Two people start at the same point One walks east at 2mph and the other walks southwest at 3mph. How fast is the distance between them changing after 30 minutes?

At a time of t hrs,

let the distance travelled by the eastbound person be 2t miles
and the distance of the other person be 3t miles
let the distance between them be d miles, and of course the angle between them is 135°
d^2 = (2t)^2 + (3t)^2 - 2(2t)(3t)cos135°
= 13t^2 - 12t^2(-√2/2)
= 13t^2 + 6√2 t^2

2d dd/dt = 26t + 12√2 t
dd/dt = (13t + 6√2 t)/d

plug in t = 1/2 or .5 into d^2 = ... to find d, then sub into dd/dt

I will let you do all that button-pushing

To find the rate at which the distance between the two people is changing, we can use the concept of relative motion.

Let's set up a coordinate system, where the starting point is the origin (0,0). The person walking east can be represented by the equation x = 2t, where x is the distance traveled by the person in the east direction, and t is the time in hours. Since they walk at a rate of 2 mph, their speed is constant.

The person walking southwest can be represented by the equation x = -3t, y = -3t, where x and y are the distances traveled by the person in the southwest direction. Again, the speed is constant at 3 mph.

To find the distance between them at any given time, we can use the distance formula:

distance = √((x2 - x1)^2 + (y2 - y1)^2)

Plugging in the values, we get:

distance = √((2t - (-3t))^2 + (0 - (-3t))^2)
= √((5t)^2 + (3t)^2)
= √(25t^2 + 9t^2)
= √(34t^2)
= √34t

Now, we can find how the distance is changing over time by taking the derivative of the distance formula with respect to time:

d(distance)/dt = d(√34t)/dt
= (1/2√34) * d(34t)/dt
= (1/2√34) * 34
= 17/√34

Therefore, the rate at which the distance between them is changing after 30 minutes (t = 0.5 hours) is:

d(distance)/dt = 17/√34 ≈ 2.93 mph

Hence, the distance between them is changing at a rate of approximately 2.93 mph after 30 minutes.