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## Analiza matematyczna funkcji wielu zmiennych

### 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:
$\vec{\textbf{v}}(s, t) = \left[ \begin{array}{c} x(s, t) \\ y(s, t) \\ z(s, t) \\ \end{array} \right]$
Its partial derivatives are computed by taking the partial derivative of each component:
\begin{aligned} \quad \dfrac{\partial \vec{\textbf{v}}}{\blueE{\partial t}}(s, t) &= \left[ \begin{array}{c} \dfrac{\partial x}{\blueE{\partial t}}(s, t) \\ \dfrac{\partial y}{\blueE{\partial t}}(s, t) \\ \dfrac{\partial z}{\blueE{\partial t}}(s, t) \\ \end{array} \right] \end{aligned}
\begin{aligned} \quad \dfrac{\partial \vec{\textbf{v}}}{\redE{\partial s}}(s, t) &= \left[ \begin{array}{c} \dfrac{\partial x}{\redE{\partial s}}(s, t) \\ \dfrac{\partial y}{\redE{\partial s}}(s, t) \\ \dfrac{\partial z}{\redE{\partial s}}(s, t) \\ \end{array} \right] \end{aligned}
• You can interpret these partial derivatives as giving vectors tangent to the parametric surface defined by start bold text, v, end bold text, with, vector, on top.

## The goal

Suppose I were to give you a function with a two-dimensional input, and a three-dimensional output, like this one:
$\vec{\textbf{v}}(t, s) = \left[ \begin{array}{c} 3\cos(t) + \cos(t)\cos(s) \\ 3\sin(t) + \sin(t)\cos(s) \\ \sin(s) \end{array} \right]$
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 parenthesis, t, comma, s, right parenthesis such that 0, is less than or equal to, t, is less than or equal to, 2, pi and 0, is less than or equal to, s, is less than or equal to, 2, pi. This can be seen as a square in the "t, s-plane". I'll draw this with a checkerboard pattern since it makes things easier to follow later on.
Two parameter input space
For any given point left parenthesis, t, comma, s, right parenthesis, the value start bold text, v, end bold text, with, vector, on top, left parenthesis, t, comma, s, right parenthesis is some point in three-dimensional space.
Concept check: Evaluate start bold text, v, end bold text, with, vector, on top, left parenthesis, pi, comma, pi, right parenthesis. In other words, where does the function start bold text, v, end bold text, with, vector, on top take the input left parenthesis, t, comma, s, right parenthesis, equals, left parenthesis, pi, comma, pi, right parenthesis?
Wybierz 1 odpowiedź:

If you imagine doing this computation for all inputs left parenthesis, t, comma, s, right parenthesis 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 $\blueE{t}$start color #0c7f99, t, end color #0c7f99

To compute a partial derivative of this function, say start fraction, \partial, start bold text, v, end bold text, with, vector, on top, divided by, \partial, t, end fraction, you take the partial derivative of each individual component.
\begin{aligned} \quad \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}}(\blueE{t}, s) &= \dfrac{\partial}{\partial \blueE{t}} \left[ \begin{array}{c} 3\cos(\blueE{t}) + \cos(\blueE{t})\cos(s) \\ \\ 3\sin(\blueE{t}) + \sin(\blueE{t})\cos(s) \\ \\ \sin(s) \end{array} \right] \\ \\ &= \left[ \begin{array}{c} \dfrac{\partial}{\partial \blueE{t}}(3\cos(\blueE{t}) + \cos(\blueE{t})\cos(s)) \\ \\ \dfrac{\partial}{\partial \blueE{t}}(3\sin(\blueE{t}) + \sin(\blueE{t})\cos(s)) \\ \\ \dfrac{\partial}{\partial \blueE{t}}(\sin(s)) \end{array} \right] \\ \\ &= \left[ \begin{array}{c} -3\sin(t) - \sin(t)\cos(s) \\ \\ 3\cos(t) + \cos(t)\cos(s) \\ \\ 0 \end{array} \right] \\ \end{aligned}
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 t, s-plane, a constant value of s corresponds with a horizontal line. Here's one such line representing s, equals, pi, slash, 2, drawn in red:
Hold s constant in the input space.
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:
Hold s constant in the output space.
The partial derivative start fraction, \partial, start bold text, v, end bold text, with, vector, on top, divided by, \partial, t, end fraction 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:
Nudge t in input space
Nudge t in output space
Specifically, the input point used for the pictures above is left parenthesis, t, start subscript, 0, end subscript, comma, s, start subscript, 0, end subscript, right parenthesis, equals, left parenthesis, start fraction, pi, divided by, 4, end fraction, comma, start fraction, pi, divided by, 2, end fraction, right parenthesis. This means the point on the torus is
\begin{aligned} \quad \vec{\textbf{v}}\left(\dfrac{\pi}{4}, \dfrac{\pi}{2} \right) &= \left[ \begin{array}{c} 3\cos(\pi/4) + \cos(\pi/4)\cos(\pi/2) \\ 3\sin(\pi/4) + \sin(\pi/4)\cos(\pi/2) \\ \sin(\pi/2) \end{array} \right] \\ \\ &= \left[ \begin{array}{c} 3\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(0) \\ 3\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(0) \\ 1 \end{array} \right] \\ \\ &= \left[ \begin{array}{c} \dfrac{3\sqrt{2}}{2} \\ \\ \dfrac{3\sqrt{2}}{2} \\ \\ 1 \end{array} \right] \\ \end{aligned}
And the tangent vector is
\begin{aligned} \quad \dfrac{\partial \vec{\textbf{v}}}{\partial t} \left(\dfrac{\pi}{4}, \dfrac{\pi}{2} \right) &= \left[ \begin{array}{c} -3\sin(\pi/4) - \sin(\pi/4)\cos(\pi/2) \\ 3\cos(\pi/4) + \cos(\pi/4)\cos(\pi/2) \\ 0 \end{array} \right] \\ \\ &= \left[ \begin{array}{c} -3\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}(0) \\ 3\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(0) \\ 0 \end{array} \right] \\ \\ &= \left[ \begin{array}{c} -\dfrac{3\sqrt{2}}{2} \\ \dfrac{3\sqrt{2}}{2} \\ 0 \end{array} \right] \\ \end{aligned}
Concept check: Why does it make sense that the z-component of this tangent vector is 0?
Wybierz 1 odpowiedź:

