from a handpicked tutor in LIVE 1-to-1 classes

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

**Solution:**

Consider x and y as the initial position of the ships A and B

Consider x as the distance between A and x

y as the distance between B and x

z as the distance between A and B

Using pythagoras theorem,

z^{2} = x^{2} + y^{2}

It can be written as

Differentiating both sides wrt t,

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

Divide the equation by 2

z dz/dt = dx/dt + dy/dt --- (1)

dx/dt = 35 km/hr

dy/dt = 25 km/hr

When t = 4 hours,

x = 40, y = 100 and

z = √(x^{2} + y^{2})

Substituting the values

z = √(40^{2} + 100^{2})

z = √(1600 + 10000)

z = √11600 = 20√29

Substituting it in equation (1)

20√29 dz/dt = 40 × 35 + 100 × 25

dz/dt = 3900/20√29

dz/dt = 195/√29 km/hr

Therefore, the distance between the ships changing at 4.00 p.m is 195/√29 km/hr.

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

**Summary:**

At noon, ship A is 100 km west of ship B. Ship A is sailing east at 35 km/hr and ship B is sailing north at 25 km/hr. The distance between the ships changing at 4.00 p.m. is 195/√29 km/hr.

visual curriculum