A-level differentiation/derivative dilemma

[Martin Harris]

Hello, and thank you for your time.

I just started my A-levels derivatives/differentiation , and I would be more than happy if you could help me clarify it.

For example I know that y is a function in terms of x right?
y=f(x)
The derivative of it is f'(x)=dy/dx .
This means it is the rate of change of y over the rate of change of x,which is also called gradient.
What is the actual difference between a function and its derivative?What is the physical meaning of that?

Let's take an example.
For the Circle:
Let's say Area $A=\pi*r^2$

then if we derivate it with respect to r it becomes:

$\frac{dA}{dr}$$=$$2*\pi*r$
Which is the circumference of the circle.

What it is the meaning of this derivative?It is the rate of change of A with respect to r.
But what does it mean ?I know that 2*pi*r it's the circumference of the circle.

Another question I have it is regarding the difference between the derivative and differentiation.

Is it right if I say
Derivative=rate of change of y with respect to y
Differentiation=find out derivative of a function.

What exactly is the difference between those 2?

I am trying to understand the physical meaning, the phenomena behind all those derivatives/differentiation problems.

Also,what about this, I know that dy/dx is the gradient,rate of change.
but I also learned that a small change can occur in the gradient which is δy/δx

we know that dy/dx~δy/δx

but what exactly is the relation between dy/dx and δy/δx?
How can δy/δx be interpreted in terms of dy/dx?
I mean I was thinking of something like $\frac{δy}{δx}$=$\frac{y+δy}{x+δx}$

Thank you very much in advance,it is much appreciated.

 

[Lucas SV]

For example I know that y is a function in terms of x right?

If you are given a formula of the form $y=f(x)$, that is what you should assume. I have to say though, that to really define a function you should specify its domain and codomain, and I hope they would at least specify the domain in an exam question. Do you know if that is the case?

What is the actual difference between a function and its derivative?What is the physical meaning of that?

For a starter, the derivative of a function is defined at a point of the function's domain. So it may be that the derivative does not exist at any point, or that it does exist in all but one point. For example, the absolute value function $y=|x|$ with the domain being the real numbers, has derivatives at all $x$ in the real numbers, except for $x=0$. If a function's derivative exists at $x$, we say the function is differentiable at $x$.

The 'physical meaning' you want to attach really depends on context. However in general, you should think of the derivative of $f$ at a point $x_0$, as the slope of the tangent to the graph $y=f(x)$ at $x_0$. More precisely, it is the limit (if it exists):
$$
f'(x_0)=\lim_{h\rightarrow 0}\left(\frac{f(x_0+h)-f(x_0)}{h}\right).
$$

What it is the meaning of this derivative?

Excellent question. I think this goes beyond what you learn in A-levels, but I will give the answer anyway. If you pursue a degree which has sufficient maths (such as maths, physics or engineering), you will learn to calculate areas in the plane using tools from calculus. In particular, you will see that the area of a circle can be calculated by multiplying two simple integrals (assuming you have come across integration yet). There are two ways to picture this.

The first picture you should have in your head, is breaking up a circle into tiny sectors, and adding up the areas of all sectors. Well, suppose each sector has a small angle $\delta\theta$. Then you find the area of the sector is $\frac{1}{2}r^2\delta \theta$. When you sum up the areas, since the sum of angles is always $2 \pi$, you will find the area of the circle is $\pi r^2$. In this case, the final integration is over angles.

The second picture is breaking up a circle into tiny annuli. Suppose each annuli has a inner circle radius $s$ and outer radius $s+\delta s$. Then, roughly, the area of the annulus is $2 \pi s \delta s$. Then adding all those areas is akin to integrating $2 \pi s$ from $0$ to $r$, the result being $\pi r^2$. In this case, the final integration is over radial distances. Now by the fundamental theorem of calculus (which tells us that differentiation is the opposite of integration), it makes sense that
$$
\frac{dA}{dr}=2\pi r.
$$If you have not done so already, you will learn about the fundamental theorem of calculus in your lessons.

You also asked if this is the rate of change of $A$ with respect to $r$ and this is correct. To see this, it suffices to multiply the rate of change by a small quantity of what you are trying to change. So
$$
\frac{dA}{dr}\delta r=2\pi r \delta r,
$$which is, as we saw, roughly the area of the annulus with inner radius $r$ and outer radius $r+\delta r$. But this area is also the difference of the area $A(r+\delta r)$ of the outer circle and the area $A(r)$ of the inner circle. So we find
$$
\frac{dA}{dr}\delta r\approx \delta A,
$$ where I have defined $\delta A=A(r+\delta r)-A(r)$ as the change in the area of the circle due to the change in radius $\delta r$.

What exactly is the difference between those 2?

They are correct but somewhat imprecise. Also you made a typo in your definition of a derivative. We talk about derivatives at points, but when someone says 'the derivative of $f$' without specifying a point, this really means a function, called $f'$, whose domain is all the points at which $f$ is differentiable. Yes, your second definition is fine, since when people talk about differentiation they are talking about the method to find $f'$, given $f$ is known.

but what exactly is the relation between dy/dx and δy/δx?

Good question. I refer to the definition I gave of the derivative at $x_0$. If we fix a specific $\delta x$ and let $h=\delta x$, we find
$$
\frac{\delta y}{\delta x}=\frac{f(x_0+h)-f(x_0)}{h}.
$$This means the slope of the line which intersects the points $(x_0,y_0)$ and $(x_0+\delta x,y_0+\delta y)$, where I have defined $y_0=f(x_0)$ and $\delta y=f(x_0+h)-f(x_0)$. Then, when you take the limit as $h\rightarrow 0$, you get the derivative. The wikipedia page on the derivative has some nice graphs to illustrate this point.

So what the definition of the derivative really says is
$$
\frac{dy}{dx}\bigg|_{x_0}=\lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x},
$$where I've used a notation for the derivative evaluated at a point $x_0$.

I hope this helps. Also hopefully, in my explanation of the area of the circle, I gave you a taste of what is to come if you decide to do a degree in maths, physics or engineering.

