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?It's to do with your claim that "The temperature of the bug does not depend on z."
*sigh*
You're only delaying the inevitable, you know. $$t = 5$$, $$a = b = c = 1$$ and $$f(t) = g(t) = \sin(\pi t)^{2}$$. What's the bug's temperature?
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]: $$\lambda$$.You see, you still fail to understand that $$\lambda \frac {\partial} {\partial x} $$ is not a vector because, contrary to your beliefs it doesn't have either sense, nor direction.
So for the bug, z(t) = 0 then.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.
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.But you set x=3 , y=2 in your earlier fumble, eh?
Why is this so difficult for you to admit to error?
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.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.?
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?
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.
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.Now that I have answered your questions, what are you attempting to accomplish with your post? Where are you going with it?
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
Finally?Finally!Pete said:If it's given that the bug doesn't move in z, then clearly:
i(t) = C
dz/dt = 0
z is constant over the bug's path. z=C. dz/dt = 0
The bug doesn't move in z, can you try the same with $$T=at+e^{-z}(bx+cy)$$ where
$$x=f(t)$$
$$y=g(t)$$ ?
Incidentally, this is not true. So what?
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.
Calculate the total derivative for each function, if you do it correctly, you should arrive to the same result: $$ab sin(2(bx+c))$$. This shows your claim to be wrong.
I didn't address dy/dx, did I.
What observer?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?Sigh. Can you make an effort and calculate the total derivative of the function?
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?
Sigh?
$$\begin{align}a, b, and c are constants.
T &= at + e^{-z}(bx+cy) \\
x &= f(t) \\
y &= g(t)
\end{align}$$
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."
You're changing the scenario. Back in post 50, you only added a z parameter to the temperature function.The bug lives in the z=0 plane. Its temperature is $$bx+cy$$ where x,t are functions of t.
Are you saying that x=f(t), y=g(t) has something to do with the observer? The observer's coordinates as they move?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 $$\frac{dT}{dt}$$.
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.Sigh. Calculate the partial derivative wrt t then.