#### Differentiate with respect to $\redE{s}$start color #bc2612, s, end color #bc2612

The partial derivative with respect to s is similar. You compute it by taking the partial derivative of each component in the definition of start bold text, v, end bold text, with, vector, on top:
\begin{aligned} \quad \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}}(t, \redE{s}) = \dfrac{\partial}{\partial \redE{s}} &\left[ \begin{array}{c} 3\cos(t) + \cos(t)\cos(\redE{s}) \\ 3\sin(t) + \sin(t)\cos(\redE{s}) \\ \sin(\redE{s}) \end{array} \right] \\ = &\left[ \begin{array}{c} -\cos(t)\sin(s) \\ -\sin(t)\sin(s) \\ \cos(s) \end{array} \right] \\ \end{aligned}
This time, we can imagine holding t constant to get some vertical line in the parameter space.
Nudge s in the input 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 start bold text, v, end bold text, with, vector, on top, the red line and the yellow velocity vector might look something like this:
Nudge s in the output space
The partial derivative start fraction, \partial, start bold text, v, end bold text, with, vector, on top, divided by, \partial, s, end fraction 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:
$\vec{\textbf{v}}(s, t) = \left[ \begin{array}{c} x(s, t) \\ y(s, t) \\ z(s, t) \\ \end{array} \right]$
Its partial derivatives are computed by taking the partial derivative of each component:
\begin{aligned} \quad \dfrac{\partial \vec{\textbf{v}}}{\blueE{\partial t}}(s, t) &= \left[ \begin{array}{c} \dfrac{\partial x}{\blueE{\partial t}}(s, t) \\ \dfrac{\partial y}{\blueE{\partial t}}(s, t) \\ \dfrac{\partial z}{\blueE{\partial t}}(s, t) \\ \end{array} \right] \end{aligned}
\begin{aligned} \quad \dfrac{\partial \vec{\textbf{v}}}{\redE{\partial s}}(s, t) &= \left[ \begin{array}{c} \dfrac{\partial x}{\redE{\partial s}}(s, t) \\ \dfrac{\partial y}{\redE{\partial s}}(s, t) \\ \dfrac{\partial z}{\redE{\partial s}}(s, t) \\ \end{array} \right] \end{aligned}
• You can interpret these partial derivatives as giving vectors tangent to the parametric surface defined by start bold text, v, end bold text, with, vector, on top.
• For example, imagine nudging a point in the input space along the start color #0c7f99, t, end color #0c7f99 direction, say from the coordinates left parenthesis, s, comma, t, right parenthesis to the coordinates left parenthesis, s, comma, t, plus, start color #bc2612, h, end color #bc2612, right parenthesis for some small start color #bc2612, h, end color #bc2612. This results in some small nudge in the output along the surface, which is represented by the vector start color #bc2612, h, end color #bc2612, start fraction, \partial, start bold text, v, end bold text, with, vector, on top, divided by, \partial, t, end fraction, left parenthesis, s, comma, t, right parenthesis.

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