 

[Martin Harris]

If you are given a formula of the form $y=f(x)$, that is what you should assume. I have to say though, that to really define a function you should specify its domain and codomain, and I hope they would at least specify the domain in an exam question. Do you know if that is the case?For a starter, the derivative of a function is defined at a point of the function's domain. So it may be that the derivative does not exist at any point, or that it does exist in all but one point. For example, the absolute value function $y=|x|$ with the domain being the real numbers, has derivatives at all $x$ in the real numbers, except for $x=0$. If a function's derivative exists at $x$, we say the function is differentiable at $x$.

The 'physical meaning' you want to attach really depends on context. However in general, you should think of the derivative of $f$ at a point $x_0$, as the slope of the tangent to the graph $y=f(x)$ at $x_0$. More precisely, it is the limit (if it exists):
$$
f'(x_0)=\lim_{h\rightarrow 0}\left(\frac{f(x_0+h)-f(x_0)}{h}\right).
$$

Excellent question. I think this goes beyond what you learn in A-levels, but I will give the answer anyway. If you pursue a degree which has sufficient maths (such as maths, physics or engineering), you will learn to calculate areas in the plane using tools from calculus. In particular, you will see that the area of a circle can be calculated by multiplying two simple integrals (assuming you have come across integration yet). There are two ways to picture this.

The first picture you should have in your head, is breaking up a circle into tiny sectors, and adding up the areas of all sectors. Well, suppose each sector has a small angle $\delta\theta$. Then you find the area of the sector is $\frac{1}{2}r^2\delta \theta$. When you sum up the areas, since the sum of angles is always $2 \pi$, you will find the area of the circle is $\pi r^2$. In this case, the final integration is over angles.

The second picture is breaking up a circle into tiny annuli. Suppose each annuli has a inner circle radius $s$ and outer radius $s+\delta s$. Then, roughly, the area of the annulus is $2 \pi s \delta s$. Then adding all those areas is akin to integrating $2 \pi s$ from $0$ to $r$, the result being $\pi r^2$. In this case, the final integration is over radial distances. Now by the fundamental theorem of calculus (which tells us that differentiation is the opposite of integration), it makes sense that
$$
\frac{dA}{dr}=2\pi r.
$$If you have not done so already, you will learn about the fundamental theorem of calculus in your lessons.

You also asked if this is the rate of change of $A$ with respect to $r$ and this is correct. To see this, it suffices to multiply the rate of change by a small quantity of what you are trying to change. So
$$
\frac{dA}{dr}\delta r=2\pi r \delta r,
$$which is, as we saw, roughly the area of the annulus with inner radius $r$ and outer radius $r+\delta r$. But this area is also the difference of the area $A(r+\delta r)$ of the outer circle and the area $A(r)$ of the inner circle. So we find
$$
\frac{dA}{dr}\delta r\approx \delta A,
$$ where I have defined $\delta A=A(r+\delta r)-A(r)$ as the change in the area of the circle due to the change in radius $\delta r$.They are correct but somewhat imprecise. Also you made a typo in your definition of a derivative. We talk about derivatives at points, but when someone says 'the derivative of $f$' without specifying a point, this really means a function, called $f'$, whose domain is all the points at which $f$ is differentiable. Yes, your second definition is fine, since when people talk about differentiation they are talking about the method to find $f'$, given $f$ is known.Good question. I refer to the definition I gave of the derivative at $x_0$. If we fix a specific $\delta x$ and let $h=\delta x$, we find
$$
\frac{\delta y}{\delta x}=\frac{f(x_0+h)-f(x_0)}{h}.
$$This means the slope of the line which intersects the points $(x_0,y_0)$ and $(x_0+\delta x,y_0+\delta y)$, where I have defined $y_0=f(x_0)$ and $\delta y=f(x_0+h)-f(x_0)$. Then, when you take the limit as $h\rightarrow 0$, you get the derivative. The wikipedia page on the derivative has some nice graphs to illustrate this point.

So what the definition of the derivative really says is
$$
\frac{dy}{dx}\bigg|_{x_0}=\lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x},
$$where I've used a notation for the derivative evaluated at a point $x_0$.

I hope this helps. Also hopefully, in my explanation of the area of the circle, I gave you a taste of what is to come if you decide to do a degree in maths, physics or engineering.

Thank you very much sir for your help and understanding.Much appreciated !

 

[Martin Harris]

The 'physical meaning' you want to attach really depends on context. However in general, you should think of the derivative of ff at a point x0x_0, as the slope of the tangent to the graph y=f(x)y=f(x) at x0x_0. More precisely, it is the limit (if it exists):

f′(x0)=limh→0(f(x0+h)−f(x0)h)​

Thank you also for this answer , I should be more clear in giving out information,sorry for that.
I was trying to think by physical meaning what exactly in terms of physics represents , like this rate of change.
For example acceleration is velocity/time
let a=accel; u=velocity; t=time
so a=u/t
da=du/dt right?
Am I right if I say that a=u/t =instant acceleration and da=du/dt as acceleration at a time t?

I want to understand the phenomena behind the derivatives, and how are they different from normal functions?
How they work, and what they mean?

 

[Martin Harris]

For example in Physics, In Thermodynamics.
We know 1st law of Thermo:
ΔU=Q-L
if we differentiate they say that we find
dU=-PdV+TdS
what is the semnification of the small "d" is it like a Δ right?Like a small change right?
U=internal energy
P=pressure
V=volume
T=temp in K
S=entropy
I assume Q=TdS=TΔS and W=PdV=PΔV
I will attach some photos, and if you don't mind could you please have a look over?
I am really keen in understanding the phenomena, and the notations .

Regarding the lower part with the Heat Capacity at constant volume Cv.
I assume that on the right corner at the bottom of the bracket that V stands for V=constant
but I quite not get the $\frac{δH}{δT}$ part
Why is the meaning of this why would divide H(enthalpy) by T(temp in K)
I also assume "δ" as a small change right? not as a "d" as a bigger difference.

I would be more than glad if you could also help me with the 2nd picture(Please have a look below if it is possible)

I quite can't understand why they used "d" and then "δ" like with $df=\frac{df}{dx}*dx+\frac{df}{dy}*dy$
Also how they got that $p=\frac{-δA}{δV}$ at constant T

And also they said we know from Calculus that in the 2nd Derivative , but they basically switched δx with δy on the bottom , in the multiplication.I am trying to understand the physics phenomena in terms of differentiation, and also the notations with those derivatives.

