334 CHAPTER 16. SPACE CURVES

In the case that t is time, the expression |r (t +h)−r (t)| is a good approximation forthe distance traveled by the object on the time interval [t, t +h]. The real distance would bethe length of the curve joining the two points but if h is very small, this is essentially equalto |r (t +h)−r (t)| as suggested by the picture below.

r(t)

r(t +h)

Therefore, |r(t+h)−r(t)|h gives for small h, the approximate distance travelled on the time

interval [t, t +h] divided by the length of time h. Therefore, this expression is really theaverage speed of the object on this small time interval and so the limit as h → 0, deservesto be called the instantaneous speed of the object. Thus |r′ (t)|T represents the speed timesa unit direction vector T which defines the direction in which the object is moving. Thusr′ (t) is the velocity of the object. This is the physical significance of the derivative whent is time. In general, r′(t) and T (t) are vectors tangent to the curve which point in thedirection of motion.

How do you go about computing r′ (t)? Letting r (t) = (r1 (t) , · · · ,rq (t)), the expres-sion

r (t0 +h)−r (t0)h

(16.1)

is equal to (r1 (t0 +h)− r1 (t0)

h, · · · ,

rq (t0 +h)− rq (t0)h

).

Then as h converges to 0, 16.1 converges to v ≡ (v1, · · · ,vq) where vk = r′k (t). This isbecause of Theorem 15.8.6 on Page 318, which says that the term in 16.1 gets close toa vector v if and only if all the coordinate functions of the term in 16.1 get close to thecorresponding coordinate functions of v.

In the case where t is time, this simply says the velocity vector equals the vector whosecomponents are the derivatives of the components of the displacement vector r (t).

Example 16.2.7 Let r (t) =(sin t, t2, t +1

)for t ∈ [0,5]. Find a tangent line to the curve

parameterized by r at the point r (2).

From the above discussion, a direction vector has the same direction as r′ (2). There-fore, it suffices to simply use r′ (2) as a direction vector for the line. r′ (2) = (cos2,4,1).Therefore, a parametric equation for the tangent line is

(sin2,4,3)+ t (cos2,4,1) = (x,y,z) .

Example 16.2.8 Let r (t) =(sin t, t2, t +1

)for t ∈ [0,5]. Find the velocity vector when

t = 1.

From the above discussion, this is simply r′ (1) = (cos1,2,1).