1550078481-Ordinary_Differential_Equations__Roberts_

154 Ordinary Differential Equations

The line segment PQ, which represents the stiff rod of length L, is tangent to

the path followed by the boat. So the slope of the line PQ is

(1)

dy

dx

JL2 -x2

x

Graph the tractrix which corresponds to a rod of length L = 10 ft. (HINT:

Solve the initial value problem consisting of the differential equation (1) and

the initial condition y(L) = 0 on the interval [.5, 10].)

Exercise 2. The y-axis and the line x = W > 0 represent the banks of a

river. The river flows in the negative y-direction with speed Sr· A boat whose

speed in still water is Sb is launched from the point (W, 0). The boat is steered

so that it is always headed toward the origin. See Figure 3.19.

x

(0, 0) (W, 0)

(x, y)

y

Figure 3.19 Path of a Boat in a River

The components of the boat's velocity in the x-direction and y-direction

are

dy.

dt =-Sr+ Sb smB.

So

dy dy/dt -sr+sbsinB -sr+sb(-y/Jx^2 +y^2 )

dx = dx/dt = -sbcose = -sb(x/Jx2 +y2)

Simplifying, we get

(2) dy sr)x

(^2) + y (^2) + SbY
dx SbX
Graph the solutions of the initial value problems consisting of the differential

equation (2) and the initial condition y(W) = 0 on the interval [O, W] where

W = .5 mi, Sr = 3 mi/hr, and (a) Sb = 2 mi/hr, (b) Sb = 3 mi/hr, and

(c) Sb= 4 mi/hr. In each case, decide if the boat lands on the opposite bank.
In those cases in which the boat does land, tell where it lands. How is the
landing point related to Sb versus Sr?