Fundamental confusions of calculus
- Thread starter arfa brane
- Start date
- Status
Once again, T does NOT represent the temperature of the bug, it is the temperature of an arbitrary point in space, someplace at z distance from the plane containing the bug. Just calculate the total derivative, would you?
But you set x=3 , y=2 in your earlier fumble, eh?
Why is this so difficult for you to admit to error?
You simply don't know what you're talking about here. It does have a direction: the x direction, and it has a magnitude[sup]*[/sup]: $$\lambda$$.
More generally, a "vector" in mathematics is any element of a vector space. It is well known and easy to show that differential operators form a vector space.
And in differential geometry, one of the standard ways of defining the notion of a vector - literally the thing with a length pointing in a specific direction - on a differentiable manifold is to identify vectors with differential operators.
[sup]*[/sup]EDIT: I should add that, strictly speaking, the notion of a magnitude for this vector is only defined on a Riemannian manifold (a differentiable manifold with a metric defined on it). The vector's magnitude is $$\lambda$$ if the magnitude of $$\frac{\partial}{\partial x}$$ is 1, which is the case if the coordinate system is Euclidean for instance.
Last edited:
So for the bug, z(t) = 0 then.
Also, T gives the temperature of the medium at arbitrary x and y coordinates in space and times t, the vast majority of which the bug never visits. So what's your point?
It isn't. I would have thought that correcting myself would in itself be an admission. But if that's not enough, have it explicitly: I hereby admit that I made an error.
You going to answer the question now?
As clearly explained in the video, the temperature of the bug is given by the value of T at the bug's location at a given time. So obviously it depends on z in the 3-dimensional case.
The total derivative dT/dt describes how the bug's temperature changes with respect to time.
Calculating it is straightforward.
In two dimensions, T = f(x, y, t)
If the bug's path is described by:
x = g(t)
y = h(t)
Then,
$$\frac{dT}{dt} = \frac{\partial T}{\partial x} \, \frac{dx}{dt} + \frac{\partial T}{\partial y} \, \frac{dy}{dt} + \frac{\partial T}{\partial t}$$
Extending to three dimensions, T = f(x, y, z, t)
If the bug's path is described by:
x = g(t)
y = h(t)
z = i(t)
Then,
$$\frac{dT}{dt} = \frac{\partial T}{\partial x} \, \frac{dx}{dt} + \frac{\partial T}{\partial y} \, \frac{dy}{dt} + \frac{\partial T}{\partial z} \, \frac{dz}{dt} + \frac{\partial T}{\partial t}$$
If it's given that the bug doesn't move in z, then clearly:
i(t) = C
dz/dt = 0
$$\frac{dT}{dt} = \frac{\partial T}{\partial x} \, \frac{dx}{dt} + \frac{\partial T}{\partial y} \, \frac{dy}{dt} + \frac{\partial T}{\partial t}$$
If the bug's path in z is not known, then the total derivative can't be calculated.
The point was for Pete to calculate a total derivative for a given function, that's all. Whether the function makes physical sense or not is secondary.
For example, the fact that the bug "never visits" any part of the space outside the plane he lives in, is irrelevant, you can easily imagine that his thermal energy is radiated in the form of em waves to any arbitrary point in space. The resulting increase in temperature decreases exponentially with the distance from the plane containing the bug.
Last edited:
Finally!
Excellent!
The bug doesn't move in z, can you try the same with $$T=at+e^{-z}(bx+cy)$$ where
$$x=f(t)$$
$$y=g(t)$$ ?
Because I am of the opinion that using anything other than the first one will lead you up the garden path when discussing the partial derivatives.
Because if you set:
u = sin(bx+c)
y = au[sup]2[/sup]
You arrive at a different answer for dy/du than you do if you set:
u = sin[sup]2[/sup](bx+c)
y = au
I was going to take it further and consider the general form of your example:
f(x) = kx + asin[sup]2[/sup](bx+c)
But it's been one and one half gross of moons since I did anything serious with differential calculas, it's not something I use on a daily basis (my job lends itself more to geospatial applications of statistics), and it's late in my day (and I've had a painful day writing science up for politicians, accountants and lawyers).
