Calculus of One and Several Variables

16.4.1 Some Simple Techniques

Recall the formula for acceleration is

a = aTT + aNN (16.13)

(16.13)

where a_T =

and a_N = κ². Of course one way to find a _T and a_N is to just find , and κ and plug in. However, there is another way which might be easier. Take the dot product of both sides with T. This gives,

a ⋅T = aTT ⋅T + aNN ⋅T = aT .

Thus

a = (a⋅T )T + aNN

and so

a− (a⋅T )T = aNN (16.14)

(16.14)

and taking norms of both sides,

|a − (a ⋅T)T | = aN.

Also from 16.14,

a-−-(a-⋅T)T--= -aN-N- = N. |a − (a ⋅T)T | aN |N|

Also recall

κ = |a×-v|,a2 + a2 = |a|2 |v|3 T N

This is usually easier than computing T^′∕

. 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.

First I need to find the velocity and acceleration. Thus

v = (− sint,1,2t),a = (− cost,0,2)

and consequently, T =

. When t = 0, this reduces to

v (0) = (0,1,0),a = (− 1,0,2), |v(0)| = 1,T = (0,1,0).

Then the tangential component of acceleration when t = 0 is

aT = (− 1,0,2)⋅(0,1,0) = 0

Now

² = 5 and so a_N = because a_T² + a_N² = ². Thus = κ² = κ⋅ 1 = κ. Next lets find N. From a = a_TT + a_NN it follows

(− 1,0,2) = 0⋅T + √5N

and so

N = √1-(− 1,0,2). 5

This was pretty easy.

You need to write this as a parametric curve. This is most easily accomplished by letting t = x. Thus a parametrization is

: t ∈.  Then you can use the formula given above. The acceleration is and the velocity is . 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.

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).

= + t = where t ∈. Note where this came from. The vector is obtained from −. 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.