Solving Related Rates Problem: Finding the Angle of Depression

[kendal12]

[SOLVED] Related Rates Problem

Here is the problem word for word:

"Angela Lansbury displayed her athletic prowess by skydiving out of a hovering helicopter 100ft away from a cliff. However, the chutes fail and she plummets to certain disaster. If her position, in feet, is given by

s(t)=-16t^2 + 15840

find the rate of change of the angle of depression in degrees/sec at t=30.9 seconds for a viewer standing at the edge of the cliff, assuming that his head is 600ft above the floor of the valley below.

Ok, I've tried to do several things, but haven't gotten anywhere. I've plugged in t to the equation and gotten 563.04, but I don't know what to do with it. I've attempted to do some trig, but my teacher then told us that even though we can solve it with trig, we need to use calculus or we won't get credit...

This problem confuses me and I don't know where to start. Any help would be appreciated.

 

[hotvette]

rate of change of the angle of depression

This should give you a hint. Rate of change means derivative. So, set up your geometry as the observer and derive an equation of the viewing angle, \Theta in terms of t and other parameters of the free falling object. Once you have the equation, take the derivative with respect to t, and you have your answer.

 

[kendal12]

I understand that this is what I need to do, but I have no idea how to even start getting the eq...

 

[hotvette]

Well, let's see. Here's what I would do.

1. Draw a picture illustrating the problem. Draw a cliff and consider the point of the cliff to be the observer. Some distance away (doesn't matter how far) draw the sky diver as a small circle, then draw the trajectory of the sky diver (straight down) as a dotted line. Draw the position of the skydiver at the start and then again at time t=30.9 (the drawing doesn't have to be to scale and the postions don't have to be exact. We're just trying to get a graphical depiction of what's going on).

2. List what you know from what you are given:
- height of the observer from the ground
- horizontal distance the sky diver is away from the observer
- height of the initial position of the skydiver (hint: at t = 0)
- height of the skydiver at the point of interest (i.e. at t = 30.9)

3. Connect the lines up and you have one or more triangles, where the vertical leg is described by the equation supplied.

4. Use geometry/trig to write an equation relating \Theta to the legs of the triangle (1 being the equation of the skydiver fall)

5. Solve the equation for \Theta in terms of t and other known quantities.

I didn't solve the problem entirely but am just describing a reasonable approach that should get you going in the right direction.

P.S. One other hint that might be valuable. I believe angle of depression means the angle from a horizontal reference line to the observer. This means you'll have a right triangle to deal with. Indeed, you've caculated that the skydiver is at a height of 563 ft at the time in question, which is lower than the height of the cliff, which supports my suspicion. Sometimes, you have to look for subtle clues in the wording.

 

[HallsofIvy]

What does your teacher have against Angela Landsbury? Yes, a few years ago it was difficult to turn on the t.v. without seeing her, but fortunately that's toned down recently. Let the poor woman live!