348 CHAPTER 16. SPACE CURVES

Example 16.6.4 Let R(t) =(cos(t) , t, t2

)for t ∈ [0,3]. Find the speed, velocity, curva-

ture, and write the acceleration in terms of normal and tangential components when t = 0.Also find N at the point where t = 0.

First I need to find the velocity and acceleration. Thus

v = (−sin t,1,2t) , a= (−cos t,0,2)

and consequently, T = (−sin t,1,2t)√sin2(t)+1+4t2

. 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) = 0Now |a|2 = 5 and so aN =

√5 because a2

T +a2N = |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 soN = 1√

5(−1,0,2) . This was pretty easy.

Example 16.6.5 Find a formula for the curvature of the curve given by the graph of 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 lettingt = x. Thus a parametrization is (t, f (t) ,0) : t ∈ [a,b] . Then you can use the formula givenabove. 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||v|3

=| f ′′(t)|

(1+ f ′(t)2)3/2 .

Sometimes curves do not come to you parametrically. This is unfortunate when itoccurs but you can sometimes find a parametric description of such curves. It should beemphasized that it is only sometimes when you can actually find a parametrization. Generalsystems of nonlinear equations cannot be solved using algebra.

Example 16.6.6 Find 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 16.6.7 Find 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 wherethis came from. The vector (−2,8,1) is obtained from (1,10,5)− (3,2,4). Now youshould check to see this works. It is usually not possible to find an explicit formula for theintersection of two surfaces as was just done.