 0}\frac{(x+\Delta x)^2-x^2}{\Delta x}$$

$$y'(x)=\lim_{\Delta x \to 0}\frac{x^2+2x\Delta x+(\Delta x)^2-x^2}{\Delta x}$$

$$y'(x)=\lim_{\Delta x \to 0}\frac{2x\Delta x+(\Delta x)^2}{\Delta x}$$

$$y'(x)=\lim_{\Delta x \to 0}\frac{\Delta x(2x+\Delta x)}{\Delta x}$$

$$y'(x)=\lim_{\Delta x \to 0}(2x+\Delta x)$$

$$y'(x)=2x$$

Which is what we already knew!

Going back to the differentials, we can write:
$$f'(x)=(x^2)'=\frac{dy}{dx}=2x$$

or:

$$dy=2xdx$$

The latter is in fact the first differential equation of this thread but we won’t dwell on that just yet.

Obtaining the derivatives of other functions:

In principle that same little limit ‘trick’ I performed on y=x² can be used to obtain the derivative of any function, including non-polynomial ones. But the good news is that we don’t have to do it because other brain boxes have already done it for us!

Plenty, plenty tables list the derivative of most functions on the Tinkerwebs. I quite like this one but by all means choose your own:

http://www.ambrsoft.com/Equations/Derivation/Derivation.htm

Note that it uses a notation we haven’t used here yet but it’s very simple. For example, what’s the derivative of cos x?

In that notation:
$$y=\cos x$$

$$y'=\frac{dy}{dx}=\frac{d}{dx}\cos x= -\sin x$$


[Edited on 10-3-2016 by blogfast25]

Derivatives, differentials and differentiation:

Nice to see you doing this Blogfast! I'm in Calculus 1 right now. but no theorems... I think there's a book in the library on advanced math for chemists and physicists, I had taken a peek at it a little while ago, quite cool.

More Rules for derivatives (and differentials):

Combining the Rules and carrying out Substitutions:

Quote: Originally posted by chemrox

Calculus courses are all about the notation. I never cared for the limit notation or the Δ forms. I like the Liebenetz dy/dx format. Of course you have to explain what they mean..and calculus is the only way to learn trig dy/dx ≠ Δy/Δx

Quote: Originally posted by blogfast25

Quote: Originally posted by chemrox   Calculus courses are all about the notation. I never cared for the limit notation or the Δ forms. I like the Liebenetz dy/dx format. Of course you have to explain what they mean..and calculus is the only way to learn trig dy/dx ≠ Δy/Δx I have to disagree somewhat here. Notation is mainly conventional. Some prefer Leibniz, others Newton ("dot y"): $$\dot{y}=\frac{dy}{dx}$$ Using calculus professionally, one has little choice but to familiarize oneself with the various notations and variations. It's a bit like musical notation: it may differ slightly from one composer to another but most will be able to read another's scribblings w/o problems. As regards limit notation, do you have an alternative? As regards "dy/dx ≠ Δy/Δx", that case has been firmly made higher up.

Quote: Originally posted by chemrox

I had to reload my browser. Now about 2/3 of the pages are gone WTF? Did all the latex disappear?

When a derivative becomes zero...

Consider the following smooth and continuous generic function y=f(x):



I’ve also drawn the tangent lines in various points, in green where the tangent has a positive gradient, in red where the tangent has a negative gradient.

As we move from left to right the positive gradient first becomes smaller and smaller, to reach zero at the top of the ‘hill’ (black tangent is horizontal).

Moving beyond that point the gradient of the tangent line becomes negative, then less and less negative to reach zero at the bottom of the ‘valley’ (black tangent is horizontal). It then becomes positive again.

The points where the tangent line’s gradient becomes zero are called optima (minimum and/or maximum) and are the points are those for which:

$$y’(x_{max})=0\:\text{and}\:y’(x_{min})=0$$

So optima occur where:

$$y’=\frac{dy}{dx}=0$$

Parabola:

An interesting case in point is the quadratic polynomial:

$$y=ax^2+bx+c$$

Graphically it shows as a parabola:



Since as:

$$y’=2ax+b$$

There always is an optimum for:

$$2ax+b=0$$

$$x_{opt}=-\frac{b}{2a}$$

The value of y_opt can be calculated from plugging into y.

Note that here the optimum is a maximum but parabolas that exhibit a minimum also exist of course (just flip the graph over 180 degrees).

This property of derivatives becoming zero where optima of the function f(x) exist is used in optimisation problems (next up).

[Edited on 19-3-2016 by blogfast25]

And related to the operator Del, is the nabla:

$$\nabla=\Big(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\Big)$$

and:

$$\nabla^2$$

But I think we're starting to 'show off', as this thread is mainly intended for the education of the calculus-impaired.

$$=-\infty$$

You tryin' to tell me somethang, mistah?

