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### Kurs: Analiza matematyczna funkcji wielu zmiennych>Rozdział 2

Lekcja 7: Pochodne cząstkowe funkcji wektorowych (artykuły)

# Partial derivatives of parametric surfaces

If you have a function representing a surface in three dimensions, you can take its partial derivative.  Here we see what that looks like, and how to interpret it.

## Do czego zmierzamy

• As setup, we have some vector-valued function with a two-dimensional input and a three-dimensional output:
$\stackrel{\to }{\mathbf{\text{v}}}\left(s,t\right)=\left[\begin{array}{c}x\left(s,t\right)\\ y\left(s,t\right)\\ z\left(s,t\right)\end{array}\right]$
Its partial derivatives are computed by taking the partial derivative of each component:
$\begin{array}{rl}\phantom{\rule{1em}{0ex}}\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial t}\left(s,t\right)& =\left[\begin{array}{c}\frac{\partial x}{\partial t}\left(s,t\right)\\ \frac{\partial y}{\partial t}\left(s,t\right)\\ \frac{\partial z}{\partial t}\left(s,t\right)\end{array}\right]\end{array}$
$\begin{array}{rl}\phantom{\rule{1em}{0ex}}\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial s}\left(s,t\right)& =\left[\begin{array}{c}\frac{\partial x}{\partial s}\left(s,t\right)\\ \frac{\partial y}{\partial s}\left(s,t\right)\\ \frac{\partial z}{\partial s}\left(s,t\right)\end{array}\right]\end{array}$
• You can interpret these partial derivatives as giving vectors tangent to the parametric surface defined by $\stackrel{\to }{\mathbf{\text{v}}}$.

## The goal

Suppose I were to give you a function with a two-dimensional input, and a three-dimensional output, like this one:
$\stackrel{\to }{\mathbf{\text{v}}}\left(t,s\right)=\left[\begin{array}{c}3\mathrm{cos}\left(t\right)+\mathrm{cos}\left(t\right)\mathrm{cos}\left(s\right)\\ 3\mathrm{sin}\left(t\right)+\mathrm{sin}\left(t\right)\mathrm{cos}\left(s\right)\\ \mathrm{sin}\left(s\right)\end{array}\right]$
Since the input is multi-dimensional, you cannot take the ordinary derivative of this function, but you can take a partial derivative. The focus of this article is on getting an intuitive feel for what those partial derivatives mean.

## Interpret the function as a surface

The function itself actually has a very nice geometric meaning. Since it has a two-coordinate input and a three-coordinate output, we can visualize it as a parametric surface.
Specifically, consider all inputs $\left(t,s\right)$ such that $0\le t\le 2\pi$ and $0\le s\le 2\pi$. This can be seen as a square in the "$ts$-plane". I'll draw this with a checkerboard pattern since it makes things easier to follow later on.
For any given point $\left(t,s\right)$, the value $\stackrel{\to }{\mathbf{\text{v}}}\left(t,s\right)$ is some point in three-dimensional space.
Concept check: Evaluate $\stackrel{\to }{\mathbf{\text{v}}}\left(\pi ,\pi \right)$. In other words, where does the function $\stackrel{\to }{\mathbf{\text{v}}}$ take the input $\left(t,s\right)=\left(\pi ,\pi \right)$?
Wybierz 1 odpowiedź:

If you imagine doing this computation for all inputs $\left(t,s\right)$ in the square, getting some point in three-dimensional space each time, all of the resulting outputs will form a two-dimensional surface in three-dimensional space. I like to imagine each point of the square moving to its appropriate location in space.
The result is a doughnut shape! Math folk call this a torus.

