Hey, I have a question on related rates and I'm a little rusty at them. I have an idea how to do this one but at a certian point i get stuck. The question is:

Two cars leave the intersection, the first travelling south at 40 km/hr and another travelling west at 90 km/hr. How fast is the distance between these cars increasing after 5 hours? Give your answer in km/hr.

So drawing the picture you get a triangle, and you can therefore say x^2 +y^2 = z^2 and we know the rates of which x and y are changing, and we are looking for the rate at which z is changing. So i know that much, the only thing that is getting me stuck is when it says after 5 hours. I thought of something like ifyou said x^2 + y^2 =5^2 then did that out and solved for one you could get something but I just got lost. I know I've done this before but I'm having a major block

z^2 = x^2 + y^2
differentiate, solve for dz/dt in terms of x, y, z, dx/dt, dy/dt You know what those are after five hours (distance=rate*time). z can be found from sqrt of sums of legs squared.

Ahhhh I knew I was missing something. Thanks a bunch, but I also have another question (I'm not doing too good one these realted rates).
A camera, located 2 km from the launch pad, is tracking the rocket that is fired straight up. When the height of the rocket is 20 km, the camera is rotating at the rate of 1/200 radians per second. What is the speed of the rocket at that instant? Give your answer in km/sec.
This is the same sort of thing right? The only thing that is stopping me is the whole radians per second. Would that mean that we would have to get trig into here somewheres. I mean I know its a triangle drawing so it would be possible, but radians really mess me up.

wow seems a bit late to answer

Yes, you're correct! The problem involves related rates, just like the previous one. The key to solving this problem is to understand that the rate at which the camera is rotating is related to the rate at which the angle is changing, which in turn is related to the rate at which the rocket's height is changing.

Let's start by drawing a diagram. We have a right triangle formed by the horizontal distance between the camera and the launch pad (2 km), the height of the rocket (20 km), and the line of sight or hypotenuse (which we'll call h). The camera is rotating, so we'll call the angle it makes with the ground at any given time θ.

Now, we need to differentiate the equation that represents the relationship between the variables. In this case, it is the trigonometric equation: tan(θ) = height/distance.

Differentiating both sides with respect to time (t), we get:
sec^2(θ) * dθ/dt = (d(height)/dt) / distance

We're looking for d(height)/dt, the rate at which the height is changing when it reaches 20 km. We're also given that dθ/dt = 1/200 radians/sec.

To find d(height)/dt, we need to find sec^2(θ). From the right triangle, we can see that sec(θ) = hypotenuse/distance, which is h/2 in this case. So, sec^2(θ) = (h/2)^2.

Now, substitute the known values and solve for d(height)/dt:
(h/2)^2 * (1/200) = d(height)/dt

Simplifying, we have:
(h^2)/400 = d(height)/dt

At the instant when the height of the rocket is 20 km (h = 20 km), we can plug in this value:
(20^2)/400 = d(height)/dt

Simplifying further gives us:
d(height)/dt = 1 km/sec

Therefore, the speed of the rocket at that instant is 1 km/sec.