So, what can we do by way of... ermm... remedial classes?

Quote: Originally posted by aga

Is that right, or did i just imagine an easy way ?

Yes, that's definitely what they're there for: make derivations easier.

Do you have problems with these:

http://www.sciencemadness.org/talk/viewthread.php?tid=65532#...

Send answers by U2U, if you prefer.

[Edited on 20-3-2016 by blogfast25]

Quote: Originally posted by blogfast25

Do you have problems with these

Quote: Originally posted by aga

Add-ups, take-aways, times-bys, share-bys, basic algebra, and possibly some trigOhNometry : they are all OK, as in i understand most of those, i.e. had a basic education.

Quote: Originally posted by yobbo I

Thanks for the bit about the infinitesimals, never knew that or understood when they were moved about.

Quote: Originally posted by blogfast25

Maybe try and explain better what you're struggling with? Here or by U2U?

Quote: Originally posted by aga

Quote: Originally posted by blogfast25   Maybe try and explain better what you're struggling with? Here or by U2U? I;m still staring at these http://www.sciencemadness.org/talk/viewthread.php?tid=65532#... and wonding what to do. If you could work through example 1, that would be enormously helpful.

Yup, history in the making: first successful derivations (2) by aga on SM! Lift off but we probably need a bit more 'lift' to come.

$$f'(x)=f(x)'=(f(x))'=[f(x)]'$$

... Are all synonymous and can all be used. I prefer f'(x): "the (first) derivative of the function f of x".

Woelen is right on all counts and as with all things, practice makes perfect! So I'll bring on some more specifically chosen examples tonight.

As you've guessed we're now all ganging up on you and further resistance is futile. You're in a corner: either hoist the white flag or fight your way out of it! We WILL teach you some calculus, whether you like it or not, hombre!

Quote: Originally posted by aga

The latex is tricky to get right 1st time (easier on paper). One by one ... $$y=3x^{-2}-7x^3+x^2-19$$ $$y'=3(-2x^{-3})-7(3x^2)+x+0$$ $$y'=-6x^{-3}-21x^2+x$$

Quote: Originally posted by aga

next ... $$u=ax^3-bx^2+c\:\text{(where a, b and c are constants)}$$ Eh ? if they're constants their derivitives are zero, so .. $$u'=0$$

Quote: Originally posted by aga

next ... $$y=x^{-2}(3x-5)$$ oooh ! i think the product rule applies $$y'=(x^{-2})'(3x-5) + (x^{-2})(3x-5)'$$ $$y'=(-2x)(3x-5)+3(x^{-2})$$ $$y'=-6x^2+10x+3x^{-2}$$

@aga:

Not sure where you're coming from with:

$$\frac{x}{y}=0$$

That would be true for:

$$x=0\:\text{OR:}y=\infty$$

But I don't see where I brought it up?

Perhaps you meant:

$$\frac{\Delta x}{\Delta y}$$

A polynomial: a sum of terms of the general form:

$$ax^2$$

E.g.:

$$y=2+3x-x^5$$

... is a polynomial.

Exercises:

$$y'=3(-2x^{-3})-7(3x^2)+x+0$$

... is a very minor error, should have been:

$$y'=3(-2x^{-3})-7(3x^2)+2x+0$$

The next one was wrong. a, b and c are constants but u is definitely a function of x and its derivative is:

$$u'=3ax^2-2bx$$

Next:

$$z'=3ax^{a-1}-5bx^{b-1}$$

... is 100 % correct.

Next:

$$y'=-6x^2+10x+3x^{-2}$$

... see woelen

Next:

$$u'=26x^{12}-14bx^6+6ax^5$$

... is 100 % correct. Well done also on simplifying it.

Next:

$$z'=\sin(x) +x \cos(x)-\cos(x)$$

... is 100 % correct.

<hr>

Looks like you're getting the hang of it.

Your prize: a bit of reminder algebra of powers: just in case.

$$x^nx^m=x^{n+m}$$

$$\frac{x^n}{x^m}=x^{n-m}$$

$$\Big(x^n\Big)^m=x^{nm}$$

$$x^0=1$$

$$\frac{1}{x^n}=x^{-n}$$

$$\sqrt{x}=x^\frac12$$

$$\sqrt[n]{x}=x^\frac1n$$


[Edited on 21-3-2016 by blogfast25]

Quote: Originally posted by aga

The next ones look harder. Here goes .. $$y=\frac{\sin x}{x}$$ Quotient rule (why they don't call it the Divide-By rule is a mystery) $$y'=\frac{x sin(x)'-sin(x)x'}{x^2}$$ $$y'=\frac{xcos(x)-sin(x)}{x^2}$$ I got a bad feeling about this one.

