A jet flying due north at 16km/min passes 2 km directly above a plan flying due east at 8km/min. How quickly are they separating when the plane is 32 km from the crossover?

What I have done (I am not sure if it is correct.):
r^2=x^2+y^2+z^2
Since the plane has gone 32 meters, 32km/8km/min=4seconds
Therefore for the jet, 16km/min(4seconds)=64km.

So, 32km=x value, 64km=y value and z would be 2 because that is how far they are from each other?

Solve for r^2=x^2+y^2+z^2
= 32^2+64^2+2^2
=5124
r=71.58

So, then,

r^2=x^2+y^2+z^2
2r d(r)/d(t)= 2x dx/dt + 2y dy/dt + 0 (since 2^2 will be a constant)

Now, we subb in the values.

71.58 dr/dt = 32 (8) + 64 (16)

dr/dt = 32 (8) + 64 (16)/71.58

=17.8km/min

Is this done correctly, or did I go wrong somewhere?

Aside from all the carelessness with the units, the math look good.

You have made some mistakes in your calculations. Let's go step-by-step to solve the problem correctly.

First, let's set up a coordinate system where the plane is at (0,0) and the jet is at (0,2) at the start.

We can find the position of the jet at any time t using the equation y = 2 + 16t since it is flying due north at 16km/min.

The position of the plane can be found using the equation x = 8t since it is flying due east at 8km/min.

When the plane is 32 km from the crossover point, its position is (32, 0). We can solve for the value of t using the equation 32 = 8t, which gives us t = 4 minutes.

Now let's find the position of the jet at t = 4 minutes. Substituting t = 4 into the equation, we get y = 2 + 16 * 4 = 66 km.

Next, we need to find the rate at which the jet is moving away from the plane, which is given by the derivative of the distance between them with respect to time.

The distance between them can be found using the Pythagorean theorem:

r^2 = (x - 0)^2 + (y - 2)^2
r^2 = (32 - 0)^2 + (66 - 2)^2
r^2 = 1024 + 4225
r^2 = 5249
r = sqrt(5249)
r ≈ 72.46 km

Taking the derivative of both sides of this equation with respect to time, we get:

2r * d(r)/d(t) = 2(x)(dx/dt) + 2(y - 2)(dy/dt)

Plugging in the known values:

2 * 72.46 * d(r)/d(t) = 2(32)(8) + 2(66 - 2)(16)

Simplifying this equation, we find:

144.92 * d(r)/d(t) = 512 + 1280

Now, solving for the rate of change of distance d(r)/d(t):

d(r)/d(t) = (512 + 1280) / 144.92
d(r)/d(t) = 1792 / 144.92
d(r)/d(t) ≈ 12.36 km/min

Therefore, when the plane is 32 km from the crossover, they are separating at a rate of approximately 12.36 km/min.

Based on the information provided, it seems like the calculations you have done are correct. Let's go through the steps to verify:

First, let's assign variables to the distances travelled by the plane and the jet:
- Let x represent the distance travelled by the plane (32 km)
- Let y represent the distance travelled by the jet (64 km)
- Let z represent the initial separation distance (2 km)

Now, we can use the formula for the distance between two points in three-dimensional space:
r^2 = x^2 + y^2 + z^2

Substituting in the given values, we have:
r^2 = 32^2 + 64^2 + 2^2
r^2 = 1024 + 4096 + 4
r^2 = 5124

Taking the square root of both sides, we get:
r = √5124
r ≈ 71.58 km

Next, we can differentiate the distance equation with respect to time:
2r d(r)/d(t) = 2x dx/dt + 2y dy/dt

Substituting in the known values:
2(71.58) d(r)/d(t) = 2(32)(8) + 2(64)(16)

Simplifying further:
143.16 d(r)/d(t) = 512 + 2048
d(r)/d(t) = (512 + 2048) / 143.16
d(r)/d(t) ≈ 17.8 km/min

Therefore, the planes are separating at a rate of approximately 17.8 km/min when the plane is 32 km from the crossover point. Your answer of 17.8 km/min is correct. Well done!