Rate of Change Question6

In this page rate of change question6 we are going to see solution of some practice question of the worksheet.

Question 6:

At noon, ship A is 100 km west of ship B. Ship A is sailing east at 35 km/hr and ship B us sailing north at 25 km/hr. How fast is the distance between the ship changing at 4.00 p.m

Solution:

Let P and Q are the starting position of the ships A and B.

To solve this problem first let us draw the rough diagram for the above situation. Let "x" be the distance between Q and A. Let "y" be the distance between Q and B. Let "z" be the distance between the points A and B.

z² = x² + y²

here the distance of x,y and z are changing with respect to time.

2 z (dz/dt) = 2 x (dx/dt) + 2 y (dy/dt)

dividing the whole equation by 2

z (dz/dt) = x (dx/dt) + y (dy/dt)

dx/dt = speed of ship A = 35 km/hr

dy/dt = speed of B = 25 km/hr

x = 40  y = 100

z = √(40)² + (100)²

z = √1600 + 10000

z = √2600

z = 20√29

here we have to find the distance between the ship changing at 4.00 p.m. From this we come to know that we have to find the value of dz/dt.

20√29 (dz/dt) = 40 (35) + (100) (25)

20√29 (dz/dt) = 1400 + 2500

20√29 (dz/dt) = 3900

          dz/dt = 3900 /(20√29)

          dz/dt = 195/√29

There the rate of change of distance between two ships is 195/√29 km/hr