## Interpreting the partial derivatives

#### Differentiate with respect to $t$‍

To compute a partial derivative of this function, say $\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial t}$, you take the partial derivative of each individual component.
$\begin{array}{rl}\phantom{\rule{1em}{0ex}}\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial t}\left(t,s\right)& =\frac{\partial }{\partial t}\left[\begin{array}{c}3\mathrm{cos}\left(t\right)+\mathrm{cos}\left(t\right)\mathrm{cos}\left(s\right)\\ \\ 3\mathrm{sin}\left(t\right)+\mathrm{sin}\left(t\right)\mathrm{cos}\left(s\right)\\ \\ \mathrm{sin}\left(s\right)\end{array}\right]\\ \\ & =\left[\begin{array}{c}\frac{\partial }{\partial t}\left(3\mathrm{cos}\left(t\right)+\mathrm{cos}\left(t\right)\mathrm{cos}\left(s\right)\right)\\ \\ \frac{\partial }{\partial t}\left(3\mathrm{sin}\left(t\right)+\mathrm{sin}\left(t\right)\mathrm{cos}\left(s\right)\right)\\ \\ \frac{\partial }{\partial t}\left(\mathrm{sin}\left(s\right)\right)\end{array}\right]\\ \\ & =\left[\begin{array}{c}-3\mathrm{sin}\left(t\right)-\mathrm{sin}\left(t\right)\mathrm{cos}\left(s\right)\\ \\ 3\mathrm{cos}\left(t\right)+\mathrm{cos}\left(t\right)\mathrm{cos}\left(s\right)\\ \\ 0\end{array}\right]\end{array}$
So...what does this new vector-valued function actually mean?
Well, computing this partial derivative requires treating the variable $s$ as if it was constant. What does this mean geometrically?
In the $ts$-plane, a constant value of $s$ corresponds with a horizontal line. Here's one such line representing $s=\pi /2$, drawn in red:
After this square gets warped and morphed into the torus, this red line gets turned in​to some circle which goes the long way around the torus:
The partial derivative $\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial t}$ tells us how the output changes slightly when we nudge the input in the $t$-direction. In this case, the vector representing that nudge (drawn in yellow below) gets transformed into a vector tangent to the red circle which represents a constant value of $s$ on the surface:
Specifically, the input point used for the pictures above is $\left({t}_{0},{s}_{0}\right)=\left(\frac{\pi }{4},\frac{\pi }{2}\right)$. This means the point on the torus is
$\begin{array}{rl}\phantom{\rule{1em}{0ex}}\stackrel{\to }{\mathbf{\text{v}}}\left(\frac{\pi }{4},\frac{\pi }{2}\right)& =\left[\begin{array}{c}3\mathrm{cos}\left(\pi /4\right)+\mathrm{cos}\left(\pi /4\right)\mathrm{cos}\left(\pi /2\right)\\ 3\mathrm{sin}\left(\pi /4\right)+\mathrm{sin}\left(\pi /4\right)\mathrm{cos}\left(\pi /2\right)\\ \mathrm{sin}\left(\pi /2\right)\end{array}\right]\\ \\ & =\left[\begin{array}{c}3\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\left(0\right)\\ 3\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\left(0\right)\\ 1\end{array}\right]\\ \\ & =\left[\begin{array}{c}\frac{3\sqrt{2}}{2}\\ \\ \frac{3\sqrt{2}}{2}\\ \\ 1\end{array}\right]\end{array}$
And the tangent vector is
$\begin{array}{rl}\phantom{\rule{1em}{0ex}}\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial t}\left(\frac{\pi }{4},\frac{\pi }{2}\right)& =\left[\begin{array}{c}-3\mathrm{sin}\left(\pi /4\right)-\mathrm{sin}\left(\pi /4\right)\mathrm{cos}\left(\pi /2\right)\\ 3\mathrm{cos}\left(\pi /4\right)+\mathrm{cos}\left(\pi /4\right)\mathrm{cos}\left(\pi /2\right)\\ 0\end{array}\right]\\ \\ & =\left[\begin{array}{c}-3\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\left(0\right)\\ 3\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\left(0\right)\\ 0\end{array}\right]\\ \\ & =\left[\begin{array}{c}-\frac{3\sqrt{2}}{2}\\ \frac{3\sqrt{2}}{2}\\ 0\end{array}\right]\end{array}$
Concept check: Why does it make sense that the $z$-component of this tangent vector is $0$?
Wybierz 1 odpowiedź:

#### Differentiate with respect to $s$‍

The partial derivative with respect to $s$ is similar. You compute it by taking the partial derivative of each component in the definition of $\stackrel{\to }{\mathbf{\text{v}}}$:
$\begin{array}{rl}\phantom{\rule{1em}{0ex}}\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial s}\left(t,s\right)=\frac{\partial }{\partial s}& \left[\begin{array}{c}3\mathrm{cos}\left(t\right)+\mathrm{cos}\left(t\right)\mathrm{cos}\left(s\right)\\ 3\mathrm{sin}\left(t\right)+\mathrm{sin}\left(t\right)\mathrm{cos}\left(s\right)\\ \mathrm{sin}\left(s\right)\end{array}\right]\\ =& \left[\begin{array}{c}-\mathrm{cos}\left(t\right)\mathrm{sin}\left(s\right)\\ -\mathrm{sin}\left(t\right)\mathrm{sin}\left(s\right)\\ \mathrm{cos}\left(s\right)\end{array}\right]\end{array}$
This time, we can imagine holding $t$ constant to get some vertical line in the parameter space.
The yellow arrow represents some velocity vector as a particle travels up along this line. Which is to say, as you vary $s$ while holding $t$ constant. After the square turns into the torus via the function $\stackrel{\to }{\mathbf{\text{v}}}$, the red line and the yellow velocity vector might look something like this:
The partial derivative $\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial s}$ can be interpreted as this resulting velocity vector on the torus.

## Podsumowanie

• As setup, we have some vector-valued function with a two-dimensional input and a three-dimensional output:
$\stackrel{\to }{\mathbf{\text{v}}}\left(s,t\right)=\left[\begin{array}{c}x\left(s,t\right)\\ y\left(s,t\right)\\ z\left(s,t\right)\end{array}\right]$
Its partial derivatives are computed by taking the partial derivative of each component:
$\begin{array}{rl}\phantom{\rule{1em}{0ex}}\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial t}\left(s,t\right)& =\left[\begin{array}{c}\frac{\partial x}{\partial t}\left(s,t\right)\\ \frac{\partial y}{\partial t}\left(s,t\right)\\ \frac{\partial z}{\partial t}\left(s,t\right)\end{array}\right]\end{array}$
$\begin{array}{rl}\phantom{\rule{1em}{0ex}}\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial s}\left(s,t\right)& =\left[\begin{array}{c}\frac{\partial x}{\partial s}\left(s,t\right)\\ \frac{\partial y}{\partial s}\left(s,t\right)\\ \frac{\partial z}{\partial s}\left(s,t\right)\end{array}\right]\end{array}$
• You can interpret these partial derivatives as giving vectors tangent to the parametric surface defined by $\stackrel{\to }{\mathbf{\text{v}}}$.
• For example, imagine nudging a point in the input space along the $t$ direction, say from the coordinates $\left(s,t\right)$ to the coordinates $\left(s,t+h\right)$ for some small $h$. This results in some small nudge in the output along the surface, which is represented by the vector $h\frac{\partial \stackrel{\to }{\mathbf{\text{v}}}}{\partial t}\left(s,t\right)$.

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