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I somehow grasp that the one graph ($f(x,y)$) is a surface and the other ($f(x(t),y(t))$) is a curve but not the manner in which they are plotted. As far as I understand, for the surface one I do calculate $z = f(x,y)$ and obtain a hight for every point $(x,y)$ that forms my surface. However, isn't the curvy one actually build the same way and thus should show up as a surface?

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You didn't talked about the domain of the functions. Lets say, for the sake of simplicity, that $f:[a,b]\times [c,d] \to \mathbb R$. Then the graph of $f(x,y)$ is the set $\{(x,y,f(x,y)) \in \mathbb R^3\mid x\in [a,b], y \in [c,d]\}$. If the function is continuous and positive, its graph would be a surface above the rectangle $[a,b]\times [c,d]$

If you choose functions $x:[0,1] \to [a,b]$, $y:[0,1] \to [c,d]$ and again $x$ and $y$ are continuous functions, then $(x(t), y(t))$ with $t \in [0,1]$ describes a curve $\Gamma$ inside the rectangle $[a,b]\times [c,d]$. So the graph of $f(x(t),y(t))$ is the set $\{(x(t),y(t),f(x(t),y(t))) \in \mathbb R^3\mid t \in [0,1]\}$, those points are on the surface, but they are only the ones that are above $\Gamma$.

Graph

In the image, the rectangle (the domain of $f$) is yellow, $\Gamma$ is black, the graph of $f(x,y)$ is green, and the graph of $f(x(t),y(t))$ is red.

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  • $\begingroup$ Right, I was confused myself because I didn't consider the domain. $\endgroup$ – Leon Apr 19 at 17:57
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The two expressions mean the same thing. The latter just specifies what the variables $x,y$ depend on. In this case time. You can think of as $x$ and $y$ vary in time, they draw out the surface $f(x,y)=f(x(t),y(t))$ as $t$ varies. I hope this helps with your understanding :).

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