Am I right?Please feel free to correct me, and also add , I am keen on learning more.

 

[Lucas SV]

Thank you also for this answer , I should be more clear in giving out information,sorry for that.
I was trying to think by physical meaning what exactly in terms of physics represents , like this rate of change.
For example acceleration is velocity/time
let a=accel; u=velocity; t=time
so a=u/t
da=du/dt right?
Am I right if I say that a=u/t =instant acceleration and da=du/dt as acceleration at a time t?

I want to understand the phenomena behind the derivatives, and how are they different from normal functions?
How they work, and what they mean?

Yes, in classical physics, we talk about motion of particles, and calculus is a heavily used tool for the understanding of kinematics, and other aspects of classical mechanics. When you are beginning to apply calculus to kinematics, the first problems you will encounter is motion in one dimension.
In this case, we typically write start by taking an inertial coordinate system. Assume the particle is restricted to move across the $x$ axis alone. Then we can ignore the $y$ and $z$ coordinates and write something like $x=f(t)$ to describe the particle's motion, where $t$ is time (you can think of $t=0$ as the time when you started your clock) and $x$ is the x-coordinate of the particle. We want $f$ to be some function, assumed to be at least twice continuously differentiable. We can take the domain to be the entire real numbers for simplicity.

Now you are right that the functions used to describe kinematics are special, and making them twice continuously differentiable is a restriction which makes them special. The intuition behind this is that you want the function to be sufficiently 'smooth' (whatever that means) in order to at least be able to define a continuous acceleration function. A function that has wild fluctuations is not very nice to describe physics, so physicists consider functions which have sufficiently nice properties, so that they can use those properties to simplify calculations. This is not done only in kinematics, but pretty much in all of physics.

Then the velocity is defined as $u=f'(t)$ and the acceleration as $a=f''(t)$. Now you can appreciate the requirement that the function be twice differentiable. With acceleration being defined, we can finally apply Newton`s second law. Now the simpler formula $a=u/t$ is only true in simpler cases, in particular this defines a differential equation $\dot{u}t=u$, whose solution is $u=At+B$ where $A$ and $B$ are constants. That is, the special case is motion under constant acceleration, and under that assumption you can solve simple kinematic problems.

Now you should realize I gave an ideal situation. When we are trying to make measurements, we can only approximate. So to measure the velocity, we can measure the position of the particle twice during a short time interval. The shorter the time interval the better approximation. In the limit as the time interval goes to zero, and as the precision and accuracy of the measurements go to infinity, you will calculate the exact velocity. Actually this statement assumes the continuity of the velocity function $u=f'(t)$ at the time $t_0$ we wish to calculate it, since continuity at $t_0$ means that $\lim_{t\rightarrow t_0}f'(t)=f'(t_0)$. Without this continuity, the limit you calculate might be different than the actual velocity. Now you can again appreciate the assumption that $f$ is twice differentiable, since the differentiability of $f'$ implies the continuity of $f'$.

These ideas can be generalized to a system of one particle in three dimensions, or even any finite number of dimensions. This is what you learn in vector calculus. You can also generalize to a system of multiple particles in any dimensions. In all of these generalizations, you will use calculus. For example, for one particle in three dimensions, the motion is described by what is called a regular curve (a generalization of the twice continuously differentiable function), which can be thought as a curve in 3D which is sufficiently 'nice'. Positions are described by vectors, curves are described by vector valued function of a real variable (still the variable is the time), and velocity and acceleration are defined as the derivatives of the curve in an analogous way.

 

[Martin Harris]

Yes, in classical physics, we talk about motion of particles, and calculus is a heavily used tool for the understanding of kinematics, and other aspects of classical mechanics. When you are beginning to apply calculus to kinematics, the first problems you will encounter is motion in one dimension.
In this case, we typically write start by taking an inertial coordinate system. Assume the particle is restricted to move across the $x$ axis alone. Then we can ignore the $y$ and $z$ coordinates and write something like $x=f(t)$ to describe the particle's motion, where $t$ is time (you can think of $t=0$ as the time when you started your clock) and $x$ is the x-coordinate of the particle. We want $f$ to be some function, assumed to be at least twice continuously differentiable. We can take the domain to be the entire real numbers for simplicity.

Now you are right that the functions used to describe kinematics are special, and making them twice continuously differentiable is a restriction which makes them special. The intuition behind this is that you want the function to be sufficiently 'smooth' (whatever that means) in order to at least be able to define a continuous acceleration function. A function that has wild fluctuations is not very nice to describe physics, so physicists consider functions which have sufficiently nice properties, so that they can use those properties to simplify calculations. This is not done only in kinematics, but pretty much in all of physics.

Then the velocity is defined as $u=f'(t)$ and the acceleration as $a=f''(t)$. Now you can appreciate the requirement that the function be twice differentiable. With acceleration being defined, we can finally apply Newton`s second law. Now the simpler formula $a=u/t$ is only true in simpler cases, in particular this defines a differential equation $\dot{u}t=u$, whose solution is $u=At+B$ where $A$ and $B$ are constants. That is, the special case is motion under constant acceleration, and under that assumption you can solve simple kinematic problems.

Now you should realize I gave an ideal situation. When we are trying to make measurements, we can only approximate. So to measure the velocity, we can measure the position of the particle twice during a short time interval. The shorter the time interval the better approximation. In the limit as the time interval goes to zero, and as the precision and accuracy of the measurements go to infinity, you will calculate the exact velocity. Actually this statement assumes the continuity of the velocity function $u=f'(t)$ at the time $t_0$ we wish to calculate it, since continuity at $t_0$ means that $\lim_{t\rightarrow t_0}f'(t)=f'(t_0)$. Without this continuity, the limit you calculate might be different than the actual velocity. Now you can again appreciate the assumption that $f$ is twice differentiable, since the differentiability of $f'$ implies the continuity of $f'$.

These ideas can be generalized to a system of one particle in three dimensions, or even any finite number of dimensions. This is what you learn in vector calculus. You can also generalize to a system of multiple particles in any dimensions. In all of these generalizations, you will use calculus. For example, for one particle in three dimensions, the motion is described by what is called a regular curve (a generalization of the twice continuously differentiable function), which can be thought as a curve in 3D which is sufficiently 'nice'. Positions are described by vectors, curves are described by vector valued function of a real variable (still the variable is the time), and velocity and acceleration are defined as the derivatives of the curve in an analogous way.