Incidentally, this is not true. So what?
Finally?
I said exactly the same thing 60-odd posts ago:
"The bug doesn't move in z" implies:
z = i(t) = C
for which we've already agreed on the answer.
If there's no relationship between z and t, if the bug's z-position is indeterminate, then clearly the bug's temperature is indeterminate and so is dt/dT.
At least have the decency to offer a constructive, or specific criticism.
One more time, z IS NOT the position of the bug, z is the distance of the observer from the plane containing the bug.
Sigh. Can you make an effort and calculate the total derivative of the function?
Calculate the total derivative for each function, if you do it correctly, you should arrive to the same result: $$ab sin(2(bx+c))$$.
Last edited:
You are right, my bad. So what? There is no effect on the calculation of the total derivative $$\frac{dy}{dx}$$.
What observer?
You seem to be discussing a scenario that is not a simple extension of that in the video. Can you please clearly explain the scenario you have in mind?
The bug's temperature is determined by the temperature at its position, not the observer's position, unless you've made some really radical change.
Sigh?
$$\begin{align}
T &= at + e^{-z}(bx+cy) \\
x &= f(t) \\
y &= g(t)
\end{align}$$
a, b, and c are constants.T &= at + e^{-z}(bx+cy) \\
x &= f(t) \\
y &= g(t)
\end{align}$$
z is an independent variable.
You want dT/dt?
Like David Metzler said in the video (4:25), "that doesn't make any sense. This isn't a one-variable function. You must mean a partial derivative."
The bug lives in the z=0 plane. Its temperature is :tex]bx+cy[/tex] where x,t are functions of t. Then, the temperature measured by an observer at altitude z is $$T=at+e^{-z}(bx+cy)$$ where $$x=f(t)$$ , $$y=g(t)$$. Calculate $$\frac{dT}{dt}$$.
Sigh. Calculate the partial derivative wrt t then.
You're changing the scenario. Back in post 50, you only added a z parameter to the temperature function.
In the video, x and y are the bug's coordinates, given by $$x=g(t)$$ and $$y=h(t)$$.
The bug's temperature matches the ambient temperature at that position and time, ie the bug's temperature is given by T = f(x,y,t).
But you've defined a new function for the bug's temperature (bx + cy), apparently unrelated to T.
Why?
Are you saying that x=f(t), y=g(t) has something to do with the observer? The observer's coordinates as they move?
In the video, dT/dt describes the rate of change of the bug's temperature wrt time.
In your scenario, I can't tell what dT/dt means, if anything. Is it meant to be the change in temperature measured by the observer wrt time?
Using the video meaning for x=g(t), y=h(t), and dT/dt, and given that "the bug lives in the z=0 plane":
$$\begin{align}
x &= g(t) \\
y &= h(t) \\
z &= 0 \\
T &= f(x, y, t) \\
&= at+e^{-z}(bx+cy) \\
\frac{dT}{dt} &= \frac{\partial T}{\partial x} \, \frac{dx}{dt} + \frac{\partial T}{\partial y} \, \frac{dy}{dt} + \frac{\partial T}{\partial z} \, \frac{dz}{dt}+ \frac{\partial T}{\partial t} \\
&= a + e^{-z}(b\frac{dx}{dt} + c\frac{dy}{dt}) \\
&= a + (b\frac{dx}{dt} + c\frac{dy}{dt})
\end{align}$$
x &= g(t) \\
y &= h(t) \\
z &= 0 \\
T &= f(x, y, t) \\
&= at+e^{-z}(bx+cy) \\
\frac{dT}{dt} &= \frac{\partial T}{\partial x} \, \frac{dx}{dt} + \frac{\partial T}{\partial y} \, \frac{dy}{dt} + \frac{\partial T}{\partial z} \, \frac{dz}{dt}+ \frac{\partial T}{\partial t} \\
&= a + e^{-z}(b\frac{dx}{dt} + c\frac{dy}{dt}) \\
&= a + (b\frac{dx}{dt} + c\frac{dy}{dt})
\end{align}$$
There's potentially some ambiguity here, so I'll present two interpretations. I'm led to believe the first is the usual interpretation, but in the context of this thread I think it's best to be explicit.
