At noon, ship A is 110 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 15 km/h. How fast is the distance between the ships changing at 4:00 PM?

Question

I tried working this out and it keeps telling me I'm wrong

Answer 1

Hey Nicole,

This problem has a lot of steps and I would encourage you to draw out the steps on paper (or even graph paper) to see the whole problem.

I started by placing (imagining?) Ship B on the origin of a coordinate plane (0,0). This step helps with directions. We can super impose our cardinal directions over the plane. In this case, we start with Ship A at point (-110,0). Because Ship A is traveling East, it comes closer and closer to the origin and thus shortens its horizontal distance to Ship B as it travels. 4 hours later, we have new values for where the ships are in the plane and thus new values for the right triangle formed between the Ships. We label the horizontal leg (on the x axis) α and the vertical leg (on the y axis) β at 4:00pm. Multiplying the speed by the time traveled, we have

α = -110 + 4(25) = -10

β= 4(15) = 60

The α value is negative because the ship is traveling eastward. We also define our values of the velocities of the ships.

(∂α/∂t) = 25 km/h

(∂β/∂t) = 15 km/h

We set up our formula for the distance between the ships using the distance formula

s = √(α² + β²)

We differentiate both sides using the chain rule and exponent rules (the square root is to the power 1/2)

(∂s/∂t)= 1/(2√(α² + β²)) * (∂/∂t)(α² + β²)

(∂s/∂t)= 1/(2√(α² + β²)) (2α(∂α/∂t) + 2β(∂β/∂t))

(∂s/∂t)= (α(∂α/∂t) + β(∂β/∂t))/(√(α² + β²))

We substitute all values in and solve:

(∂s/∂t)= (α(∂α/∂t) + β(∂β/∂t))/(√(α² + β²))

(∂s/∂t)= ((-10)(25) + (60)(15))/(√((-10)² + (60)²))

(∂s/∂t)= (650)/(√(3700))

(∂s/∂t) ≈ 10.6859 km/h

I hope this helped.

-Pete