MATH 251 Lecture 5

« previous | Friday, January 27, 2012 | next »

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

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

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

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:

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

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

${\vec {v}}\left(t\right)={\vec {r}}'\left(t\right)=\left(-2\sin {t},\,\cos {t},\,1\right)$

The second derivative will be the acceleration:

${\vec {a}}\left(t\right)={\vec {v}}'\left(t\right)={\vec {r}}''\left(t\right)=\left(-2\cos {t},\,-\sin {t},\,0\right)$

Planes

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

Given:

Arbitrary point on plane:	${\vec {X}}=\left\langle x,y,z\right\rangle$
Point (on plane):	${\vec {P}}=\left\langle x_{0},y_{0},z_{0}\right\rangle$
Normal Vector:	${\vec {n}}=\left\langle p,q,r\right\rangle$

• The vector form of a plane is: $\left({\vec {X}}-{\vec {P}}\right)\cdot {\vec {n}}=0$
• the normal form equation of a plane is: $p\left(x-x_{0}\right)+q\left(y-y_{0}\right)+r\left(z-z_{0}\right)=0$

Distance between point and Plane

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

${\vec {n}}=\left\langle 4,\,2,\,-1\right\rangle$

Given any point on the plane—$Q=(0,0,-2)$, for example—the distance is given by: $\mathrm {comp} _{\vec {n}}\left(P-Q\right)=\left(P-Q\right)\cdot {\hat {n}}$.