High School Basic doubts in vector and multi variable calculus

[Hamiltonian]

TL;DR

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If say we have a scalar function $T(x,y,z)$ (say the temperature in a room). then the rate at which T changes in a particular direction is given by the above equation)
say You move in the $Y$direction then $T$ does not change in the $x$ and $z$ directions hence $dT = \frac{\partial T}{\partial y}dy$
I don't understand why there is a $dy$ in this equation.
basically, I don't understand how the above equation gives the change of the function $T$ in any particular direction. My books says a theorem on partial derivatives states the above equation but I am unable to find any such theorem(maybe because it is obvious? but if it is I don't see why).

 

[etotheipi]

This is essentially the definition of the total derivative. I'll give a Physicsy explanation... you can think of it like, $\frac{\partial T}{\partial x} \Delta x$ is the approximate change in $T$ when moving a small distance $\Delta x$ in the $x$ direction, and the same for $y$ and $z$. If that's not clear, then think back to this statement of Taylor's theorem in one-dimension, $$f(x + \Delta x) \approx f(x) + \Delta x f'(x) \implies \Delta f(x) \approx f'(x) \Delta x$$The total change in $T$ along a small displacement $\Delta \vec{r} = \Delta x\hat{x} + \Delta y\hat{y} + \Delta z\hat{z}$ just going to be the sum of the changes due to moving in all three directions, i.e. approximately $\Delta T = \frac{\partial T}{\partial x} \Delta x + \frac{\partial T}{\partial y} \Delta y + \frac{\partial T}{\partial z} \Delta z$. The exact statement is$$dT = \frac{\partial T}{\partial x} dx + \frac{\partial T}{\partial y} dy + \frac{\partial T}{\partial z} dz = \nabla T \cdot d\vec{r}$$

 

[PeroK]

Summary:: -

View attachment 271587
If say we have a scalar function $T(x,y,z)$ (say the temperature in a room). then the rate at which T changes in a particular direction is given by the above equation)
say You move in the $Y$direction then $T$ does not change in the $x$ and $z$ directions hence $dT = \frac{\partial T}{\partial y}dy$
I don't understand why there is a $dy$ in this equation.
basically, I don't understand how the above equation gives the change of the function $T$ in any particular direction. My books says a theorem on partial derivatives states the above equation but I am unable to find any such theorem(maybe because it is obvious? but if it is I don't see why).

You could take an example function $T$ and a small change in $x, y, z$ and compute the change in $T$ and check the above equation is a good approximation for small $dx, dy, dz$.

PS to keep things simple, do it for a function of two variables.

 

[Hamiltonian]

You could take an example function $T$ and a small change in $x, y, z$ and compute the change in $T$ and check the above equation is a good approximation for small $dx, dy, dz$.

PS to keep things simple, do it for a function of two variables.

I am a bit confused about how exactly I can prove the above equation using this method.

say I try to prove it for $T(x,y) = x + y$ are you suggesting I take numerical values and then prove the above equation?

 

[PeroK]

say $$T(x, y) = x + y$$
$$\Delta T = \Delta x\hat x + \Delta y\hat y$$
how exactly should I proceed from here to prove the above equation?

First you have to understand what the equation in your OP is saying:

If we have $T$ at some point $T(x_0, y_0)$ and we look at $T(x_0 + dx, y_0 + dy)$, where we will take $dx, dy$ to be small and finite. Then:
$$dT \equiv T(x_0 + dx, y_0 + dy) -T(x_0, y_0) \approx \frac{\partial T(x_0, y_0)}{\partial x} dx + \frac{\partial T(x_0, y_0)}{\partial y} dy$$ where that means the partial derivatives are evaluated at $(x_0, y_0)$.

And, when $dx, dy$ are differentials, we have equality, rather than a finite approximation.

 

[Hamiltonian]

$$T(x_0 + dx, y_0 + dy) -T(x_0, y_0) \approx \frac{\partial T(x_0, y_0)}{\partial x} dx + \frac{\partial T(x_0, y_0)}{\partial y} dy$$
how exactly are u getting this?

 

[PeroK]

$$T(x_0 + dx, y_0 + dy) -T(x_0, y_0) \approx \frac{\partial T(x_0, y_0)}{\partial x} dx + \frac{\partial T(x_0, y_0)}{\partial y} dy$$
how exactly are u getting this?

I'm suggesting you test it out as a numerical approximation. And get a feel for why it works.

 

[Hamiltonian]

I'm suggesting you test it out as a numerical approximation. And get a feel for why it works.

ok,
so if $$T(x,y) = x + y$$
$x_0 = 1$ $y_0 = 1$
$$T(1+dx, 1+dy) - T(1, 1) = 1 + dx + 1 + dy -1 -1 = dx + dy$$
$$\frac {\partial T}{\partial x}dx + \frac{\partial T}{\partial y} dy = dx + dy$$

ok so the above equation is true but how exactly did you derive it?

 

[PeroK]

ok,
so if $$T(x,y) = x + y$$
$x_0 = 1$ $y_0 = 1$
$$T(1+dx, 1+dy) - T(1, 1) = 1 + dx + 1 + dy -1 -1 = dx + dy$$
$$\frac {\partial T}{\partial x}dx + \frac{\partial T}{\partial y} dy = dx + dy$$

ok so the above equation is true but how exactly did you derive it?

It's not much more than using the concept of a partial derivative as a rate of change. It should be geometrically intuitive.

 

[PeroK]

For example, the single variable case falls out of the definition of the derivative:
$$f'(x_0) = \lim_{\Delta x \rightarrow 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}$$ Which means that for small $\Delta x$ we have $$\Delta f \equiv f(x_0 + \Delta x) - f(x_0) \approx f'(x_0)\Delta x$$ Hence the definition of the single-variable differential: $$df = f'(x)dx$$