Question

An airplane is flying in a horizontal, straight-line path. The speed of the airplane is 100 m/s, and it's altitude is 1000m. What is the rate of change of the angle of elevation, theta, when the horizontal distance from a reference point P on the ground is 2000m?

Answer 1

differentiate both sides with respect to time...
use the givens (dx/dt = 100 m/s and x=2000m to sub.s in - you'll also need to use x to find the value of theta at that particular moment)

Answer 2

Can you explain what you mean by differntiating both sides? are yousupposed to end up with something like 2x(dx/dt) + (2)2000(dy/dt) = (hypotenuse)?

Answer 3

not really. we don't care about the hypotenuse here. just the angle

Answer 4

\[\frac{ d }{ dt} (\tan \theta = 1000/x)\]

Answer 5

\[\sec^2 \theta \frac{ d \theta }{ dt} = \frac{ -1 }{ x^{2}} \frac{ dx }{dt }\]

Answer 6

dx/dt is given, x is given.... what's theta when x =2000?

Answer 7

\[\sec^2\theta(d \theta/ dt) = -1/4000\] I got this far, but I am confused on how to solve for theta without knowing what dtheta/dt is

Answer 8

you're asked to *find* d(theta) /dt

Answer 9

to do that you'll need to find theta...

Answer 10

pretty easy...:

Answer 11

RHS should be -1/40000 btw....