Question
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.)
Somehow i got to the answer 20562/641 which is approx. 32 radians per second.
Somehow i got to the answer 20562/641 which is approx. 32 radians per second.
Answers
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
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
what did you use pythagoras for and how?
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
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