Interpretation one:
$$\begin{align}
x &= f(t) \\
y &= g(t) \\
T &= at + e^{-z}(bx+cy) \\
\frac{\partial T}{\partial t} &= \left.\frac{\partial T}{\partial t}\right|_{x,y,z} \\
&= a
\end{align}$$
x &= f(t) \\
y &= g(t) \\
T &= at + e^{-z}(bx+cy) \\
\frac{\partial T}{\partial t} &= \left.\frac{\partial T}{\partial t}\right|_{x,y,z} \\
&= a
\end{align}$$
Interpretation two:
$$\begin{align}
x &= f(t) \\
y &= g(t) \\
T &= at + e^{-z}(bf(t)+cg(t)) \\
\frac{\partial T}{\partial t} &= \left.\frac{\partial T}{\partial t}\right|_z \\
&= a + e^{-z}(b\frac{dx}{dt} + c\frac{dy}{dt})
\end{align}$$
x &= f(t) \\
y &= g(t) \\
T &= at + e^{-z}(bf(t)+cg(t)) \\
\frac{\partial T}{\partial t} &= \left.\frac{\partial T}{\partial t}\right|_z \\
&= a + e^{-z}(b\frac{dx}{dt} + c\frac{dy}{dt})
\end{align}$$
Last edited:
$$T(a,b,c,x,y,z,t) = a t + (bx + cy) e^{-z}$$ is one function, defined over seven variables.
$$T(a,b,c,z,t) = a t + (b f(t) + c g(t) ) e^{-z}$$ is another, defined over five variables.
$$T_{\tiny a,b,c,z}(t) = a t + (b f(t) + c g(t) ) e^{-z}$$ is another, a function of a single variable.
The partial derivative of the function T that most people see this discussion about is all of these:
$$\frac{d \quad}{dt} T_{\tiny a,b,c,z}(t) = \frac{\partial \quad}{\partial t} T(a,b,c,z,t) = \left. \frac{\partial \quad}{\partial t} T(a,b,c,x,y,z,t) \right| _{ x = f(t), \; y = g(t) } = \frac{\partial \quad}{\partial t} T(a,b,c,f(t),g(t),z,t) = \frac{\partial \quad}{\partial t} \left( a t + (b f(t) + c g(t) ) e^{-z} \right) = a + b e^{-z} f'(t) + c e^{-z} g'(t)$$
The total derivative is what you get when you assume that all the non-constant parameters are potentially functions of t.
$$\frac{d \quad}{dt} T(a,b,c,z,t) \\ = \frac{\partial \quad}{\partial a} T(a,b,c,z,t) \frac{d a}{d t} + \frac{\partial \quad}{\partial b} T(a,b,c,z,t) \frac{d b}{d t} + \frac{\partial \quad}{\partial c} T(a,b,c,z,t) \frac{d c}{d t} + \frac{\partial \quad}{\partial z} T(a,b,c,z,t) \frac{d z}{d t} + \frac{\partial \quad}{\partial t} T(a,b,c,z,t) \\ = t a'(t) + f(t) e^{-z} b'(t) + g(t) e^{-z} c'(t) - ( b f(t) + c g(t) ) e^{-z} z'(t) + a + b e^{-z} f'(t) + c e^{-z} g'(t)$$
But this is a ridiculous way to go about teaching it. Here is a better way:
$$ f(x,y) = xy \\ \frac{\partial \quad}{\partial x} f(x,y) = y \\ \frac{\partial \quad}{\partial y} f(x,y) = x \\ \frac{d \quad}{d x} f(x,y) = \frac{\partial \quad}{\partial x} f(x,y) + \frac{\partial \quad}{\partial y} f(x,y) \frac{d y}{d x} = y + x \frac{d y}{d x} $$
If we learn that $$y(x) = x$$, then we can talk about the partial and total derivatives of the new function of a single variable: $$g(x) = f(x,y(x)) = x^2$$
and we get $$\frac{d \quad}{d x} g(x) = \frac{\partial \quad}{\partial x} g(x) = \frac{\partial \quad}{\partial x} x^2 = 2x$$.
