To determine
To find: The quantities in the given problem.
Answer
The quantities in the given problem is dxdt=35 km/h and dydy=25 km/h.
Explanation
Given that at the afternoon the ship A is 150 km west of the ship B; the ship A is sailing East at 35 km/h and the ship B is sailing North at 25 km/h.
Let time t (in hour) and x is the distance traveled by ship A (in km), and y is the distance traveled by the ship B (in km), then we are given that dxdt=35 km/h and dydy=25 km/h.
To determine
To find: The unknown.
Explanation
Formula used:
Pythagorean Theorem
Calculation:
Unknown: The rate at which the distance between the ships changing at 4:00 pm.
Suppose till 4:00 pm the ship A has been covered x km distance toward West, then (150−x) is the remaining distance between ship A and West end. And in the same the ship B covered y km distance towards North.
To determine
To draw: The picture of the plane the position of the ships at 4 pm.
Explanation
Draw the picture of the plane position of the ships at 4 pm is given below.
From Figure 1, it is observed that the position of the ships at 4 pm where A and B is the position of ships at 4pm.
To determine
To write: The equation that relates the quantities.
Answer
The Equation is dzdt =1z[−(150−x)dxdt+ydydt].
Explanation
Relates the given quantities with an equation:
Apply Pythagorean Theorem on the above right angle triangle,
z2=(150−x)2+y2 (1)
Differentiate equation (1) with respect to the time t,
ddt(z2)=ddt[(150−x)2+y2]2zdzdt =ddt[(150−x)2]+ddt(y2)2zdzdt =−2(150−x)dxdt+2ydydtdzdt =1z[−(150−x)dxdt+ydydt]
To determine
To find: How fast is the distance between the ships changes at 4:00 pm.
Answer
The required rate at which the distance between the ships changing at 4:00pm is dzdt=215101 km/h≈21.4 km/h.
Explanation
At 4pm,
The distance covered by the ship A:
x=4×35=140
The distance covered by the ship B:
y=4×25=100
And the distance between the ships:
z=(150−140)2+1002=102+1002=100+10000=10,100
Substitute 140 for x and 100 for y and 35 for dxdt and 25 for dydt and 10,100 for z in dzdt.
dzdt=110,100[−(150−140)(35)+(100)(25)]=110,100[−(10)(35)+(100)(25)]=110,100[−350+2500]=110,100(2150)
Therefore, on further simplification the required rate of change between the distance of ships A and B as follows.
dzdt=2150(101)(100)=215010101=215101 km/h≈21.4 km/h
Therefore, the required rate at which the distance between the ships changing at 4:00 PM is dzdt=215101 km/h≈21.4 km/h.