# Properties of Curves (continued)

Lecture 12 examined some properties of curves on which we will now make a further expansion, looking at tangential and normal components of acceleration, oscillatory and exponential motion, and the distance from a point to a line.

v = dR/dt = dR/ds * ds/dt = T * ds/dt
a = dv/dt = dT/dt * ds/dt + T * =
= dT/dt * (ds/dt)^2 + T *
dT/ds = K*N,
where K is the radius of curvature.
a = K(ds/dt)^2 * N + * T
a = aN * N + aT * T
aN = K*(ds/dt)^2 = K |v|^2 = (|v|^2)/
rho, where rho is the radius.
aT = = d|v|/dt
aN
from above is the centripetal acceleration.

Example:
Circle:
for constant speed:

### Diff. Equations with complex roots.

x'' + 2ax' + bx = 0
x = C1 exp(r1*t) + C2 exp(r2*t),
where r1, and r2 are both real
or, x = (C1 + C2*x) exp(r*t), where, r1 = r2 = r
or the oscillatory solution:
x = exp(-r*t) (C1 cos(w*t) + C2 sin(w*t))
This would look similar to the following:

Example:

m*x'' + C*x' + k*x = 0
x'' + (C/m)x' + (k/m)x = 0

We will now define w^2 = k/m, and the damping factor, C = 2 m z w, where z represents the Greek letter Zeta.
So, x'' + 2zw * x' + w^2 * x = 0
In order to solve a differential equation we can use the same method as earlier:
(D^2 + 2zw*D + w^2) x = 0
or, make the assumption that x = exp(r*t)
=> (r^2 + 2zw*r + w^2) exp(r*t) = 0
These two equations are equivalent.
>From this we can solve for the different values of r:

For z^2 < 1,
=> r1 = - zw + i w(1-z^2)^(1/2)
r2 = - zw - i w(1-z^2)^(1/2)

Let wd = w (1-z^2)^(1/2)
x = d1 exp(-z w t) exp(i wd t) +
+ d2 exp(-z w t) exp(-i wd t) =
= exp(-z w t) [d1 (cos(wd t) + i sin(wd t)) +
+ d2 (cos(wd t) - i sin(wd t))] =
= exp(-z w t) [C1 cos(wd t) + C2 sin(wd t)]

The Natural Frequency (w) can be represented as:

and the Damped Frequency (wd) as:

In the case where z = 1, the system is called critically damped, and the equation takes the form:
x = (A + Bt) exp(-z w t)
It is called overdamped for z>1.

Example 2:
Find the distance (d) from a point (S), to a line (PB).

u is the unit vector in the direction of the line.

Example 3:
Find the distance between the point S(1,1,1), and the line: x = -t, y = 1 + t, and z = t
v = -i + j + k
u = (-i + j + k)/3^(1/2)
Chose t = 0
P = (0,1,0), => PS = i + k
=> d = 2^(1/2)

Created by Milos Borojevic on 3/8/95
Edited by on Lawrence C. Weintraub 3/13/95
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