Question

An aircraft is flying horizontally at a constant height of 4,000 ft above a fixed observation point. At a certain instant, the speed of the aircraft is 300 mi/hr and the angle of elevation is 30 degrees.

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? Express the result in unites of ft/s. Remember, 1 mile is 5280 ft.

Answer 1

Related rates, right?

Answer 2

Yes :)

Answer 3

\[
\tan \theta = \frac{4,000}{x}
\]

Answer 4

Differentiate both sides.

Answer 5

Then you get this tanΘ=4000x^−1

Answer 6

Yeah. That is our equation. Now just differentiate.

Answer 7

With respect to \(t\).

Answer 8

sec^2θ (dθ/dt)=−4000x^−2 (dx/dt)

Answer 9

Good.

Answer 10

I got
dθ/dt=−11/400

Answer 11

"the speed of the aircraft is 300 mi/hr and the angle of elevation is 30 degrees. "

This means \(\theta=30^\circ\) and \(x'=300\).

Answer 12

What did you get for \(x\)?

Answer 13

x=4000√3

Answer 14

Great

Answer 15

So is dθ/dt correct?

Answer 16

http://www.wolframalpha.com/input/?i=sec%5E2%2830%29+*y+%3D+300*+-4000%2F%284000*sqrt%283%29%29%5E2+

I'm getting \(-3/160\)

Answer 17

Don't you change 300 mi into feet because the height is 4000 ft

Answer 18

So that the units are the same

Answer 19

Oh, I didn't notice.

Answer 20

I guess it would be:
http://www.wolframalpha.com/input/?i=sec%5E2%2830%29+*y+%3D+%281584000%29%2F%2860%5E2%29+*+-4000%284000*sqrt%283%29%29%5E%28-2%29+

Answer 21

-11/400 in radians.

Answer 22

ok

Answer 23

I think you can do these problems.

Answer 24

I do not know how to do the next part

Answer 25

I'm not sure if I go the right answer

Answer 26

*got

Answer 27

\[
z^2=x^2+y^2
\]Differentiate.

Answer 28

oh ok

Answer 29

Thanks! :)