Quote: Originally posted by aga

That's 4 out of 38 which is 100% ! Woohoo !

@aga:

Apart from a few silly mistakes they're all good!

Wrong:

$$v'=2\Big(a+\frac1x)\Big.-x^{-2}$$

I think you just forgot to put in the multiplication sign. It should have been:


$$v'=2\Big(a+\frac1x\Big).-x^{-2}$$
Simplified:

$$v'=-\frac{2}{\sqrt{x}}\Big(a+\frac1x\Big)$$

Wrong but it was a hard one:


$$z'=e^{\sqrt{2-x^3}}.\frac12(2-x^3)^{-\frac12}$$

You still need to derive:

$$(2-x^3)'=-3x^2$$

and tag it onto your result, so it becomes:

$$z'=e^{\sqrt{2-x^3}}.\frac12(2-x^3)^{-\frac12}.(-3x^2)$$

Cleaned up:

$$z'=-\frac{3x^2}{2\sqrt{2-x^3}}e^{\sqrt{2-x^3}}$$

Wrong and very silly mistake:

$$f'=e^{0.5231x}.0$$

0.5231 is just a constant, so the derivative should have been:

$$f'=0.5231e^{0.5231x}$$

All in all fantastic progress, also on the LaTex!

Last smallish batch of mixed exercises tomorrow, before normal service is resumed.

Stay tuned!

[Edited on 23-3-2016 by blogfast25]

The last five. They're not particularly difficult but I have 'mixed things up a bit':

$$y=\frac{4\cos x}{\sin 2x}+3x$$
$$u=x^3e^{2-x}$$
$$v=(x^2+x+1)e^x+\frac{1}{x-1}$$
$$z=[\ln(x-2)]^3$$
$$f=x^2\sqrt [3] {x^3-1}$$

Remember: loose talk costs lives, keep calm and carry on and a faulty derivative will set a missile on its creator rather than take to the skies!

Quote: Originally posted by j_sum1

Nope. The 2 is just a constant. It probably looks more familiar written: y= -x^3 + 2 y' = -3x^2

Quote: Originally posted by aga

Christ ! so the tiniest cock-up leads to enormously different results. woelen is right : 100% meticulous or do not bother when it comes to these calculations. Now might be a good time to ask what they are used for, as in what they apply to in the real world.

Quote: Originally posted by aga

So deltaV=vomit ?

Maybe but what about :

$$z'$$

and

$$f'$$

??

Not really:

$$z'=3[\ln(x-2)']^2$$

Should be:

$$z'=3[\ln(x-2)]^2 \times [\ln(x-2)]'$$

$$z'=3[\ln (x-2)]^2 \frac{1}{x-2}(x-2)'$$

$$z'=3[\ln (x-2)]^2 \frac{1}{x-2}$$

Also:

$$f'=(x^2)'(x^3-1)^\frac13+(x^2)[(x^3-1)^\frac13]'$$

is correct but should become:

$$f'=2x(x^3-1)^\frac13+x^2\frac13(x^3-1)^{-2/3}(x^3-1)'$$

Etc...

Now have a look this, before we resume 'normal service':

<hr>
A little supplement for aga, about rate of change...

Imagine an object (symbolised by the star), moving along a horizontal x axis:



We don’t know at what speed the object moves or whether that speed is constant or not. Now assume the object changes position from x₁ to x₂, between time t₁ and t₂. Then with:

$$\Delta x=x_2-x_1$$
and:
$$\Delta t=t_2-t_1$$
The average speed between these points would be:

$$v_{average}=\frac{\Delta x}{\Delta t}$$

But v_average tells us almost nothing about how the speed differs from one instant to another.

To find the instantaneous speed in every instant t we need to calculate:

$$v(t)=\lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}=\frac{dx}{dt}$$

v(t) is the rate of position (x) change in time (syn.: as a function of time). If:

$$\frac{\Delta x}{\Delta t}=C$$

... with C a constant, then obviously the speed is constant: the rate of position change in time is then:

$$v(t)=C\:\mathrm{m/s}$$

But if:

$$v(t) \neq C$$

Then v(t) will be some function of time.

In that case we can define another rate of change: the rate at which v changes in time and it's called the acceleration a. It can be defined analogously to the above reasoning as:

$$a=\frac{dv(t)}{dt}$$

In the (frequent) case where the acceleration is constant, say:

$$a=\frac{dv}{dt}=3\:\mathrm{m/s^2}$$

... then speed will increase by 3 m/s every second. The rate of change of speed then is 3 m/s² or 3 m/s per second.