Another, equally valid route is to use $$y = x, \; \frac{d y}{d x} = 1$$ and armed with the total derivative, we get directly $$ \left. \frac{\partial \quad}{\partial x} f(x,y) \right| _{y=x} = \left. y + x \frac{d y}{d x} \right| _{y=x} = x + x = 2x$$
So if you are going to wax pedantic, you need to be extremely formal about what the function is and which parameters are variables or not.
$$T(a,b,c,z,t) = a t + (b f(t) + c g(t) ) e^{-z}$$ is another, defined over five variables.
$$T_{\tiny a,b,c,z}(t) = a t + (b f(t) + c g(t) ) e^{-z}$$ is another, a function of a single variable.
The partial derivative of the function T that most people see this discussion about is all of these:
$$\frac{d \quad}{dt} T_{\tiny a,b,c,z}(t) = \frac{\partial \quad}{\partial t} T(a,b,c,z,t) = \left. \frac{\partial \quad}{\partial t} T(a,b,c,x,y,z,t) \right| _{ x = f(t), \; y = g(t) } = \frac{\partial \quad}{\partial t} T(a,b,c,f(t),g(t),z,t) = \frac{\partial \quad}{\partial t} \left( a t + (b f(t) + c g(t) ) e^{-z} \right) = a + b e^{-z} f'(t) + c e^{-z} g'(t)$$
The total derivative is what you get when you assume that all the non-constant parameters are potentially functions of t.
$$\frac{d \quad}{dt} T(a,b,c,z,t) \\ = \frac{\partial \quad}{\partial a} T(a,b,c,z,t) \frac{d a}{d t} + \frac{\partial \quad}{\partial b} T(a,b,c,z,t) \frac{d b}{d t} + \frac{\partial \quad}{\partial c} T(a,b,c,z,t) \frac{d c}{d t} + \frac{\partial \quad}{\partial z} T(a,b,c,z,t) \frac{d z}{d t} + \frac{\partial \quad}{\partial t} T(a,b,c,z,t) \\ = t a'(t) + f(t) e^{-z} b'(t) + g(t) e^{-z} c'(t) - ( b f(t) + c g(t) ) e^{-z} z'(t) + a + b e^{-z} f'(t) + c e^{-z} g'(t)$$
But this is a ridiculous way to go about teaching it. Here is a better way:
$$ f(x,y) = xy \\ \frac{\partial \quad}{\partial x} f(x,y) = y \\ \frac{\partial \quad}{\partial y} f(x,y) = x \\ \frac{d \quad}{d x} f(x,y) = \frac{\partial \quad}{\partial x} f(x,y) + \frac{\partial \quad}{\partial y} f(x,y) \frac{d y}{d x} = y + x \frac{d y}{d x} $$
If we learn that $$y(x) = x$$, then we can talk about the partial and total derivatives of the new function of a single variable: $$g(x) = f(x,y(x)) = x^2$$
and we get $$\frac{d \quad}{d x} g(x) = \frac{\partial \quad}{\partial x} g(x) = \frac{\partial \quad}{\partial x} x^2 = 2x$$.
Another, equally valid route is to use $$y = x, \; \frac{d y}{d x} = 1$$ and armed with the total derivative, we get directly $$ \left. \frac{\partial \quad}{\partial x} f(x,y) \right| _{y=x} = \left. y + x \frac{d y}{d x} \right| _{y=x} = x + x = 2x$$
So if you are going to wax pedantic, you need to be extremely formal about what the function is and which parameters are variables or not.
Last edited:
- Status