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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:
    v(s,t)=[x(s,t)y(s,t)z(s,t)]
    Its partial derivatives are computed by taking the partial derivative of each component:
    vt(s,t)=[xt(s,t)yt(s,t)zt(s,t)]
    vs(s,t)=[xs(s,t)ys(s,t)zs(s,t)]
  • You can interpret these partial derivatives as giving vectors tangent to the parametric surface defined by v.

The goal

Suppose I were to give you a function with a two-dimensional input, and a three-dimensional output, like this one:
v(t,s)=[3cos(t)+cos(t)cos(s)3sin(t)+sin(t)cos(s)sin(s)]
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 0t2π and 0s2π. 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.
Two parameter input space
For any given point (t,s), the value v(t,s) is some point in three-dimensional space.
Concept check: Evaluate v(π,π). In other words, where does the function v take the input (t,s)=(π,π)?
Wybierz 1 odpowiedź:

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
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 vt, you take the partial derivative of each individual component.
vt(t,s)=t[3cos(t)+cos(t)cos(s)3sin(t)+sin(t)cos(s)sin(s)]=[t(3cos(t)+cos(t)cos(s))t(3sin(t)+sin(t)cos(s))t(sin(s))]=[3sin(t)sin(t)cos(s)3cos(t)+cos(t)cos(s)0]
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=π/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 vt 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 (t0,s0)=(π4,π2). This means the point on the torus is
v(π4,π2)=[3cos(π/4)+cos(π/4)cos(π/2)3sin(π/4)+sin(π/4)cos(π/2)sin(π/2)]=[322+22(0)322+22(0)1]=[3223221]
And the tangent vector is
vt(π4,π2)=[3sin(π/4)sin(π/4)cos(π/2)3cos(π/4)+cos(π/4)cos(π/2)0]=[32222(0)322+22(0)0]=[3223220]
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 v:
vs(t,s)=s[3cos(t)+cos(t)cos(s)3sin(t)+sin(t)cos(s)sin(s)]=[cos(t)sin(s)sin(t)sin(s)cos(s)]
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 v, the red line and the yellow velocity vector might look something like this:
Nudge s in the output space
The partial derivative vs 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:
    v(s,t)=[x(s,t)y(s,t)z(s,t)]
    Its partial derivatives are computed by taking the partial derivative of each component:
    vt(s,t)=[xt(s,t)yt(s,t)zt(s,t)]
    vs(s,t)=[xs(s,t)ys(s,t)zs(s,t)]
  • You can interpret these partial derivatives as giving vectors tangent to the parametric surface defined by 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 hvt(s,t).

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