# 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.

Here you know the angle, and you are given dTheta/dt. You are looking for dh/dt

You know TanTheta=h/2, and you know dtheta/dt when h=2-

start with TanTheta=h/2
take the derivative with respect to time. It will come out rather simply.

sec^2theta * dTheta/dt = 1/2 dh/dt

You know sec from the triangle. (Pyth theorem)

## To solve this related rates problem, we need to use trigonometry and apply the concept of derivatives.

In this case, we have a right triangle formed by the camera, the rocket, and the height of the rocket. Let's label the sides of the triangle as follows:
- The opposite side (h) represents the height of the rocket.
- The adjacent side (2 km) represents the distance between the camera and the launch pad.

First, notice that the tangent of the angle theta (TanTheta) is equal to the opposite side (h) divided by the adjacent side (2 km). So, we have TanTheta = h/2.

Now, let's differentiate both sides of this equation with respect to time. This is because we are given the rate of change of theta with respect to time (dTheta/dt), and we want to find the rate of change of h with respect to time (dh/dt). Differentiating both sides using the chain rule, we get:

sec^2(theta) * dTheta/dt = 1/2 * dh/dt

Since we know TanTheta = h/2, we can use the Pythagorean theorem to find sec(theta), which is equal to sqrt((h/2)^2 + (2 km)^2).

So, we have sec(theta) = sqrt((h/2)^2 + (2 km)^2).

Substituting this value into the equation above, we get:

(sqrt((h/2)^2 + (2 km)^2))^2 * dTheta/dt = 1/2 * dh/dt

Simplifying further, we have:

((h/2)^2 + (2 km)^2) * dTheta/dt = (1/2) * dh/dt

Now, we are given that dTheta/dt = 1/200 radians per second. Let's substitute this value into the equation:

((h/2)^2 + (2 km)^2) * (1/200) = (1/2) * dh/dt

Next, we are given that h = 20 km. Let's substitute this value into the equation:

((20 km/2)^2 + (2 km)^2) * (1/200) = (1/2) * dh/dt

Simplifying the equation further:

((10 km)^2 + (2 km)^2) * (1/200) = (1/2) * dh/dt

Solving for dh/dt, we have:

((100 km^2) + (4 km^2)) * (1/200) = (1/2) * dh/dt

(104 km^2) * (1/200) = (1/2) * dh/dt

104 km^2 / 200 = (1/2) * dh/dt

0.52 km = (1/2) * dh/dt

Therefore, the speed of the rocket at that instant is 0.52 km/sec.