Thank you very much for your answer.
I also have 1 dilemma left regarding maths differentiation.
We know a=u/t ; accel=vel/time (eq1); This is instant acceleration

but we also know if we differentiate eq 1 that da=du/dt (eq2)
What eq2 means? Well we know the rate of change of accel = rate of change of velocity/rate of change of time

But what physically means this?
Like a small difference in accel=small difference in velocity/small difference in time

OK, this is also understood, but what about a practical/physical use.

Let's take a car example
a2=6.5 m/s²
a1=0m/s²

u2=100m/s
u1=0m/s

t2=11s
t1=0s

where da=a2-a1; du=u2-u1; and dt=t2-t1;

This implies that $6.5-0m/s^2=\frac{100-0m/s}{11-0s}$
=>6.5=100/11
6.5=9.09
It is obvious false, but I want to understand the physical meaning behind it ,practical use, and also the notation.

Thank you very much in advance it is much appreciated.

 

[Lucas SV]

da=du/dt

I'm not sure what this means, where did you get this equation from? You might be trying to use similar notation to thermodynamics, but you have to be cautious that differentials are used a bit differently in thermodynamics, for instance, if you see $dW$, there is usually no function $W$ such that $dW$ is its differential, although I will need to check this.

This implies that 6.5−0m/s2=100−0m/s11−0s6.5−0m/s2=100−0m/s11−0s6.5-0m/s^2=\frac{100-0m/s}{11-0s}

That is not correct. As I stated in #6, equation 1, $a=u/t$ is only true for a particle in constant acceleration (take $u(t)=At+B$ and differentiate). When you wrote $a_1$ and $a_2$ with different values, you already claimed that the particle does not have constant acceleration so that equation 1 becomes invalid. That is why you got a contradiction.

Also you should specify the motion, otherwise there is not enough information to solve the problem. In this case, I would assume this is a problem where the acceleration depends linearly on time, that is $a(t)=Kt$, where $K$ is a constant. This is the simplest assumption you can make, consistent with $a(0 s)=0 m/s^2$ and $a(T)=6.5 m/s^2$, with $T=11 s$. From this information you can compute $K$ (do it!). At this point I should ask if you know integration? If so, you can integrate $a(t)$ and use the velocity initial conditions to find the velocity function. If not, you should try to find a function whose derivative is $a(t)$.

Now a note about notation. I was tying to be clear by using $f(t)$, which would be used by a mathematician (mathematicians like clarity), but this is not standard in physics, at least at the introductory classical mechanics level. The standard notation is $x=x(t)$, $v=v(t)=\dot{x}(t)$ and $a=a(t)=\dot{v}(t)=\ddot{x}(t)$, which is actually really handy. That is, often we give the function $f$ the same name as the dependent variable $y$, even though technically they are different. You should learn to be flexible about notation, since in real world problems people use different notations and conventions. Actually I'm still learning this nowadays.

 

[Martin Harris]

I'm not sure what this means, where did you get this equation from? You might be trying to use similar notation to thermodynamics, but you have to be cautious that differentials are used a bit differently in thermodynamics, for instance, if you see $dW$, there is usually no function $W$ such that $dW$ is its differential, although I will need to check this.

That is not correct. As I stated in #6, equation 1, $a=u/t$ is only true for a particle in constant acceleration (take $u(t)=At+B$ and differentiate). When you wrote $a_1$ and $a_2$ with different values, you already claimed that the particle does not have constant acceleration so that equation 1 becomes invalid. That is why you got a contradiction.

Also you should specify the motion, otherwise there is not enough information to solve the problem. In this case, I would assume this is a problem where the acceleration depends linearly on time, that is $a(t)=Kt$, where $K$ is a constant. This is the simplest assumption you can make, consistent with $a(0 s)=0 m/s^2$ and $a(T)=6.5 m/s^2$, with $T=11 s$. From this information you can compute $K$ (do it!). At this point I should ask if you know integration? If so, you can integrate $a(t)$ and use the velocity initial conditions to find the velocity function. If not, you should try to find a function whose derivative is $a(t)$.

Now a note about notation. I was tying to be clear by using $f(t)$, which would be used by a mathematician (mathematicians like clarity), but this is not standard in physics, at least at the introductory classical mechanics level. The standard notation is $x=x(t)$, $v=v(t)=\dot{x}(t)$ and $a=a(t)=\dot{v}(t)=\ddot{x}(t)$, which is actually really handy. That is, often we give the function $f$ the same name as the dependent variable $y$, even though technically they are different. You should learn to be flexible about notation, since in real world problems people use different notations and conventions. Actually I'm still learning this nowadays.

Thank you for your answer.
Yes the acceleration is time dependent in this case, linear.
So if I say a=u/t ;I can't apply differentiation on both sides?
Thus da=du/dt .Is it wrong ?Why?
I mean I know a=u/t ; it is instantaneous acceleration
But what about the acceleration when small changes occur in velocity and time?

 

[Lucas SV]

I mean I know a=u/t ; it is instantaneous acceleration

I think you're confusing average and instantaneous acceleration. This webpage defines the terms: http://physics.info/acceleration/.

Thus da=du/dt .Is it wrong ?Why?

As I said $a=u/t$ is not always right, but let us, for the sake of argument say it is (and I'm once again assuming $a$ and $u$ are functions of time. Then you $da$ is called a differential and is given by $da=\frac{da}{dt}dt$. Now to calculate the derivative of $a$, we use the product rule.
$$
\frac{da}{dt}=\frac{d}{dt}(u/t)=\frac{1}{t}\frac{du}{dt}-\frac{u}{t^2}=\frac{a}{t}-\frac{u}{t^2}
$$

 

[Martin Harris]

I think you're confusing average and instantaneous acceleration. This webpage defines the terms: http://physics.info/acceleration/.

Alright, I understood now, makes sense.
Cheers for that.

