Główna zawartość

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.

Kontekst

Do czego zmierzamy

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

The goal

Suppose I were to give you a function with a two-dimensional input, and a three-dimensional output, like this one:

$\vec{v} (t, s) = [\begin{matrix} 3 \cos (t) + \cos (t) \cos (s) \\ 3 \sin (t) + \sin (t) \cos (s) \\ \sin (s) \end{matrix}]$ ‍

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

(t, s)

such that

0 \leq t \leq 2 π

and

0 \leq s \leq 2 π

. 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.

For any given point

(t, s)

, the value

\vec{v} (t, s)

is some point in three-dimensional space.

Concept check: Evaluate

\vec{v} (π, π)

. In other words, where does the function

\vec{v}

take the input

(t, s) = (π, π)

If you imagine doing this computation for all inputs

(t, s)

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.

Filmy wideo na Khan Academy

Zobacz transkrypcję filmu

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 \vec{v}}{\partial t}

, you take the partial derivative of each individual component.

\begin{aligned} \frac{\partial \vec{v}}{\partial t} (t, s) & = \frac{\partial}{\partial t} [\begin{array}{c} 3 \cos (t) + \cos (t) \cos (s) \\ 3 \sin (t) + \sin (t) \cos (s) \\ \sin (s) \end{array}] \\ = [\begin{array}{c} \frac{\partial}{\partial t} (3 \cos (t) + \cos (t) \cos (s)) \\ \frac{\partial}{\partial t} (3 \sin (t) + \sin (t) \cos (s)) \\ \frac{\partial}{\partial t} (\sin (s)) \end{array}] \\ = [\begin{array}{c} - 3 \sin (t) - \sin (t) \cos (s) \\ 3 \cos (t) + \cos (t) \cos (s) \\ 0 \end{array}] \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 = π / 2

, drawn in red:

After this square gets warped and morphed into the torus, this red line gets turned into some circle which goes the long way around the torus:

The partial derivative

\frac{\partial \vec{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

(t_{0}, s_{0}) = (\frac{π}{4}, \frac{π}{2})

. This means the point on the torus is

\begin{aligned} \vec{v} (\frac{π}{4}, \frac{π}{2}) & = [\begin{array}{c} 3 \cos (π / 4) + \cos (π / 4) \cos (π / 2) \\ 3 \sin (π / 4) + \sin (π / 4) \cos (π / 2) \\ \sin (π / 2) \end{array}] \\ = [\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}] \\ = [\begin{array}{c} \frac{3 \sqrt{2}}{2} \\ \frac{3 \sqrt{2}}{2} \\ 1 \end{array}] \end{aligned}

And the tangent vector is

\begin{aligned} \frac{\partial \vec{v}}{\partial t} (\frac{π}{4}, \frac{π}{2}) & = [\begin{array}{c} - 3 \sin (π / 4) - \sin (π / 4) \cos (π / 2) \\ 3 \cos (π / 4) + \cos (π / 4) \cos (π / 2) \\ 0 \end{array}] \\ = [\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}] \\ = [\begin{array}{c} - \frac{3 \sqrt{2}}{2} \\ \frac{3 \sqrt{2}}{2} \\ 0 \end{array}] \end{aligned}

Concept check: Why does it make sense that the

z

-component of this tangent vector is

0

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

\vec{v}

\begin{aligned} \frac{\partial \vec{v}}{\partial s} (t, s) = \frac{\partial}{\partial s} & [\begin{array}{c} 3 \cos (t) + \cos (t) \cos (s) \\ 3 \sin (t) + \sin (t) \cos (s) \\ \sin (s) \end{array}] \\ = & [\begin{array}{c} - \cos (t) \sin (s) \\ - \sin (t) \sin (s) \\ \cos (s) \end{array}] \end{aligned}

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

\vec{v}

, the red line and the yellow velocity vector might look something like this:

The partial derivative

\frac{\partial \vec{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:
$\vec{v} (s, t) = [\begin{matrix} x (s, t) \\ y (s, t) \\ z (s, t) \end{matrix}]$ ‍
Its partial derivatives are computed by taking the partial derivative of each component:
$\begin{aligned} \frac{\partial \vec{v}}{\partial t} (s, t) & = [\begin{array}{c} \frac{\partial x}{\partial t} (s, t) \\ \frac{\partial y}{\partial t} (s, t) \\ \frac{\partial z}{\partial t} (s, t) \end{array}] \end{aligned}$ ‍
$\begin{aligned} \frac{\partial \vec{v}}{\partial s} (s, t) & = [\begin{array}{c} \frac{\partial x}{\partial s} (s, t) \\ \frac{\partial y}{\partial s} (s, t) \\ \frac{\partial z}{\partial s} (s, t) \end{array}] \end{aligned}$ ‍
You can interpret these partial derivatives as giving vectors tangent to the parametric surface defined by $\vec{v}$ ‍.
For example, imagine nudging a point in the input space along the $t$ ‍ direction, say from the coordinates $(s, t)$ ‍ to the coordinates $(s, t + h)$ ‍ for some small $h$ ‍. This results in some small nudge in the output along the surface, which is represented by the vector $h \frac{\partial \vec{v}}{\partial t} (s, t)$ ‍.

Chcesz dołączyć do dyskusji?

Zaloguj się

Sortuj według

Na razie brak głosów w dyskusji

Rozumiesz angielski? Kliknij tutaj, aby zobaczyć więcej dyskusji na angielskiej wersji strony Khan Academy.

Analiza matematyczna funkcji wielu zmiennych