#### 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.