Let's take a car example.
A car goes on the highway.
we have u2=80mph and u1=60mph
t2=13 and t1 =10

So the average accel is $\frac{80-60}{13-10}$ =$20/3$=6.666m/s²

Now 2nd scenario for instant accel same car
a=du/dt

Can we say that d=Δ? like small difference

a=du/dt right?

I suppose we need velocity as a function of time to find out instant accel right?We cannot compute it with the above numbers.

 

[Lucas SV]

So the average accel is 80−6013−1080−6013−10\frac{80-60}{13-10} =20/320/320/3=6.666m/s2

That is correct. Notice that as long as you have the initial and final velocity and the time difference, you can always compute this, irrespective of the body's motion.

Can we say that d=Δ? like small difference
a=du/dt right?

Let me give you some history on this. Both Newton and Leibniz are considered to have been the inventors of calculus. The notation $\frac{du}{dt}$ is due to Leibniz. He called $da$ and $dt$ 'infinitesimal' quantities, but it was never clear what this meant. So the definition you should be looking at is, quoting #2, just changing the variable names and the symbol for the difference:
$$\frac{du}{dt}\bigg|_{t_0}=\lim_{\Delta t\rightarrow 0}\frac{\Delta u}{\Delta t}.$$Actually I should have used $\Delta$ all along in #2 since it is more standard, instead of using $\delta$.

You can picture this as choosing smaller and smaller time differences $\Delta t$ and by doing so improving the approximation on the limit. For example suppose the error in the approximation becomes a 10th of the previous error, every time you improve the approximation. Then, in the limit, the error will become zero, and the computed value is exact. However, this is not how you usually compute derivatives. Instead you learn rules of differentiation, and the justification of such rules rely on the limit definition of the derivative.

So first of all, from the data you just gave, it is not possible to compute the instantaneous acceleration. The only way to compute it, is to give me a graph of the velocity as a function of time, or give me the formula. The very important point to be made is that, you can draw many different graphs, whose curves pass through $(t_1,u_1)$ and $(t_2,u_2)$, yet these graphs can have very different instantaneous accelerations at each point.

I highly suggest looking at the pictures and animations found in https://en.wikipedia.org/wiki/Derivative, if you have not done already, to give you an intuitive picture of the derivative.

I suppose we need velocity as a function of time to find out instant accel right?We cannot compute it with the above numbers.

That is correct, you need the velocity as a function of time to compute the instantaneous acceleration.

 

[Martin Harris]

That is correct. Notice that as long as you have the initial and final velocity and the time difference, you can always compute this, irrespective of the body's motion.Let me give you some history on this. Both Newton and Leibniz are considered to have been the inventors of calculus. The notation $\frac{du}{dt}$ is due to Leibniz. He called $da$ and $dt$ 'infinitesimal' quantities, but it was never clear what this meant. So the definition you should be looking at is, quoting #2, just changing the variable names and the symbol for the difference:
$$\frac{du}{dt}\bigg|_{t_0}=\lim_{\Delta t\rightarrow 0}\frac{\Delta u}{\Delta t}.$$Actually I should have used $\Delta$ all along in #2 since it is more standard, instead of using $\delta$.

You can picture this as choosing smaller and smaller time differences $\Delta t$ and by doing so improving the approximation on the limit. For example suppose the error in the approximation becomes a 10th of the previous error, every time you improve the approximation. Then, in the limit, the error will become zero, and the computed value is exact. However, this is not how you usually compute derivatives. Instead you learn rules of differentiation, and the justification of such rules rely on the limit definition of the derivative.

So first of all, from the data you just gave, it is not possible to compute the instantaneous acceleration. The only way to compute it, is to give me a graph of the velocity as a function of time, or give me the formula. The very important point to be made is that, you can draw many different graphs, whose curves pass through $(t_1,u_1)$ and $(t_2,u_2)$, yet these graphs can have very different instantaneous accelerations at each point.

I highly suggest looking at the pictures and animations found in https://en.wikipedia.org/wiki/Derivative, if you have not done already, to give you an intuitive picture of the derivative.That is correct, you need the velocity as a function of time to compute the instantaneous acceleration.

Alright ,thank you very much for answer.It is much appreciated!
Could you also please explain me if you have time what is about that partial derivative in post #5?

Thank you very much in advance.

 

[Lucas SV]

Ok, I got around to reading #5, turns out I didn't need to brush up on thermodynamics so much for a simple reason: you are putting the cart before the horse. Let me explain. You are trying to learn material that uses partial derivatives and functions of two variables, before you even understand and are familiar to derivatives of functions of one variable. My suggestion is make sure you understand the derivative of a function of one variable well and that you have a lot of practise with differentiating, before moving on to partial derivatives.

dU=-PdV+TdS
what is the semnification of the small "d" is it like a Δ right?Like a small change right?

Not quite. $\Delta$ is used to talk about differences, whereas $d$ is used to talked about differentials. I will define this in the case of one variable. Let $f$ be a differentiable function (of one real variable). Fix a real number $x_0$. Then the differential $df$ is a variable that can be written as a function of the variable $x$ (in the same way $y$ can be written as a function of $x$, or $y=f(x)$), as defined by this equation:
$$df=f'(x_0)(x-x_0).$$ We can make a nicer looking equation by defining $dx=x-x_0$. Then
$$df=f'(x_0)dx.$$ Intuitively, since $df$ depends linearly on $x$, it can be visualized by a line. Which line? Precisely the tangent line to $f$ at point $x_0$.

So if you want to compute differentials, you must compute derivatives. The case $dU$ is based on the same principle, but $U$ is not a function of one variable, so I will not go into it.

I assume that on the right corner at the bottom of the bracket that V stands for V=constant

That is correct.

but I quite not get the δHδTδHδT\frac{δH}{δT} part

The notation that you are referring to, used in the picture, is called partial derivative, $\frac{\partial H}{\partial T}$.

I quite can't understand why they used "d" and then "δ" like with df=dfdx∗dx+dfdy∗dydf=dfdx∗dx+dfdy∗dydf=\frac{df}{dx}*dx+\frac{df}{dy}*dy

This comes from the chain rule for functions of two variables. You will learn this if you do a course that has multivariable calculus.

