A rocket is launched straight up. At t seconds after being launched, its altitude is given by h = 4t^2 metres. You are on the ground 400 metres from the launch site, watching the rocket. The line of sight from you to the rocket makes an angle theta� radians with the horizontal. Four seconds after launch, by how many radians per second is theta changing? (Give an exact answer.)

Question

Somehow i got to the answer 20562/641 which is approx. 32 radians per second.

Answer 1

let the angle be Ø, so we have

tanØ = h/400
h = 400 tanØ
dh/dt = 400 sec^2 Ø dØ/dt

but dh/dt = 8t

400 sec^2 Ø dØ/dt = 8t
dØ/dt = t/(50sec^2 Ø)

so for the given case of t = 4
h = 4(4^2) = 64
tanØ = 64/400 = 4/25
then using Pythagoras
cosØ = 25/√641
secØ = √641/25
sec^2 Ø = 641/625

so dØ/dt = 4/(50(641/625)) = 50/641
= appr .078 rad/sec

Your answer seems totally unreasonable
that would be 1833° per second, making no sense

Answer 2

what did you use pythagoras for and how?

Answer 3

we were given that t = 4
so we had to "freeze" the situation at t = 4
That gave us tan O = 4/25
so we have a right-angled triangle with opposite 4 and adjacent 25.
Since we need the secant of that angle
and secant = 1/cosine , and cosine = adjacent/hypotenuse, so ...
hypotenuse^2 = 25^2 + 4^2 = 641
etc