MATH 251 Lecture 5

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Curves and Surfaces

curves

2 equations in (x,y,z)

1 independent variable (parameterized)

surfaces

1 equation in (x,y,z)

2 independent variables (2 parameters)

Parameterizing a curve/line

1. a point $P$ and a direction vector $\overrightarrow{v}$
2. two points (convert to point and direction vector with $\overrightarrow{v}=\overrightarrow{Q-P})$

Given a point $P$ and a direction vector $\overrightarrow{v}$, the parametric form of a line is

$\overrightarrow{r}\left(t\right)=P+t\overrightarrow{v}$

In general, a parameterized curve consists of 3 functions that give a position in space.

Tangent Line

To get a vector tangent to a curve in 3D space, take the derivative of each component in the parametric form.

Here's an elliptical helix:

$\overrightarrow{r}\left(t\right)=(2\cos t,\sin t,t)$

The tangent vector will represent the velocity at time $t$

$\overrightarrow{v}\left(t\right)={\overrightarrow{r}}^{\prime }\left(t\right)=(-2\sin t,\cos t,1)$

The second derivative will be the acceleration:

$\overrightarrow{a}\left(t\right)={\overrightarrow{v}}^{\prime }\left(t\right)={\overrightarrow{r}}^{″}\left(t\right)=(-2\cos t,-\sin t,0)$

Planes

• Three noncolinear points determine a plane.
• Point and a normal vector

Given:

Arbitrary point on plane:	$\overrightarrow{X}=⟨x,y,z⟩$
Point (on plane):	$\overrightarrow{P}=⟨x_0,y_0,z_0⟩$
Normal Vector:	$\overrightarrow{n}=⟨p,q,r⟩$

• The vector form of a plane is: $(\overrightarrow{X}-\overrightarrow{P})\cdot \overrightarrow{n}=0$
• the normal form equation of a plane is: $p(x-x_0)+q(y-y_0)+r(z-z_0)=0$

Distance between point and Plane

Find the distance between $P=(1,-1,3)$ and $4x+2y-z=2$.

$\overrightarrow{n}=⟨4,2,-1⟩$

Given any point on the plane—$Q=(0,0,-2)$, for example—the distance is given by: ${\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}}_{\overrightarrow{n}}(P-Q)=(P-Q)\cdot \hat{n}$.