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Analiza matematyczna funkcji wielu zmiennych

Kurs: Analiza matematyczna funkcji wielu zmiennych > Rozdział 2

Lekcja 7: Pochodne cząstkowe funkcji wektorowych (artykuły)

Derivatives of vector-valued functions

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How to compute, and more importantly how to interpret, the derivative of a function with a vector output.

Kontekst

Do czego zmierzamy

To take the derivative of a vector-valued function, take the derivative of each component:
$\begin{aligned} \dfrac{d}{dt} \left[ \begin{array}{c} x(t) \\ y(t) \end{array} \right] = \left[ \begin{array}{c} x'(t) \\ y'(t) \end{array} \right] \end{aligned}$
If you interpret the initial function as giving the position of a particle as a function of time, the derivative gives the velocity vector of that particle as a function of time.

The derivative of a vector-valued function

Good news! Computing the derivative of a vector-valued function is nothing really new. As such, I'll keep this article pretty short. The main new takeaway is interpreting the vector derivative.

Dive in with an example

Let's start with a relatively simple vector-valued function start bold text, s, end bold text, with, vector, on top, left parenthesis, t, right parenthesis, with only two components,

$\begin{aligned} \vec{\textbf{s}}(t) = \left[ \begin{array}{c} 2 \sin(t) \\ 2 \cos(t/3)t \end{array} \right] \end{aligned}$

To take the derivative of start bold text, s, end bold text, with, vector, on top, just take the derivative of each component:

$\begin{aligned} \dfrac{d \vec{\textbf{s}}}{dt}(t) &= \left[ \begin{array}{c} \frac{d}{dt}(2 \sin(t)) \\ \frac{d}{dt}(2 \cos(t/3))t \end{array} \right] \\ \\ &= \left[ \begin{array}{c} 2\cos(t) \\ 2 \cos(t/3) - \frac{2}{3}\sin(t/3)t \end{array} \right] \end{aligned}$

You might also write this derivative as start bold text, s, end bold text, with, vector, on top, prime, left parenthesis, t, right parenthesis. This derivative is a new vector-valued function, with the same input t that start bold text, s, end bold text, with, vector, on top has, and whose output has the same number of dimensions.

More generally, if we write the components of start bold text, s, end bold text, with, vector, on top as x, left parenthesis, t, right parenthesis and y, left parenthesis, t, right parenthesis, we write its derivative like this:

$\begin{aligned} \vec{\textbf{s}}'(t) = \left[ \begin{array}{c} x'(t) \\ y'(t) \end{array} \right] \end{aligned}$

Derivative gives a velocity vector.

For the example above, how can we visualize what the derivative means? First, to visualize

$\begin{aligned} \vec{\textbf{s}}(t) = \left[ \begin{array}{c} 2 \sin(t) \\ 2 \cos(t/3)t \end{array} \right] \end{aligned}$

we note that the output has more dimensions than the input, so it is well-suited to be viewed as a parametric function.

Each point on the curve represents the tip of a vector

\left[ \begin{array}{c} 2 \sin(t_0) \\ 2 \cos(t_0/3)t_0 \end{array} \right]

for some specific number t, start subscript, 0, end subscript. For instance, when t, start subscript, 0, end subscript, equals, 2 we draw a vector to the point

\begin{aligned} \quad \vec{\textbf{s}}(2) = \left[ \begin{array}{c} 2 \sin(2) \\ 2 \cos(2/3)\cdot 2 \end{array} \right] \approx \left[ \begin{array}{c} 1{,}819 \\ 3{,}144 \end{array} \right] \end{aligned}

When we do this for all possible inputs t, the tips of the vectors start bold text, s, end bold text, with, vector, on top, left parenthesis, t, right parenthesis will trace out a certain curve:

What do we get when we plug in some value of t, perhaps 2 again, to the derivative?

\begin{aligned} \quad \dfrac{d \vec{\textbf{s}}}{dt}(2) &= \left[ \begin{array}{c} 2\cos(2) \\ 2 \cos(2/3) - \frac{2}{3}\sin(2/3)\cdot 2 \end{array} \right]\\ &\approx \left[ \begin{array}{c} -0{,}832 \\ 0{,}747 \end{array} \right] \end{aligned}

This is also some two-dimensional vector.

It's hard to see what this derivative vector represents when it just sits at the origin, but if we shift it so that its tail sits on the tip of the vector start bold text, s, end bold text, with, vector, on top, left parenthesis, 2, right parenthesis, it has a wonderful interpretation:

If start bold text, s, end bold text, with, vector, on top, left parenthesis, t, right parenthesis represents the position of a traveling particle as a function of time, start fraction, d, start bold text, s, end bold text, with, vector, on top, divided by, d, t, end fraction, left parenthesis, t, start subscript, 0, end subscript, right parenthesis is the velocity vector of that particle at time t, start subscript, 0, end subscript.
Derivative is a velocity vector tangent to the curve.

In particular, this means the direction of the vector is tangent to the curve, and its magnitude indicates the speed at which one travels along this curve as t increases at a constant rate (as time tends to do).

Concept Check: Suppose the position in two-dimensional space of a particle, as a function of time, is given by the function

\begin{aligned} \quad \vec{\textbf{s}}(t) = \left[ \begin{array}{c} t^2 \\ t^3 \end{array} \right] \end{aligned}

Ile wynosi start fraction, d, start bold text, s, end bold text, with, vector, on top, divided by, d, t, end fraction?

(Choice A)
$\begin{aligned} \quad \left[ \begin{array}{c} 3t^2 \\ 2t \end{array} \right] \end{aligned}$
(Choice B)
$\begin{aligned} \quad \left[ \begin{array}{c} 2t \\ 3t^2 \end{array} \right] \end{aligned}$
(Choice C)
$\begin{aligned} \quad \left[ \begin{array}{c} t \\ t^2 \end{array} \right] \end{aligned}$

[Odpowiedź]

What is the speed of the particle at time t, equals, 3?

(Choice A)
square root of, 6, squared, plus, 27, squared, end square root, approximately equals, 27, comma, 659
(Choice B)
6, squared, plus, 27, squared, equals, 765

[Odpowiedź]

Podsumowanie

To take the derivative of a vector-valued function, take the derivative of each component.
If you interpret the initial function as giving the position of a particle as a function of time, the derivative gives the velocity vector of that particle as a function of time.

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