Are you saying you have to learn these equations at A-levels? I'd be very surprised if that was the case. This looks like material from an undergraduate module in thermodynamics. I mean if you want to learn it by all means do so, I always support people who want to go further (I'm one of those people). But I cannot stress enough how important is calculus of one variable to really understand this stuff.

 

[Martin Harris]

Ok, I got around to reading #5, turns out I didn't need to brush up on thermodynamics so much for a simple reason: you are putting the cart before the horse. Let me explain. You are trying to learn material that uses partial derivatives and functions of two variables, before you even understand and are familiar to derivatives of functions of one variable. My suggestion is make sure you understand the derivative of a function of one variable well and that you have a lot of practise with differentiating, before moving on to partial derivatives.Not quite. $\Delta$ is used to talk about differences, whereas $d$ is used to talked about differentials. I will define this in the case of one variable. Let $f$ be a differentiable function (of one real variable). Fix a real number $x_0$. Then the differential $df$ is a variable that can be written as a function of the variable $x$ (in the same way $y$ can be written as a function of $x$, or $y=f(x)$), as defined by this equation:
$$df=f'(x_0)(x-x_0).$$ We can make a nicer looking equation by defining $dx=x-x_0$. Then
$$df=f'(x_0)dx.$$ Intuitively, since $df$ depends linearly on $x$, it can be visualized by a line. Which line? Precisely the tangent line to $f$ at point $x_0$.

So if you want to compute differentials, you must compute derivatives. The case $dU$ is based on the same principle, but $U$ is not a function of one variable, so I will not go into it.That is correct.The notation that you are referring to, used in the picture, is called partial derivative, $\frac{\partial H}{\partial T}$.This comes from the chain rule for functions of two variables. You will learn this if you do a course that has multivariable calculus.

Are you saying you have to learn these equations at A-levels? I'd be very surprised if that was the case. This looks like material from an undergraduate module in thermodynamics. I mean if you want to learn it by all means do so, I always support people who want to go further (I'm one of those people). But I cannot stress enough how important is calculus of one variable to really understand this stuff.

Thank you very much for your answer it is much appreciated.
Yes, you are right, I am always keen in finding new stuff.
My area of interest is Thermodynamics, and I want to learn more about it , however I need to understand the phenomena behind it, and also those derivatives and notations.

 

[Lucas SV]

My area of interest is Thermodynamics, and I want to learn more about it , however I need to understand the phenomena behind it, and also those derivatives and notations.

That is nice, it is a huge area. My suggestion is to continue learning calculus, aiming for functions of multiple variables. Meanwhile read about thermodynamics from more basic material that does not rely as much on calculus, and also read on the history.

Personally, I first looked into multivariable calculus while in high school, way before I actually had a course about it, and I used the following lecture series.

There are some newer ones out there I would guess, now in 2016. But I can recommend Berkeley because I watched it. It starts with one variable calculus, explaining the concepts of limits, derivatives and integrals. Then it goes to vector calculus, and then functions of two variables... You won't need all of it for thermodynamics; my guess is that you can stop about halfway.

 

[Martin Harris]

That is nice, it is a huge area. My suggestion is to continue learning calculus, aiming for functions of multiple variables. Meanwhile read about thermodynamics from more basic material that does not rely as much on calculus, and also read on the history.

Personally, I first looked into multivariable calculus while in high school, way before I actually had a course about it, and I used the following lecture series.

There are some newer ones out there I would guess, now in 2016. But I can recommend Berkeley because I watched it. It starts with one variable calculus, explaining the concepts of limits, derivatives and integrals. Then it goes to vector calculus, and then functions of two variables... You won't need all of it for thermodynamics; my guess is that you can stop about halfway.

Thank you very much for your support! It is much appreciated!

 

[Stephen Tashi]

I want to understand the phenomena behind the derivatives, and how are they different from normal functions?

A derivative is a function, so you shouldn't say that derivatives are "not normal' functions. The definition of the derivative f'(x) of a function f(x) tells how to find f'(x) at a specific value of x, but since you can attempt to do this this at all possible values of x, the derivative is also a function "of x", although there may be particular values where f(x) exists and f'(x) does not.

 

[Martin Harris]

I did the multivariable calculus and understood the partial derivatives.
Basically you derivate with regards to 1 variable and the other is constant.

For example z=x²+xy+y²

now if we derivate with respect to x and y is constant we have
$\frac{δz}{δx}=2x+y$

However the thing that confused me in post #5 in 1st picture let's say is that he did partial and outside the (δU/δT) he put V which means V=ct.
This means that we will differentiate the function U with respect to T at V=ct.
I know at V=ct that P1*V1=P2*V2 isothermal

Also 2nd picture in post #5 is confusing .

So how exactly to approach that partial derivative?

Thank you very much in advance!

 

[Mark44]

I did the multivariable calculus and understood the partial derivatives.
Basically you derivate with regards to 1 variable and the other is constant.

We say "you differentiate with respect to one variable." That's called the partial derivative with respect to, say, x. "Derivate" is also a word, but it's used in financial contexts, not math contexts.

For example z=x²+xy+y²

now if we derivate with respect to x and y is constant we have
$\frac{δz}{δx}=2x+y$

The usual notation is $\frac{\partial z}{\partial x}$, but it can also be written as $z_x$.

However the thing that confused me in post #5 in 1st picture let's say is that he did partial and outside the (δU/δT) he put V which means V=ct.
This means that we will differentiate the function U with respect to T at V=ct.
I know at V=ct that P1*V1=P2*V2 isothermal

Also 2nd picture in post #5 is confusing .

So how exactly to approach that partial derivative?

Thank you very much in advance!

 

[Mark44]

However the thing that confused me in post #5 in 1st picture let's say is that he did partial and outside the (δU/δT) he put V which means V=ct.
This means that we will differentiate the function U with respect to T at V=ct.

No, it means the partial of U with respect to T, with V held constant.
Here's the notation I think you're referring to:
$$\left(\frac{\partial U}{\partial T}\right)_V$$

I know at V=ct that P1*V1=P2*V2 isothermal

Also 2nd picture in post #5 is confusing .

So how exactly to approach that partial derivative?

I'm not sure which one you mean.
In the middle of that page is this equation:
$$dA = -p~dV - S~ dT$$

In the first equation below that, they are taking the partial of A with respect to V, with T held constant. If T is constant, dT = 0, and they are essentially taking partials wrt (with respect to) V of both sides of this equation: $dA = -p~dV$, and solving it for p.

