calculus - related rates

Question

A police car traveling south toward Sioux Falls at 160 km/h pursues a truck traveling east away from Sioux Falls, Iowa, at 140 km/h.

At time t = 0, he police car is 42 km north and the truck is 55 km east of Sioux Falls. Calculate the rate at which the distance between the vehicles is increasing when t = 10

min. Round your answer to one decimal place and include correct units.

Answer 1

Let's start with what we know and what we want to find out. We know the velocities of the police car and the truck. Let's call those p'(t) and t'(t). And we want to find out the rate at which their distance is increasing at t=10 minutes. Let's call that d'(1/6) (1/6 because t is in hours). I'm also going to assume that Sioux Falls is at the origin, the police car is initially at (0,42), and the truck is initially at (55,0). Now let's write everything down.

p'(t) = -160

t'(t) = 140

d'(1/6) = ???

Note that I'm using -160 because the police car is initially north of the city and is moving south (meaning it's y-position is decreasing).

First, let's integrate p'(t) and t'(t) to get their position functions:

p(t) = -160t + C

t(t) = 140t + C

Since we are also given their initial positions, we can fill in what C is:

p(t) = -160t + 42

t(t) = 140t + 55

Now we can use the distance formula to create a function for d(t):

d(t) = √(p(t)² + t(t)²)

d(t) = √((-160t + 42)² + (140t + 55)²)

d(t) = √(45200t² + 1960t + 4789)

Now we can differentiate d(t) using the chain rule to get d'(t):

d'(t) = 1/2 * (45200t² + 1960t + 4789)^-1/2 * (90400t + 1960)

d'(t) = (45200t + 980) / (45200t² + 1960t + 4789)^1/2

And finally we can plug in 1/6 for t to get our answer (I just used a calculator here):

d'(1/6) ≈ 106.6567

So at t = 10 minutes, the two cars are moving away from each other at a rate of 106.7 km/h. Hopefully I haven't made any silly mistakes along the way.