Chapter 3: Applications of Differentiation
Section 3.6: Related Rates
Example 3.6.1
At 1:00 PM a ship sets sail due north at a speed of 14 knots, and an hour later a second ship sets sail due east at a speed of 19 knots. At what rate is the distance between these two ships increasing at 7:00 PM?
Solution
Mathematical Solution
Let the ships traverse the coordinate system shown in Figure 3.6.1(a).
Let t=0 be the time at which the second ship starts. (This would be 2:00. By 7:00, the second ship will have sailed for 5 hours.)
Then, make the analysis shown in Table 3.6.1(a).
x.=19
y.=14
xt=19 t
yt=14t+1
dt=19 t2+14t+12
Find d.5
p1:=Student:-VectorCalculus:-PlotVector([<3,0>,<0,2>],color=black,width=.1,head_length=.5): p2:=plot([[3,0],[0,2]],style=line,color=black): p3:=plots:-textplot({[1.43,1.2,typeset(d)],[.16,.8,typeset(y)],[1.2,.2,typeset(x)]},font=[Times,12]): plots:-display(p1,p2,p3,scaling=constrained,axes=none); unassign('p1','p2','p3'):
Table 3.6.1(a) Data for the given problem
Figure 3.6.1(a) Coordinate system for the ships
The quantity to be found is the derivative of dt evaluated at t=5, that is,
d.tx=a|f(x)t=5 = 121114t+392557t2+392t+196x=a|f(x)t=5 = 29811608116081 ≐ 23.51 knots
Thus, the rate of separation five hours after the second ship starts is approximately 23.51 knots (nautical miles per hour).
This problem is atypical in that the specific time at which to make the measurement of the related rate is actually given.
Some authors would start the clock when the first ship departs. Under this approach, the distance traveled by the second ship would be xt=19t−1. The choice made in Table 3.6.1(a) exchanges the subtraction for an addition.
Maple Solution
Control-drag dt from Table 3.6.1(a). Press the Enter key.
Context Panel: Differentiate≻With Respect To≻t
Context Panel: Evaluate at a Point (t=5)
Context Panel: Approximate≻5
19 t2+14t+12
557t2+392t+196
→differentiate w.r.t. t
121114t+392557t2+392t+196
→evaluate at point
29811608116081
→at 5 digits
M2≔MutableSeta,b,c
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