Similar idea for the equation just below that one.

 

[Martin Harris]

No, it means the partial of U with respect to T, with V held constant.
Here's the notation I think you're referring to:
$$\left(\frac{\partial U}{\partial T}\right)_V$$
I'm not sure which one you mean.
In the middle of that page is this equation:
$$dA = -p~dV - S~ dT$$

In the first equation below that, they are taking the partial of A with respect to V, with T held constant. If T is constant, dT = 0, and they are essentially taking partials wrt (with respect to) V of both sides of this equation: $dA = -p~dV$, and solving it for p.

Similar idea for the equation just below that one.

Thank you.
Yes indeed I was reffering to δU/δT at V=ct
So what exactly is the difference between δz/δx and δz/δx at y=ct?
where z=x²+xy+y²

I am asking because I know how to do partial derivative in multivariable calculus, but I didn't quite got the part when they say for example V=ct outside the bracket, how does that influence?

Thank you in advance.

 

[Mark44]

Thank you.
Yes indeed I was reffering to δU/δT at V=ct

It's not "at V = ct". The V subscript means "with V held constant."

BTW, you are using δ (lowercase Greek delta). You should be using this character ∂, as in ∂U/∂T. Click on the ∑ menu item above. It's one of the characters that are available.

So what exactly is the difference between δz/δx and δz/δx at y=ct?
where z=x²+xy+y²

None that I'm aware of, assuming that you mean "with y fixed" instead of "at y = ct."
For your example $\frac {\partial z}{\partial x}$ and $\left(\frac {\partial z}{\partial x}\right)_y$ mean exactly the same thing.

I am asking because I know how to do partial derivative in multivariable calculus, but I didn't quite got the part when they say for example V=ct outside the bracket, how does that influence?

I'm not sure where you're getting this. You posted two images of some pages with partial derivatives. In neither of those did it have V = ct.

 

[Martin Harris]

It's not "at V = ct". The V subscript means "with V held constant."

BTW, you are using δ (lowercase Greek delta). You should be using this character ∂, as in ∂U/∂T. Click on the ∑ menu item above. It's one of the characters that are available.
None that I'm aware of, assuming that you mean "with y fixed" instead of "at y = ct."
For your example $\frac {\partial z}{\partial x}$ and $\left(\frac {\partial z}{\partial x}\right)_y$ mean exactly the same thing.
I'm not sure where you're getting this. You posted two images of some pages with partial derivatives. In neither of those did it have V = ct.

Thank you for your reply.

In post #5 in 1st image at heat capacity at V held ct ,outside the brackets of the partial derivatives.
They say Cv=(δU/δT)V held ct
what is the difference between Cv=(δU/δT)V held ct and this Cv=(δU/δT)
Thank you

 

[Mark44]

In post #5 in 1st image at heat capacity at V held ct ,outside the brackets of the partial derivatives.
They say Cv=(δU/δT)V held ct

What I said before that you apparently didn't see:

You are using δ (lowercase Greek delta). You should be using this character ∂, as in ∂U/∂T. Click on the ∑ menu item above. It's one of the characters that are available.

Earlier you wrote "at V = ct" which I assumed meant that V was set to some constant c times t, presumably time. "ct" is not a recognized abbreviation for "constant" so you threw me off by using "ct" in this way

what is the difference between Cv=(δU/δT)V held ct and this Cv=(δU/δT)

Not much, really. Here the assumption is that U is a function of V and T; i.e., U = f(V, T).
∂U/∂T means the partial of U with respect to T. It is understood that V is held constant. The notation (∂U/∂T)_V just emphasizes that V is held constant.

 

[Martin Harris]

What I said before that you apparently didn't see:Earlier you wrote "at V = ct" which I assumed meant that V was set to some constant c times t, presumably time. "ct" is not a recognized abbreviation for "constant" so you threw me off by using "ct" in this way
Not much, really. Here the assumption is that U is a function of V and T; i.e., U = f(V, T).
∂U/∂T means the partial of U with respect to T. It is understood that V is held constant. The notation (∂U/∂T)_V just emphasizes that V is held constant.

Thank you for your answer! Much appreciated
Now I have another issue ,regarding this picture(please have a look below)

So, we have that Helmholtz Energy called A id.Then he says that pressure p=partial of A with respect to V at T=ct;
So we need to derive the A id expression with respect to V.

I quite don't get it how exactly he got $N*Kb*T*\frac{d*lnV}{dV}$ I don't get from where he got d*lnV/dV
where Kb=Boltzmann constant
N=number of particles
T=temperature
V=VOLUME
Aid=Helmholtz Energy
n=number of moles
R=8.314

Could you please to the derivation step by step so I can understand better?Or explain on steps what he did there?

I mean it is obvious that in our case our variable will be V, and all the constant will be treated as constants in our derivation.
We were expecting our P to be = ϑRT/V

Thank you in advance sir!

 

[Mark44]

In the equation for A^id that's the first equation in the image, the Nε₀ term and the last term on the right (-Nk_B[...]) are either constant or involve T. With T held constant (please don't write this as T = ct -- this is very confusing), he ignores all of the terms that are independent of V.

This gives $A = -Nk_BT[\ln(\frac V {N \Lambda^3}) + 1]$, which can be simplified to $-Nk_BT[\ln(V) - \ln(N \Lambda^3) + 1]$
Next, he takes the partial derivative of A with respect to V, getting
$\frac{\partial A}{\partial V} = -Nk_BT \cdot \frac {\partial} {\partial V}[\ln(V) - \ln(N \Lambda^3) + 1]$
$= -Nk_BT \frac {\partial \ln(V)} {\partial V}$
Note that the 2nd and 3rd terms in brackets above are constants, so their partial derivatives are both 0.
$\frac{\partial (\ln(V))}{\partial V}$ is the same as $\frac {d(\ln(V))}{dV}$, since we're taking the partial with respect to V of a function only of V, so he rewrites the equation above as $\frac{\partial A}{\partial V} = -Nk_BT \frac{d(\ln(V))}{dV}$
$= -Nk_BT\frac 1 V = -\frac {Nk_BT} V$

 

[Martin Harris]

In the equation for A^id that's the first equation in the image, the Nε₀ term and the last term on the right (-Nk_B[...]) are either constant or involve T. With T held constant (please don't write this as T = ct -- this is very confusing), he ignores all of the terms that are independent of V.

