Reading: Introduction to Curves & Surfaces (Due 10/24)

For your next reading assignment, you will read a few pages about curves and surfaces from the course notes: Chapter 2, pages 7–23. This material should be enough to get you started on the written/coding exercises NOW, rather than waiting until we are done with the full set of lectures. We will cover these topics in greater depth during lecture (especially the topic of curvature).

Assignment: Read the pages above, and write 2–3 sentences summarizing what you read, plus at least one question about something you didn’t understand, or some thought/idea that occurred to you while reading the article.

Handin instructions can be found in the “Readings” section of the Assignments page.  Note that you must send your summary in no later than 10am Eastern on the date of the next lecture (October 24, 2017).


11 thoughts on “Reading: Introduction to Curves & Surfaces (Due 10/24)”

  1. The calendar says that assignment 2 is due 10/21 (i.e. before 10/24, when the reading assignment is due). Did you perhaps mean that the reading assignment is due 10/17 or 10/19?

  2. I am lost in the first topic of the reading here.
    It says that the geometry of the surface can be described via a map $f: M \rightarrow \mathbb{R}^{3}$ from a region $M$ in the Euclidean plane $\mathbb{R}^{3}$.

    It also says that $df$ is a map from a vector in the plane to the surface. However, it does not make sense for me.

    1) It seems that It takes an anchored vector at a point in $M$ and finds the correspondent vector on the surface $f(M)$, but tangent to the surface. Is this interpretation correct?

    2)To map from a vector anchored in space we need both the position of the vector and the vector itself (as in the illustration of p.7). However, $f$ itself only have $u$ and $v$ as arguments.

    3) When we solve $df$ for a $f(u, v)$ should it result in two components $a.dx + b.dy$? In this case, the first component (a) would be related to the $\frac{\partial f}{\partial u}$ and the second with the $\frac{\partial f}{\partial v}$?

  3. Very very very basic question about vector\exterior calculus.

    Let’s say I have a function that generates a cylinder:
    $f \space\mathbb{R^2} \rightarrow \mathbb{R^3}$
    $f(u, v) \rightarrow \begin{bmatrix} cos(u) \\ sin(u) \\ v \end{bmatrix} $.

    I know how to find the partial derivatives
    $\tfrac{\partial}{\partial u}f(u, v) = \begin{bmatrix} -sin(u) \\ cos(u) \\ 0 \end{bmatrix} $
    $\tfrac{\partial}{\partial v}f(u, v) = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} $

    Step 2
    However, we have the exterior derivative $df$ that is a function with both partials… something like this
    $df(u, v) = \begin{bmatrix} -sin(u) \\ cos(u) \\ 0 \end{bmatrix}\wedge du + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \wedge dv$

    Step 3
    I understand that the $df$ is a function that would accept a vector $(u, v)$ as input and would generate the tangent of the cylinder relative to the position $f(u, v)$. I am trying to compute this by hand and there is a problem. When $u$ and/or $v$ is 0, the partial related to that component should not be evaluated at all. For instance,

    $df(1,0) = \tfrac{\partial}{\partial u}f(1, 0) = \begin{bmatrix} -sin(1) \\ cos(1) \\ 0 \end{bmatrix} $

    Step 4
    If I only use $u$ and $v$ as the input and treat $df$ as a regular function, it does not cancel the second vector, and I end up with the wrong result:

    $df(1, 0) = \begin{bmatrix} -sin(1) \\ cos(1) \\ 0 \end{bmatrix}\wedge du + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \wedge dv$

    How does it cancel the second vector? I think it is related to that part of exterior calculus in the written assignment 1. In this case, I should be able not only to use the $u, v$ as a numerical argument for df, but there also should be some way to use the $\wedge$ to filter which components are going to be computed.

    1. The confusion may come from the fact that (for brevity) we often drop the point at which the function is being evaluated, and only talk about the vector on which it’s being evaluated. A more explicit way to write the differential would be:

      \[ \left. df \right|_{(u,v)} = \left. \left( \frac{\partial f}{\partial u} du + \frac{\partial f}{\partial v} dv \right) \right|_{(u,v)}, \]

      where the bar means “evaluate this expression at the point $(u,v)$.” For instance, if we now wanted to apply \(df\) to a tangent vector field \(X(u,v) := a(u,v) \tfrac{\partial}{\partial u} + b(u,v) \tfrac{\partial}{\partial v}\), we would write

      \[ \left. df(X(u,v)) \right|_{(u,v)} = \left. \left( \frac{\partial f}{\partial u} du(X(u,v)) + \frac{\partial f}{\partial v} dv(X(u,v)) \right) \right|_{(u,v)}, \]

      which we could write more directly as

      \[ \left. df(X(u,v)) \right|_{(u,v)} = \left. \left( \frac{\partial f}{\partial u} a + \frac{\partial f}{\partial v} b \right) \right|_{(u,v)}. \]

      At some point you start to see the value of dropping some of the arguments! But it does take a little while to get adjusted and understand exactly what is going on from context.

  4. I’m not sure I understand why “When u and/or v is 0, the partial related to that component should not be evaluated at all.” The derivative of a function can be zero even if the argument of the function itself is zero (for example, the derivative of $$\sin{x}$$ is non-zero at $$x = 0$$). I think the result you get using the rules (i.e. the second result) is correct.

    1. Think about the intuition. If I measure a change in $u$ (i.e. if I have a vector with a $v=0$), I am moving on the surface of the cylinder horizontally (parallel to the xy plane). In this case, the derivative should have $z = 0$. This is what the $\frac{partial}{partial u}f$ expresses.

      I understand $df$ as a directional derivative. In this case, if I plug a vector (1, 0), I should have something like a dot product of the vector with the gradient. Therefore the vector (1, 0) does not represent the coordinates, but the change I am applying. In this case, it will result in the same thing as the $\tfrac{\partial}{\partial u}$ as the the equivalent of the dot operation will cancel the second component of the function. This is my interpretation of this process, using the notation of exterior calculus from assignment 1:
      $df(1, 0) = (\begin{bmatrix} -sin(u) \\ cos(u) \\ 0 \end{bmatrix}\wedge du + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \wedge dv) \wedge (1 \tfrac{\partial}{\partial u} + 0 \tfrac{\partial}{\partial v}) = (\begin{bmatrix} -sin(u) \\ cos(u) \\ 0 \end{bmatrix}\wedge du \wedge \tfrac{\partial}{\partial u} + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \wedge dv \wedge \tfrac{\partial}{\partial u} =\begin{bmatrix} -sin(u) \\ cos(u) \\ 0 \end{bmatrix}$

      Can anyone tell me if this is the right way to think about it. Thanks.

  5. Since we know that in pair [T N B], T is for tangent, N is for normal, and what’s the intuition of B? It seems B is not covered much in the reading material. Thanks.

    1. The vector $B$ is the binormal, which is orthogonal to $N$ and $T$, i.e., it is the third vector needed to complete an orthonormal frame (the Frenet frame). The intuition is that the change in $T$ tells you how much the curve is bending in plane (quantified by the curvature $\kappa$), whereas the change in $B$ tells you how much the curve is bending out of plane (quantified by the torsion $\tau$).

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