√a2+ x2

1

––

m1

rise

––

run

rise

––

run

468 Snell’s law

Snell’s law

Two different lines with the same slope are

PARAL

-

LEL

. For example, the lines y= 2x+ 3 and y= 2x– 5 are

parallel. If two

PERPENDICULAR

lines have slopes m1and

m2respectively, then the following relationship holds:

m1· m2= –1

One can see this by drawing the horizontal and verti-

cal line segments for the run and the rise for the first

line (and so m1= ), and then rotating that picture

counterclockwise 90°. In this new picture, what was

the run is now a rise for a perpendicular line, and

what was a rise is now a run pointing in the opposite

direction. Thus m2= – = – .

D

IFFERENTIAL CALCULUS

deals with the issue of

defining slope for curves that are not straight lines.

See also

COLLINEAR

;

GRADIENT

;

TANGENT

.

Snell’s law This principle, first described in 1621 by

Dutch mathematician Willebrord van Roijen Snell, is

best explained through the study of an

OPTIMIZATION

problem. We phrase it here in a modern setting.

A lifesaver sets off to rescue a swimmer in dis-

tress. The drowning swimmer, however, is not

directly in front of the lifesaver but off at an

angle along the shore. To reach the swimmer,

the lifesaver must run across the sand and then

dive into the water and swim to the rescue. A

question arises: as the lifesaver can run much

faster than she can swim, toward which point

along the shore should she run so that the total

amount of time it takes to reach the swimmer

is at a minimum?

Surprisingly, the straight-line path from the lifesaver sta-

tion to the swimmer does not provide the quickest route.

To analyze this problem, let a, b, α, and βbe the dis-

tances and angles shown. For ease, assume that the hori-

zontal displacement between the lifesaver and the

swimmer is 1 unit. Suppose also that the speed of the life-

saver on shore is rmeters per second, and in the water, s

meters per second. We wish to find the length xalong the

shore that produces a path requiring the least amount of

time to follow.

According to P

YTHAGORAS

’

S THEOREM

, the total

distance the lifesaver runs on shore is meters.

As speed equals distance over time, this takes a total of

seconds to complete. In the same way, we see

that the lifesaver will be in the water

before reaching the swimmer. Thus, for this position x

along the shore, the total time T(x) taken to reach the

swimmer is given as:

seconds. We now use the techniques of

CALCULUS

to

find a local minimum for this function. (See

MAXIMUM

/

MINIMUM

.)

A local minimum of this function can only occur

if the first derivative of this function is zero. This

yields:

The second-derivative test shows that this does indeed

correspond to a minimum. Thus the optimal solution

to this problem occurs when the quantity

equals . Notice in the diagram that

sinα= and sinβ= Thus, the

lifesaver should run to the position on shore such that