Of course an object may also be decelerating, in that case:

$$a=\frac{dv}{dt}<0$$
<hr>
Exercise:

If:

$$x(t)=3t^2+2t+4$$
Then what is:

$$v(t)$$

and:

$$a(t)$$


[Edited on 25-3-2016 by blogfast25]

Quote: Originally posted by aga

Sorry, still fumbling about in the earlier errors, trying to see where i went disastrously wrong. Very confused about what Rule applies where, and to which bits. This is all new, so no experience to draw on. I do wish somebody else would wish to learn this stuff. It would be helpful to see how others approach a problem like these and how they'd solve it. I guess all the nacent genii are all afraid that they'd come second to a drunkard. sigh. unlikely !

Optimisation problems:

In this post it was shown that the derivative y’ of a function y can become zero for specific values of x, known as optima:

$$\Big(\frac{dy}{dx}\Big)_{x_{opt}}=y'(x_{opt})=0$$

This opens up the possibility of solving a whole range of optimisation problems: business owners want to maximise their profit function, rocket scientists look for the longest possible range of their missiles, industrial chemists want to maximise yield of their processes, car makers try to minimise fuel consumption of their creations, etc etc.

Most real world optimisation problems are multi-variable problems and mathematically outside the scope of this thread but below we’ll look at a few simple single variable problems.

1. Simple function:
$$y=(x-2)^2+5$$

To find the optimum:

$$y'=2(x-2)(x-2)'+0$$
$$y'=2(x-2)(1)=2(x-2)$$

Optimum:

$$y'=2(x-2)=0 \implies x=2$$
And:
$$y=5$$

This is in fact a minimum of the function y.

2. Window size optimisation problem:

A window of the following shape needs to be constructed:

The window maker has only 12 m of frame material to work with. What are the dimensions x and y so that the window has maximum surface area S?

The circumference of the window is:
$$L=2x+2y+\pi x=12$$
The surface area is given by:
$$S=2xy+\frac{\pi}{2} x^2$$
Extract y from the first equation:
$$y=6-x-\frac{\pi}{2}x$$
And insert it into the second one:
$$S=2x(6-x-\frac{\pi}{2}x)+\frac{\pi}{2}x^2$$
$$S=12x-2x^2-\pi x^2+\frac{\pi}{2}x^2$$
If we assume this function has an optimum, then:
$$\frac{dS}{dx}=12-4x-2\pi x+\pi x=0$$

$$x=\frac{12}{4+\pi}=1.68\:\mathrm{m}$$

$$y=1.68\:\mathrm{m}$$

3. Can optimisation problem:

A manufacturer wants to produce a 1500 cm³ capacity can, using as little material as possible:



Total surface area of the can:
$$A=2\pi r^2+2 \pi rh$$
Total volume of the can:
$$V=\pi r^2h=1500\:\mathrm{cm^3}$$
Extract h from the second equation:
$$h=\frac{1500}{\pi r^2}$$
And insert it into the second one:
$$A=2\pi r^2+2 \pi r \frac{1500}{\pi r^2}=2\pi r^2+\frac{3000}{r}$$
To see if there’s an optimum, derive A to r:
$$\frac{dA}{dr}=4\pi r+\frac{0.r-3000}{r^2}=4\pi r-\frac{3000}{r^2}$$
Set the derivative to zero:
$$\frac{dA}{dr}=\frac{4\pi r^3-3000}{r^2}=0$$

We know that (always): $$r>0$$

So: $$4\pi r^3-3000=0$$
$$r=\sqrt[3]{\frac{3000}{4\pi}}=6.20\:\mathrm{cm}$$
$$h=12.4\:\mathrm{cm}$$

<hr>

Exercise:

It can be shown that the heat loss q (in Watt per unit of length) of a homogeneously insulated cylinder is given by:

$$q=\frac{\pi \Delta T}{\frac{1}{2k}\ln \frac{D}{d}+\frac{1}{hD}}$$



(Source)

Where T_c is cylinder temperature, T_a is ambient temperature, 2D is total diameter (insulation included), 2d is cylinder diameter. Note that:

$$\Delta T=T_c-T_a=\text{constant}$$

$$k\:\text{and}\:h\:\text{are both constants}$$

$$d=\text constant$$

Determine for which value of D, in function of k and h, q is minimised.

Hint:

Set:

$$q=\frac{\pi \Delta T}{u}$$

Then see if:

$$\frac{1}{u}$$

... has an optimum.


[Edited on 26-3-2016 by blogfast25]

Quote: Originally posted by blogfast25

Let's call: $$p=\ln(x-2)$$ So: $$z=p^3$$ $$z'=3p^2 \times p'$$

Quote: Originally posted by aga

So obvious once you showed it ! It's hard to think outside the bin-bag.

Quote: Originally posted by blogfast25

To find the instantaneous speed in every instant t we need to calculate: $$v(t)=\lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}=\frac{dx}{dt}$$ v(t) is the rate of position (x) change in time (syn.: as a function of time).