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

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.

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

Answer 4

To find the rate at which theta is changing, we can use the relationship between the angle theta and the altitude h of the rocket.

Given that the altitude of the rocket is given by h = 4t^2, we can find the value of t when the rocket is 400 meters away from us.

Since the rocket is 400 meters away from the launch site, and the altitude at time t is given by h = 4t^2, we can set up the following equation:

400 = 4t^2

Dividing both sides by 4, we get:

100 = t^2

Taking the square root of both sides, we find:

t = 10

So, when the rocket is 400 meters away from us, the time elapsed is 10 seconds.

To find the rate at which theta is changing at that particular moment, we need to find d(theta)/dt, the derivative of theta with respect to time t.

Since we know the altitude of the rocket is h = 4t^2, we can determine the value of theta by using the tangent function:

tan(theta) = h / 400

Substituting h = 4t^2 and t = 10, we have:

tan(theta) = (4(10)^2) / 400

Simplifying, we get:

tan(theta) = 1

To find the derivative d(theta)/dt, we can differentiate both sides of the equation tan(theta) = 1 with respect to t:

sec^2(theta) * d(theta)/dt = 0

Since sec^2(theta) is never zero, we can conclude that d(theta)/dt must be zero.

Therefore, the rate at which theta is changing at that particular moment is zero radians per second, not 32 radians per second as you mentioned.

Please double-check your calculations or the given question to ensure the accuracy of the answer.