Chapter 3: Applications of Differentiation
Section 3.6: Related Rates
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?
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
=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.
a.t=9 ⇒at=9 t+1
Table 3.6.2(a) Data for the given problem
Figure 3.6.2(a) Coordinate system for the ships
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.
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
→differentiate w.r.t. t
→evaluate at point
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
Approximate the second solution
Control-drag the second solution
Context Panel: Approximate≻5
−114⁢990731+970⁢10491851/3+177914⁢990731+970⁢10491851/3+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.
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