This gives $A = -Nk_BT[\ln(\frac V {N \Lambda^3}) + 1]$, which can be simplified to $-Nk_BT[\ln(V) - \ln(N \Lambda^3) + 1]$
Next, he takes the partial derivative of A with respect to V, getting
$\frac{\partial A}{\partial V} = -Nk_BT \cdot \frac {\partial} {\partial V}[\ln(V) - \ln(N \Lambda^3) + 1]$
$= -Nk_BT \frac {\partial \ln(V)} {\partial V}$
Note that the 2nd and 3rd terms in brackets above are constants, so their partial derivatives are both 0.
$\frac{\partial (\ln(V))}{\partial V}$ is the same as $\frac {d(\ln(V))}{dV}$, since we're taking the partial with respect to V of a function only of V, so he rewrites the equation above as $\frac{\partial A}{\partial V} = -Nk_BT \frac{d(\ln(V))}{dV}$
$= -Nk_BT\frac 1 V = -\frac {Nk_BT} V$

Thank you sir, I understood until here :

$\frac{\partial A}{\partial V} = -Nk_BT \cdot \frac {\partial} {\partial V}[\ln(V) - \ln(N \Lambda^3) + 1]$

I understood that at T held constant the 2nd and 3rd term in brackets will be 0 because they are constant.
However from here:

=$-Nk_BT \frac {\partial \ln(V)} {\partial V}$

I did not understand from where the dln(V)/dV comes from, until the end of the solution.

I would be more than grateful if you could lend me a hand.

Thank you very much in advance.

 

[Mark44]

Thank you sir, I understood until here :

$\frac{\partial A}{\partial V} = -Nk_BT \cdot \frac {\partial} {\partial V}[\ln(V) - \ln(N \Lambda^3) + 1]$

I understood that at T held constant the 2nd and 3rd term in brackets will be 0 because they are constant.
However from here:

=$-Nk_BT \frac {\partial \ln(V)} {\partial V}$

The 2nd and 3rd terms I referred are contants, relative to V. They aren't 0 -- their partial derivatives are 0.

I did not understand from where the dln(V)/dV comes from, until the end of the solution.

I wrote something incorrect in my previous post, which is now corrected. What I meant was that, since ln(V) is a function only of V, then $\frac{\partial (\ln(V))} {\partial V}$ is the same as $\frac {d(\ln(V))}{dV}$

 

[Martin Harris]

The 2nd and 3rd terms I referred are contants, relative to V. They aren't 0 -- their partial derivatives are 0.
I wrote something incorrect in my previous post, which is now corrected. What I meant was that, since ln(V) is a function only of V, then $\frac{\partial (\ln(V))} {\partial V}$ is the same as $\frac {d(\ln(V))}{dV}$

Thank you very much for your quick reply , I appreciate it.
I understood that the partial derivatives of constants are 0.
I understood that d(ln(V)) comes from its partial derivative.
However I still don't get from where that partial derivative of V appeared from?I mean this ∂V
How was the transistion made from this −ln(NΛ3)+1 to the ∂V

Thank you in advance

 

[Mark44]

However I still don't get from where that partial derivative of V appeared from?I mean this ∂V
How was the transistion made from this −ln(NΛ3)+1 to the ∂V

From an earlier post.

This gives $A = -Nk_BT[\ln(\frac V {N \Lambda^3}) + 1]$, which can be simplified to $-Nk_BT[\ln(V) - \ln(N \Lambda^3) + 1]$
Next, he takes the partial derivative of A with respect to V, getting
$\frac{\partial A}{\partial V} = -Nk_BT \cdot \frac {\partial} {\partial V}[\ln(V) - \ln(N \Lambda^3) + 1]$
$= -Nk_BT \frac {\partial \ln(V)} {\partial V}$

He's taking the partial with respect to V of both sides of the equation.
$\frac {\partial} {\partial V}[\ln(V) - \ln(N \Lambda^3) + 1] = \frac {\partial} {\partial V}[\ln(V)]$. Those two other terms drop out because their partial derivatives are zero.

 

[Martin Harris]

From an earlier post.He's taking the partial with respect to V of both sides of the equation.
$\frac {\partial} {\partial V}[\ln(V) - \ln(N \Lambda^3) + 1] = \frac {\partial} {\partial V}[\ln(V)]$. Those two other terms drop out because their partial derivatives are zero.

Riiiight, makes sense thank you very much , now understood this bit.
Also looking at picture, could you please tell me how he made transition from nkBT* dlnV/dv to NKbT/V ?

Thank you sir very much

 

[Mark44]

Also looking at picture, could you please tell me how he made transition from nkBT* dlnV/dv to NKbT/V ?

What is $\frac d {dx} [\ln(x)]$?
This is one of the first differentiation rules that you learned in calculus. You won't be able to follow a discourse in thermodynamics using partial derivatives if their are gaps in you knowledge of single-variable calculus.

 

[Martin Harris]

What is $\frac d {dx} [\ln(x)]$?
This is one of the first differentiation rules that you learned in calculus. You won't be able to follow a discourse in thermodynamics using partial derivatives if their are gaps in you knowledge of single-variable calculus.

Yes, I understood is 1/V because of the ln derivation.
In this 2nd picture could you please tell me how he recognized , how did he found out that those terms are 1st term 2nd term 3rd term ?What was the criteria?How he identified which term is 1st which is 2nd and how?
Please see below

Thank you in advance

 

[Martin Harris]

And last confusion I have is regarding this problem.
Please have a look below if possible.

In 1st picture we have the formulas for various energies.
In 2nd picture I solved the exercise plugging in numbers in the formula and I got right answer.
My problem is in the 3rd picture.
that 35335 comes from ΔU=ϑCvΔT I guess, but why?
And also when I try to compute the other 2 terms from the brackets I cannot manage to get that values?
I use same formulas as in 1st picture but instead of T=300K I put T=2000K and still don't get the values.

Thank you very much for your understanding and cooperation.