Chapter 3: Applications of Differentiation
Section 3.6: Related Rates
Example 3.6.2
At 1:00 PM a ship traveling at 9 knots sets sail north-east along a line that makes a 30° angle with a line running due east. An hour later, a second ship sets sail due north, and at 11:00 PM, the distance between the ships is observed to be increasing at a rate of 97/7 knots. Assuming it travels at constant speed, how fast is the north-bound ship traveling?
Solution
Mathematical Solution
In the labeled diagram in Figure 3.6.2(a), at is the distance sailed by the first ship; bt, the distance sailed by the second. The distance between the ships is dt.
Let t=0 be the time at which the second ship starts. This is 2:00, so at 11:00, this second ship will have sailed for 9 hours. (The first ship will have sailed for 10 hours.)
Since b. is not known, call this speed s.
At time t, the first ship will have sailed at=9t+1 nautical miles; the second ship, bt=s t.
See Table 3.6.2(a) for a summary of this data.
Apply the law of cosines to the triangle shown in Figure 3.6.2(a), obtaining
dt
=at2+bt2−2 at bt cosπ/3
=81 t+12+s t2−29⋅t+1⋅s⋅t⋅cosπ/3
Set d.9=97/7 and solve for s=16.
Given:
a.t=9 ⇒at=9 t+1
b.t=s⇒bt=s t
d.9=97/7
Find s=b.t=b.9
p1:=Student:-VectorCalculus:-PlotVector([<0,2>,<3*cos(Pi/6),3*sin(Pi/6)>],color=black,width=.1,head_length=.5): p2:=plot([[3*cos(Pi/6),3*sin(Pi/6)],[0,2]],style=line,color=black): p3:=plots:-textplot({[1.17,1.9,typeset(d(t))],[.19,1,typeset(b(t))],[1.1,.89,typeset(a(t))]},font=[Times,12]): p4:=plots:-textplot({[.7,.14,typeset(Pi/6)],[.19,.37,typeset(Pi/3)]}): plots:-display(p1,p2,p3,p4,scaling=constrained,tickmarks=[0,0]); unassign('p1','p2','p3','p4'):
Table 3.6.2(a) Data for the given problem
Figure 3.6.2(a) Coordinate system for the ships
Maple Solution
Construct dt and obtain d.9
Construct the radical on the right-hand side of the law of cosines, replacing at and bt with the expressions listed in Table 3.6.2(a).
Press the Enter key.
Context Panel: Differentiate≻With Respect To≻t
Context Panel: Evaluate at a Point≻t=9
81 t+12+s t2−29⋅t+1⋅s⋅t⋅cosπ/3
81t2+162t+81+s2t2−9st2−9st
→differentiate w.r.t. t
12162t+162+2s2t−18st−9s81t2+162t+81+s2t2−9st2−9st
→evaluate at point
121620+18s2−171s8100+81s2−810s
Solve d.9=97/7 for s
Using its equation label, set the derivative equal to 97/7 and press the Enter key.
Context Panel: Solve≻Solve
=97/7
121620+18s2−171s8100+81s2−810s=977
→solve
s=16,s=−114990731+97010491851/3+177914990731+97010491851/3+1
Approximate the second solution
Control-drag the second solution
Context Panel: Approximate≻5
−114990731+97010491851/3+177914990731+97010491851/3+1→at 5 digits−6.9646
The second solution is negative, and is therefore rejected. The first solution, namely, s=16, gives the speed of the north-bound ship.
<< Previous Example Section 3.6 Next Example >>
© Maplesoft, a division of Waterloo Maple Inc., 2024. All rights reserved. This product is protected by copyright and distributed under licenses restricting its use, copying, distribution, and decompilation.
For more information on Maplesoft products and services, visit www.maplesoft.com
Download Help Document
