Question

An aircraft is flying horizontally at a constant height of 4000 ft above a fixed observation point P. At a certain instant the angle of elevation theta is 30 degrees and decreasing, and the speed of the aircraft is 300 mi/hr.

A) How fast is theta decreasing at this instant?

B) How fast is the distance between the aircraft and the observation point changing at this instant?

Answer 1

How do you do b?

Answer 2

For part a I got -11/ 400

Answer 3

You need to differentiate implicitly to find d(theta)/dt and dh/dt.

Answer 4

For d(theta)/dt, I got -11/400

Answer 5

I already did part A, I just need help on part B, please

Answer 6

Is that correct or is that just what you got?

Answer 7

@Psymon -11/ 400 is what I got for part A. It could also be -0.0275 rad/ sec. I think this is correct

Answer 8

Nvm.

Answer 9

@Psymon I found the question online and they got that answer

Answer 10

That was dumb, lol x_x

Answer 11

What was dumb?

Answer 12

My suggestion for part B, lol. But yeah, Im not going to be able to get part A, not sure how ya gotit. Part B I might have. Just dont know if its right.

Answer 13

I did part A like this:
tan(theta)=4000/x
tan(theta)=4000X^-1
do deriv
(sec^2 theta)(dtheta/dt)=-4000x^-2 (dx/dt)
(4/3)(d theta/dt)=-4000(sqrt{3))^-2 (440)
d theta/dt=-11/400\

Answer 14

Im not sure where you even got sqrt(3) and the 440 from.

Answer 15

You turn 300 mi/h into 440 ft/sec so that the units are the same because of the height 4000 ft

Answer 16

Yeah, my mind wouldve never caught that. Alright, I wont bother you with my part B then.

Answer 17

Noobs xD

Answer 18

A solution using Mathematica for the calculations is attached.