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Analiza matematyczna funkcji wielu zmiennych

Kurs: Analiza matematyczna funkcji wielu zmiennych > Rozdział 3

Lekcja 1: Płaszczyzny styczne i linearyzacja lokalna

Tangent planes

Just as the single variable derivative can be used to find tangent lines to a curve, partial derivatives can be used to find the tangent plane to a surface.

Kontekst

Pochodne cząstkowe

Do czego zmierzamy

A tangent plane to a two-variable function $f (x, y)$ ‍ is, well, a plane that's tangent to its graph.

The equation for the tangent plane of the graph of a two-variable function $f (x, y)$ ‍ at a particular point $(x_{0}, y_{0})$ ‍ looks like this:
$T (x, y) = f (x_{0}, y_{0}) + f_{x} (x_{0}, y_{0}) (x - x_{0}) + f_{y} (x_{0}, y_{0}) (y - y_{0})$ ‍

The task at hand

Think of a scalar-valued function with a two-coordinate input, like this one:

f (x, y) = - x^{2} - y^{2} + 3

Intuitively, it's common to visualize a function like this with its three-dimensional graph.

Remember, you can describe this graph more technically by describing it as a certain set of points in three-dimensional space. Specifically, it is all the points that look like this:

(x, y, f (x, y)) = (x, y, - x^{2} - y^{2} + 3)

Here,

x

and

y

can range over all possible real numbers.

A tangent plane to this graph is a plane which is tangent to the graph. Hmmm, that's not a good definition. This is hard to describe with words, so I'll just show a video with various different tangent planes.

Filmy wideo na Khan Academy

Zobacz transkrypcję filmu

Key question: How do you find an equation representing the plane tangent to the graph of the function at some specific point

(x_{0}, y_{0}, f (x_{0}, y_{0}))

in three-dimensional space?

Representing planes as graphs

Well, first of all, which functions

g (x, y)

have graphs that look like planes?

The slope of a plane in any direction is constant over all input values, so both partial derivatives

g_{x}

and

g_{y}

would have to be constants. The functions with constant partial derivatives look like this:

g (x, y) = a x + b y + c

Here,

a

b

, and

c

are each some constant. These are called linear functions. Well, technically speaking they are affine functions since linear functions must pass through the origin, but it's common to call them linear functions anyway.

Question: How can you guarantee that the graph of a linear function passes through a particular point

(x_{0}, y_{0}, z_{0})

in space?

One clean way to do this is to write our linear function as

g (x, y) = a (x - x_{0}) + b (y - y_{0}) + z_{0}

[Wait, this doesn't look like ax+by+c]

Concept check: With

g

defined this way, compute

g (x_{0}, y_{0})

[Wyjaśnij, o co chodzi]

Writing

g (x, y)

like this makes it clear that

g (x_{0}, y_{0}) = z_{0}

. This guarantees that the graph of

g

must pass through

(x_{0}, y_{0}, z_{0})

(x_{0}, y_{0}, g (x_{0}, y_{0})) = (x_{0}, y_{0}, z_{0})

The other constants

a

and

b

are free to be whatever we want. Different choices for

a

and

b

result in different planes passing through the point

(x_{0}, y_{0}, z_{0})

. The video below shows how those planes change as we tweak

a

and

b

Filmy wideo na Khan Academy

Zobacz transkrypcję filmu

Equation for a tangent plane

Back to the task at hand. We want a function

T (x, y)

that represents a plane tangent to the graph of some function

f (x, y)

at a point

(x_{0}, y_{0}, f (x_{0}, y_{0}))

, so we substitute

f (x_{0}, y_{0})

for

z_{0}

in the general equation for a plane.

T (x, y) = f (x_{0}, y_{0}) + a (x - x_{0}) + b (y - y_{0})

As you tweak the values of

a

and

b

, this equation will give various planes passing through the graph of

f

at the desired point, but only one of them will be a tangent plane.

Of all the planes passing through

(x_{0}, y_{0}, f (x_{0}, y_{0}))

, the one tangent to the graph of

f

will have the same partial derivatives as $f$ ‍. Pleasingly, the partial derivatives of our linear function are given by the constants

a

and

b

Try it! Take the partial derivatives of the equation for $T (x, y)$ ‍ above.
[Rozwiązanie]

Therefore setting

a = f_{x} (x_{0}, y_{0})

and

b = f_{y} (x_{0}, y_{0})

will guarantee that the partial derivatives of our linear function

T

match the partial derivatives of

f

. Well, at least they will match for the input

(x_{0}, y_{0})

, but that's the only point we care about. Putting this together, we get a usable formula for the tangent plane.

\begin{array}{r} T (x, y) = f (x_{0}, y_{0}) + f_{x} (x_{0}, y_{0}) (x - x_{0}) + f_{y} (x_{0}, y_{0}) (y - y_{0}) \end{array}

Example: Finding a tangent plane

Filmy wideo na Khan Academy

Zobacz transkrypcję filmu

Problem:

Given the function

f (x, y) = \sin (x) \cos (y)

find the equation for a plane tangent to the graph of

f

above the point

(\frac{π}{6}, \frac{π}{4})

The tangent plane will have the form

\begin{array}{r} T (x, y) = f_{x} (x_{0}, y_{0}) (x - x_{0}) + f_{y} (x_{0}, y_{0}) (y - y_{0}) + f (x_{0}, y_{0}) \end{array}

Krok 1: Find both partial derivatives of

f

f_{x} (x, y) =

f_{y} (x, y) =

[Odpowiedź]

Krok 2: Evaluate the function

f

as well as both these partial derivatives at the point

(\frac{π}{6}, \frac{π}{4})

f (π / 6, π / 4) =

f_{x} (π / 6, π / 4) =

f_{y} (π / 6, π / 4) =

[Odpowiedź]

Putting these three numbers into the general equation for a tangent plane, you can get the final answer

T (x, y) =

[Odpowiedź]

Podsumowanie

A tangent plane to a two-variable function $f (x, y)$ ‍ is, well, a plane that's tangent to its graph.

The equation for the tangent plane of the graph of a two-variable function $f (x, y)$ ‍ at a particular point $(x_{0}, y_{0})$ ‍ looks like this:

T (x, y) = f (x_{0}, y_{0}) + f_{x} (x_{0}, y_{0}) (x - x_{0}) + f_{y} (x_{0}, y_{0}) (y - y_{0})

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