calculus
The physical meaning of the differentiation of $x$ with respect to $y$ is the rate of change of $x$ with respect to $y$. But, I am finding it difficult in understanding the geometrical interpretation of the expression $\dfrac{\mathrm{d}x}{\mathrm{d}y}$. On a given curve on any given points, what does the expression actually tells us?
I currently know that differentiating the equation of a curve gives the tangent of that curve. But how can we understand that tangent line in terms of $y$ and $x$?
The gradient defines a direction; the magnitude of the gradient is the slope of your surface in that direction.
This direction just so happens to be the one in which you have to go to get the maximum slope.
Let's say you take the gradient of an N surface in N+1 space. For instance, the gradient of a 2D surface in 3D space. The gradient will point in the direction that you have to go in to get the biggest increase in "height" (that +1 dimension). So, in other words, if you go in the direction in which the gradient points, you'll see the largest increase.
The magnitude of the gradient is the rate at which that increase happens. Literally, it is the slope of the surface at that point along the axis defined by the gradient's direction. Consequently, the magnitude of the gradient of some point on a surface is the steepest slope you can find on that surface!
Gradients are actually defined to behave as described above, but, perhaps you are like me, and want a little bit of mathematical proof of this.
Let's start with the tangent plane to your surface at some point. If the above is true, then the magnitude of your gradient should equal the slope of the plane along the direction defined by your gradient! To start, let's quickly define the slope of a plane in a certain direction:
If we want to know the slope of a plane in a certain direction, we simply find the slope of said plane between the point (0, 0), and the point represented by our arbitrary direction. To simplify this procedure, we can shift our plane so that it passes through the point (0, 0, 0), since shifting a plane does not change its slope.
To recap, the slope of a plane in some 2D direction D is the same as the slope of a similar plane (which passes through (0,0,0))) between the points (0,0) and D.
The slope of a plane between 2D points (0,0) and D is given by:
$$ \frac {\frac{\partial z}{\partial x}D_x + \frac{\partial z}{\partial y}D_y}{||D||} $$
Or:
$$ \frac {\frac{\partial z}{\partial x}D_x + \frac{\partial z}{\partial y}D_y}{\sqrt{D_x^2 + D_y^2}} $$
Since this slope-in-a-direction is defined only in terms of the partial derivatives of our plane, and since the partial derivatives of any tangent plane to a surface are the same as those of the surface at the tangent point, we may make the claim that:
The slope of a surface at some point (x, y) in the direction D is given by the expression: $$ \frac {\frac{\partial f(x,y)}{\partial x}D_x + \frac{\partial f(x,y)}{\partial y}D_y}{\sqrt{D_x^2 + D_y^2}} $$
Meanwhile, the gradient of our surface at this point is given by the expression:
$$ {\frac{\partial f(x,y)}{\partial x} \hat{x} + \frac{\partial f(x,y)}{\partial y} \hat{y}} $$
which means that the slope of our surface at some point (x, y) in the direction of our gradient at that point is (as defined by our slope-in-direction argument earlier) given by:
$$ \frac {\frac{\partial f(x,y)}{\partial x} \cdot \frac{\partial f(x,y)}{\partial x} + \frac{\partial f(x,y)}{\partial y} \cdot \frac{\partial f(x,y)}{\partial y}}{\sqrt{\left(\frac{\partial f(x,y)}{\partial x}\right)^2 + \left(\frac{\partial f(x,y)}{\partial y} \right)^2}} $$
This, of course, simplifies to:
$$ \frac {\left(\frac{\partial f(x,y)}{\partial x}\right)^2 + \left(\frac{\partial f(x,y)}{\partial y} \right)^2}{\sqrt{\left(\frac{\partial f(x,y)}{\partial x}\right)^2 + \left(\frac{\partial f(x,y)}{\partial y} \right)^2}} $$
Now for the fun part! Let's define some temporary variable J as:
$$ J = \left(\frac{\partial f(x,y)}{\partial x}\right)^2 + \left(\frac{\partial f(x,y)}{\partial y} \right)^2 $$
Then our "simplified" expression for the slope of our curve at any point (x,y) along its gradient at (x,y) becomes:
$$ \frac {J}{\sqrt{J}} $$
Which is the same thing as simply
$$ \sqrt{J} $$
Which is the same thing as
$$ \sqrt{\left(\frac{\partial f(x,y)}{\partial x}\right)^2 + \left(\frac{\partial f(x,y)}{\partial y} \right)^2} $$
Which, but of course, by pythagorean theorem, is the same expression as the magnitude of our gradient at the point (x,y)!
And so, we have just proven that, yes, the magnitude of the gradient of a surface at some point is the same as the slope of that surface along said gradient.
I hope this helped!
Side note: Hans Lundmark's answer touches on this a bit; those curves made from the intersections of your volume in 4D space with planes at evenly spaced positions on the 4th axis (equivalent to the contour lines on a surface in 3D space) will, indeed, be closer together when the slope of your volume is steeper; this is because, as one might expect, we cross through more height-surfaces (and therefore more "vertical" distance) in an area which is more steep than in an area which is less steep. Hopefully the above slope explanation makes why that is (specifically, how it relates to the magnitude of the gradient) a bit clearer.
As you said the curve back and forth in $y$ direction. Let's assume the curve $C$ does not have that little twist in $x$ direction at the beginning. This means the curve $C$ is a function of $x$.
Then the line integral with respect to $x$ would be $$\int_{x_1}^{x_2} f(x,y(x))\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$$
where $\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$ is the infinitesimal arc length of each subinterval. So it is exactly as in the picture.
Now for the $y$ direction, since $C$ is not really a function in this direction, we have to separate it into parts. In this case, we have three parts $C_1,C_2,C_3$, as shown below
On each subcurve, we can do the same thing as before:
$$\int_{y_1}^{y_2} f(x(y),y)\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy$$
where $\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy$ is the arc length of an subinterval.
Best Answer
Differentiating $x$ with respect to $y$ gives you the gradient of the tangent at the point $\left(x, y(x)\right)$ on the curve $y(x)$. The gradient of the tangent at the point $\left(x, y(x)\right)$ indicates the rate of change of $y(x)$ at that specific point. Because on a curve, the gradient varies, differentiating is basically taking the limit of the $\dfrac{\Delta y}{\Delta x}$ as $\Delta x \to 0$ and hence finding the gradient or rate of change of a function at a specific point on a curve.