Two people start from the same point. One walks east at 1 mi/h and the other walks northeast at 2 mi/h. How fast is the distance between the people changing after 15 minutes?

Incorrect

If you had checked my arithmetic like I suggested you would have found my error to be in

d^2 = 5(.25)^2 - 2√2(.25)^2 = .489277
It should have been
d^2 = 5(.25)^2 - 2√2(.25)^2 = .135723
then
d = .3684

I will let you make the rest of the corrections.

BTW, I also used km/h instead of mi/h, but that has no effect on the calculation

Well, if I were to guess, I'd say the distance between the two people is definitely changing at a speed of "too slow for a drag race." But in all seriousness, let's calculate it!

To solve this, we can use the Pythagorean theorem. After 15 minutes, which is 1/4 of an hour, the person walking east would have traveled 1/4 mile (since 1 mile per hour * 1/4 hour = 1/4 mile). The person walking northeast would have traveled 2/4 mile (since 2 miles per hour * 1/4 hour = 2/4 mile).

Now, we can use the Pythagorean theorem to find the distance between the two people after 15 minutes:
Distance = √((1/4)^2 + (2/4)^2)

After some calculations, we find that the distance between the two people is 1/2 mile.

Now, we need to find how fast this distance is changing. To do that, we can differentiate the distance equation with respect to time (t):

d(Distance)/dt = d(√((1/4)^2 + (2/4)^2))/dt

After more calculations, we find that the rate of change of the distance between the two people after 15 minutes is 0.

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

Let's assume that the starting point is the origin (0,0) on a coordinate plane. We can represent the person walking east as moving along the x-axis and the person walking northeast as moving along a path that forms a 45-degree angle with the positive x-axis.

The position of the person walking east at any given time t (in hours) can be represented by the coordinates (1t, 0), while the position of the person walking northeast can be represented by the coordinates (2t cos(45°), 2t sin(45°)). Here, we use the fact that the person walking northeast is moving at a speed of 2 mi/h, and the angle between their path and the positive x-axis is 45 degrees.

To find the distance between the two people, we can use the distance formula:

Distance^2 = [(2t cos(45°) - 1t)^2 + (2t sin(45°) - 0)^2]

= (2t cos(45°) - 1t)^2 + (2t sin(45°))^2

= (2t cos(45°))^2 - 2t cos(45°) * 1t + (1t)^2 + (2t sin(45°))^2

= 4t^2(cos^2(45°)) - 2t^2cos(45°) + t^2 + 4t^2(sin^2(45°))

= 4t^2(0.5) - 2t^2(0.7071) + t^2 + 4t^2(0.5)

= t^2 - 0.7071t^2 + t^2 + t^2

= 3t^2 - 0.7071t^2

= 2.2929t^2

Now, we can find the rate at which the distance is changing by taking the derivative of the distance with respect to time (t), and then substitute t=0.25 (15 minutes = 0.25 hours):

d(Distance^2)/dt = d(2.2929t^2)/dt

= 2 * 2.2929 * t

= 4.5858t

Plugging in t = 0.25:

d(Distance^2)/dt = 4.5858 * 0.25

= 1.1465

Therefore, the rate at which the distance between the two people is changing after 15 minutes is approximately 1.1465 mi/h.

At a time of t hours

let the distance covered by the eastbound person be 1t km
let the distance covered by the northeastbound person be 2t km
let the distance between them be d km
by the Cosine Law
d^2 = t^2 + 4t2 - 2(t)2t)cos45°
= 5t^2 - 4t^2(√2/2)
= 5t^2 - 2√2 t^2
when t = 15 min or .25 hours
d^2 = 5(.25)^2 - 2√2(.25)^2 = .489277
d = .6995

2d dd/dt = 10t -4√2t
dd/dt = (5t - 2√2t)/d
= (5(.25) - 2√2(.25))/.6995 = .776

at that moment they are separating at .776 km/h

check my arithmetic