# 22. Parametric Surfaces and Surface Integrals

## b. Tangent and Normal Vectors

## 2. Tangent Vectors

For a 2D curvilinear coordinate system \(\vec R(u,v)\), the coordinate tangent vectors are the tangent vectors to the coordinate curves. \[ \vec e_u=\dfrac{\partial\vec R}{\partial u} \qquad \text{and} \qquad \vec e_v=\dfrac{\partial\vec R}{\partial v} \] \(\vec e_u\) is tangent to the \(u\)-curves and points in the direction of increasing \(u\). \(\vec e_v\) is tangent to the \(v\)-curves and points in the direction of increasing \(v\). We do the same thing for parametric surfaces:

The tangent vector to the \(u\)-curves is
\[
\vec{e}_u=\dfrac{\partial\vec R}{\partial u}
=\left\langle
\dfrac{\partial x}{\partial u},
\dfrac{\partial y}{\partial u},
\dfrac{\partial z}{\partial u}
\right\rangle
\]
and points in the direction of increasing \(u\).

The tangent vector to the \(v\)-curves is
\[
\vec{e}_v=\dfrac{\partial\vec R}{\partial v}
=\left\langle
\dfrac{\partial x}{\partial v},
\dfrac{\partial y}{\partial v},
\dfrac{\partial z}{\partial v}
\right\rangle
\]
and points in the direction of increasing \(v\).

PY: Make the blue vector shorter.

The vector \(\vec{e}_u\) (as well as \(\vec{e}_v\)) is analogous to the to tangent (or velocity) vector to a parametric curve, \(\vec r(t)=\left\langle x(t),y(t),z(t)\right\rangle\), given by: \[ \vec{v}=\dfrac{d\vec r}{dt} =\left\langle \dfrac{dx}{dt},\dfrac{dy}{dt},\dfrac{dz}{dt}\right\rangle \] with the following exceptions:

- The parameter for \(\vec R(u,v_0)\) is \(u\) rather than \(t\); so we need to differentiate the with respect to \(u\).
- Since the \(v\)-coordinate is held constant at \(v_0\), the derivative needs to be a partial derivative with respect to \(u\) instead of a total derivative with respect to \(t\).
- We do not refer to the tangent vector as a velocity because the parameter, \(u\), represents a coordinate on the surface, not a time.
- We use the symbols \(\vec e_u\) and \(\vec e_v\) rather than the letter \(\vec{v}\) for the tangent vectors, because they are not velocities and also because we have already used \(v\) for a parameter.

Find the tangent vectors to the sphere of radius \(\rho=2\), parametrized by: \[ \vec R(\phi,\theta) =\left\langle 2\sin\phi\cos\theta,2\sin\phi\sin\theta,2\cos\phi\right\rangle \] PY: Add the tangent vectors to the plot.

The tangent vectors to the \(\phi\) and \(\theta\) curves are \[\begin{aligned} \vec{e}_\phi&=\dfrac{\partial\vec R}{\partial\phi} =\left\langle 2\cos\phi\cos\theta,2\cos\phi\sin\theta,-2\sin\phi\right\rangle \\ \vec{e}_\theta&=\dfrac{\partial\vec R}{\partial\theta} =\left\langle -2\sin\phi\sin\theta,2\sin\phi\cos\theta,\quad 0\quad \right\rangle \end{aligned}\] Notice that \(\vec{e}_\theta\) is horizontal since it is tangent to a line of latitude and it points East in the direction of increasing \(\theta\). Similarly, \(\vec{e}_\phi\) points South, tangent to a line of longitude, in the direction of increasing \(\phi\).

Compute the tangent vectors to the elliptic paraboloid \(z=x^2+y^2\) parametrized by: \[ \vec R(r,\theta)=\left\langle r\cos\theta,r\sin\theta,r^2\right\rangle \] PY: Change the grid on the paraboloid to polar. Add tangent vectors.

\(\begin{aligned} \vec e_r&=\left\langle \cos\theta,\sin\theta,2r\right\rangle \\ \vec e_\theta&=\left\langle -r\sin\theta,r\cos\theta,0\right\rangle \end{aligned}\)

We simply compute: \[\begin{aligned} \vec e_r&=\dfrac{\partial\vec R}{\partial r} =\left\langle \cos\theta,\sin\theta,2r\right\rangle \\ \vec e_\theta&=\dfrac{\partial\vec R}{\partial\theta} =\left\langle -r\sin\theta,r\cos\theta,0\right\rangle \end{aligned}\]

Find the tangent vectors to the piece of the cylinder \(x^2+z^2=4\) between \(y=0\) and \(y=6+2x+3z\) parametrized by: \[ \vec R(\theta,y)=\left\langle 2\cos\theta,y,2\sin\theta\right\rangle \] PY: Fix the grid to y theta. Add tangent vectors.

\(\begin{aligned} \vec e_\theta&=\left\langle -2\sin\theta,0,2\cos\theta\right\rangle \\ \vec e_y&=\left\langle 0,1,0\right\rangle \end{aligned}\)

We simply compute: \[\begin{aligned} \vec e_\theta&=\dfrac{\partial\vec R}{\partial\theta} =\left\langle -2\sin\theta,0,2\cos\theta\right\rangle \\ \vec e_y&=\dfrac{\partial\vec R}{\partial y} =\left\langle 0,1,0\right\rangle \end{aligned}\]

Notice the parameter ranges have absolutely nothing to do with the tangent vectors. They only affect the limits of integration.

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