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3.2 First fundamental form I (metric)

The differential arc length of a parametric curve is given by (2.2). Now if we replace the parametric curve by a curve $ u=u(t)$ , $ v=v(t)$ which lies on the parametric surface $ {\bf r}={\bf r}(u,v)$ , then
$\displaystyle ds = \left\vert\frac{d{\bf r}}{dt}\right\vert dt = \left\vert{\bf... ...ot({\bf r}_{u}\dot{u}+{\bf r}_{v}\dot{v})}dt = \sqrt{Edu^2 + 2Fdudv + Gdv^2}\;,$     (3.11)

where
$\displaystyle E = {\bf r}_u \cdot {\bf r}_u, \;\;\; F = {\bf r}_u \cdot {\bf r}_v, \;\;\; G = {\bf r}_v \cdot {\bf r}_v\;.$     (3.12)

The first fundamental form is defined as

$\displaystyle I = ds^2 = d{\bf r} \cdot d{\bf r} = E du^2 + 2Fdudv + Gdv^2\;,$     (3.13)

and $ E$ , $ F$ , $ G$ are called the first fundamental form coefficients and play important roles in many intrinsic properties of a surface. The first fundamental form $ I$ can be rewritten as
$\displaystyle I = \frac{1}{E}(E \; du + F \; dv)^2 + \frac{EG-F^2}{E} {dv}^2\;.$     (3.14)

Since $ ({\bf r}_u \times {\bf r}_v)^2 = ({\bf r}_u \times {\bf r}_v) \cdot({\bf r}_u ... ...bf r}_u) ({\bf r}_v \cdot {\bf r}_v) - ({\bf r}_u \cdot {\bf r}_v)^2=EG - F^2>0$ 3.1 and $ E = {\bf r}_u \cdot {\bf r}_u>0$ , $ I$ is positive definite, provided that the surface is regular. That is $ I\geq 0$ and $ I=0$ if and only if $ du=0$ and $ dv=0$ .

Example 3.2.1. Let us compute the arc length of a curve $ u=t$ , $ v=t$ for $ 0\leq t \leq 1$ on a hyperbolic paraboloid $ {\bf r}(u,v) = (u,v,uv)^T$ where $ 0\leq u,v \leq 1$ as shown in Fig. 3.4 (a). We have

$\displaystyle {\bf r}_u = (1,0,v)^T, \;\;\;{\bf r}_v = (0,1,u)^T,\;\;\; E = {\b... ...{\bf r}_u \cdot {\bf r}_v = uv, \;\;\; G = {\bf r}_v \cdot {\bf r}_v = 1+u^2\;,$      

and along the curve the first fundamental form coefficients are
$\displaystyle E = 1 + t^2,\;\;\;F=t^2,\;\;\;G=1+t^2\;,$      

thus,
$\displaystyle ds = \sqrt{E\dot{u}^2 + 2F\dot{u}\dot{v} + G\dot{v}^2}dt = 2\sqrt{t^2 + \frac{1}{2}}dt\;.$      

Finally the arc length for $ 0\leq t \leq 1$ is given by
$\displaystyle s = 2\int_0^1\sqrt{t^2 + \frac{1}{2}}dt = \left[t\sqrt{t^2 + \fra... ...right)\right]_0^1 = \sqrt{\frac{3}{2}} + \frac{1}{2}log(\sqrt{2} + \sqrt{3})\;.$      

Figure 3.4: Hyperbolic paraboloid: (a) arc length along $ u=t$ , $ v=t$ , (b) area bounded by positive $ u$ and $ v$ axes and a quarter circle
\begin{figure}\vspace*{-5mm} \centerline{ \psfig{file=fig/bilinear_diag.ps,height=2.5in} \psfig{file=fig/bilinear_arc.ps,height=2.7in}}\end{figure}

The angle between two curves on a parametric surface $ {\bf r}_1={\bf r}(u_1(t), v_1(t))$ and $ {\bf r}_2={\bf r}(u_2(t), v_2(t))$ can be evaluated by taking the inner product of the tangent vectors of $ {\bf r}_1$ and $ {\bf r}_2$ , yielding

$\displaystyle \cos\omega = \frac{Edu_1du_2 + F(du_1dv_2 + dv_1du_2) + Gdv_1dv_2... ...ac{dv_1}{ds_1}\frac{du_2}{ds_2}\right) + G\frac{dv_1}{ds_1}\frac{dv_2}{ds_2}\;.$     (3.17)

As a result of the above equation, the orthogonality condition for the two tangent vectors $ \dot{\bf r}_1$ and $ \dot{\bf r}_2$ is:
$\displaystyle Edu_1du_2 + F(du_1dv_2 + dv_1du_2) + Gdv_1dv_2=0\;.$     (3.18)

In particular when the two curves are the $ u$ and $ v$ iso-parametric curves, (3.17) reduces to
$\displaystyle \cos\omega = \frac{{\bf r}_u\cdot{\bf r}_v}{\vert{\bf r}_u\vert\v... ...{\bf r}_u\cdot{\bf r}_u} \sqrt{{\bf r}_v\cdot{\bf r}_v}}=\frac{F}{\sqrt{EG}}\;.$     (3.19)

Thus the iso-parametric curves are orthogonal if $ F=0$ .

The area of a small parallelogram with vertices $ {\bf r}(u,v)$ , $ {\bf r}(u+\delta u,v)$ , $ {\bf r}(u,v+\delta v)$ and $ {\bf r}(u+\delta u,v+ \delta v)$ as illustrated in Fig. 3.5, is approximated by

$\displaystyle \delta A = \vert{\bf r}_u\delta u\times {\bf r}_v\delta v\vert=\sqrt{EG-F^2}\delta u\delta v\;,$     (3.20)

or in differential form
$\displaystyle dA = \sqrt{EG-F^2}dudv\;.$     (3.21)

Figure 3.5: Area of small surface patch
\begin{figure}\centerline{ \psfig{file=fig/area.eps,height=1.6in}}\end{figure}

Example 3.2.2. Let us compute the area of a region of the hyperbolic paraboloid that is used in Example 3.2.1. The region is bounded by positive $ u$ and $ v$ axes and a quarter circle $ u^2 + v^2 = 1 $ as shown in Fig. 3.4 (b). Substituting $ EG-F^2 = (1+v^2)(1+u^2)-u^2 v^2 = 1+u^2+v^2$ into  (3.21), we obtain

$\displaystyle A = \int_D \sqrt{1+u^2+v^2}dudv\;.$      

To perform the integration it is easier to change variables, $ u = r \cos \theta$ , $ v = r \sin\theta$ , so that
$\displaystyle A = \int_0^{ \frac{\pi}{2}} \int_0^1 \sqrt{1+ r^2} \; r \; d\theta \; dr = \frac{ \pi}{6} (\sqrt{8} -1)\;.$      


Footnotes

...3.1
Here the vector identity
$\displaystyle ({\bf a}\times {\bf b})\cdot ({\bf c}\times {\bf d})= ({\bf a}\cd... ...f c}) ({\bf b}\cdot {\bf d}) - ({\bf a}\cdot {\bf d}) ({\bf b}\cdot {\bf c})\;,$     (3.15)

with the special case
$\displaystyle ({\bf a}\times {\bf b})\cdot ({\bf a}\times {\bf b})= ({\bf a}\cdot {\bf a}) ({\bf b}\cdot {\bf b}) - ({\bf a}\cdot {\bf b})^2\;,$     (3.16)

is used.
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Next: 3.3 Second fundamental form Up: 3. Differential Geometry of Previous: 3.1 Tangent plane and   Contents   Index
December 2009