. To illustrate the use of these
simple observations, consider the example worked above which was fairly messy.
I will make it easier by selecting a value of t and by using the above simplifying
techniques.
Example 16.4.5Let R
(t)
=
( )
cos(t),t,t2
for t ∈
[0,3]
. Find the speed, velocity,curvature, and write the acceleration in terms of normal and tangential componentswhen t = 0. Also find N at the point where t = 0.
First I need to find the velocity and acceleration. Thus
Then the tangential component of acceleration when t = 0 is
aT = (− 1,0,2)⋅(0,1,0) = 0
Now
|a|
2 = 5 and so aN =
√5-
because aT2 + aN2 =
|a|
2. Thus
√5
= κ
|v(0)|
2 = κ⋅ 1 = κ.
Next lets find N. From a = aTT + aNN it follows
(− 1,0,2) = 0⋅T + √5N
and so
N = √1-(− 1,0,2).
5
This was pretty easy.
Example 16.4.6Find a formula for the curvature of the curve given by the graphof y = f
(x)
for x ∈
[a,b]
. Assume whatever you like about smoothness of f.
You need to write this as a parametric curve. This is most easily accomplished by
letting t = x. Thus a parametrization is
(t,f (t),0)
: t ∈
[a,b]
. Then you can use the
formula given above. The acceleration is
(0,f′′(t),0)
and the velocity is
(1,f′(t) ,0)
.
Therefore,
a × v = (0,f′′(t),0)× (1,f ′(t),0) = (0,0,− f′′(t)).
Therefore, the curvature is given by
|a× v| |f′′(t)|
-|v|3- = (-------2)3∕2.
1+ f′(t)
Sometimes curves do not come to you parametrically. This is unfortunate when it
occurs but you can sometimes find a parametric description of such curves. It
should be emphasized that it is only sometimes when you can actually find a
parametrization. General systems of nonlinear equations cannot be solved using
algebra.
Example 16.4.7Find a parametrization for the intersection of the surfaces
y+ 3z = 2x2 + 4 and y +2z = x+ 1.
You need to solve for x and y in terms of x. This yields
z = 2x2 − x+ 3,y = − 4x2 + 3x − 5.
Therefore, letting t = x, the parametrization is
(x,y,z) = (t,− 4t2 − 5+ 3t,− t+ 3 + 2t2).
Example 16.4.8Find a parametrization for the straight line joining
(3,2,4)
and
(1,10,5)
.
(x,y,z)
=
(3,2,4)
+ t
(− 2,8,1)
=
(3− 2t,2+ 8t,4 + t)
where t ∈
[0,1]
. Note where
this came from. The vector
(− 2,8,1)
is obtained from
(1,10,5)
−
(3,2,4)
. Now you
should check to see this works. It is usually not possible to find an explicit formula for the
intersection of two surfaces as was just done.