Calculus: Analytic Geometry and Calculus, with Vectors

3.9 Increments and differentials 193

Even though we have written (1) in such a way that t does not appear, we rise
to the occasionand differentiate with respect to t to obtain

du dxl dv_dyl dw dzl

(2) sdsdt =(u-x)(dt dt/+(v-y)Cdt dt/+(w-z)Cdt dt/

To obtain the answer, we determine s from (1) and then ds/dt from (2) when

(3) U X2, V 2, W Z2, x=x1, y =yl, z=zl

du do dw dx dy

(4) = q2a2,

_

• g2b2,

_

dt dt dt - 72C2, Tt = Qlal, Tt = flbl,

dz

dt =41c1

Remark: The formula (2) is trying to tell us something. The vector PQ in the

formula

(5) PQ = (u-x)i+(v-y)j+(w-z)k

is the displacement of Q relative to P. The velocity v of Q relative to P is

obtained by differentiating this with respect to t. Thus

(6) °dn a Tt)i+(dt

-at)k.

Hence (2) tells us that

(7) s dt =

that is, the left side is the scalar product of the displacement vector and the

velocity of Q relative to P. Since IPQI = s, the vector PQ/s is a unit vector in

the direction of PQ. The relations (2) and (7) therefore bear a simple message.

They tell us that the rate of change of the distance between two bodies is the scalar

component of the relative velocity of the bodies in the direction of the line joining the

bodies. This agrees with and is perhaps even a consequence of the fact that if

one body moves in a circle having its center at the other body, then the distance

between the bodies is always the radius of the circle.

(^13) Kitty was riding a horse on a merry-go-round of radius R. When she was
south of the center pole and going east with speed s, she exuberantly threw a ball
toward the pole. Kitty expected to hit the pole, but unfortunately Chester was
riding gallantly ahead of her and the ball hit him on the chin when he was east of
the center. Sketch a figure showing the east and north vector components of the
velocity (relative to terra firma) of the ball and mark the place where Chester was
sitting when Kitty threw the ball.

3.9 Increments and differentials As we become educated, we pick

up assorted ideas akin to the idea that we never try to find the weight

(1 avoirdupois grain or 0.0648 gram) of a kernel of medieval wheat by

finding the weight of a truckload of the stuff and subtracting the weight of

the decreased load resulting from removal of the kernel. The difficulty