Rate of change of an angle...?

Question

A helicopter rises at the rate of 8 feet per second from a point on the ground 60 feet from an observer.

Find the rate of change of the angle of elevation when the helicopter is 25 feet above the ground.

Explain your work, and good luck! :)

Answer 1

let dh/dt = 8 ft/sec

angle = x and so change in angle is dx/dt

From right triangle trig, tan(x) = h/60 so
taking the derivative wrt t we get sec^2(x) dx/dt = 1/60 dh/dt

and we know dh/dt = 8 ft/sec so we have

sec^2 (x) dx/dt = 8/60

Now when h= 25 the direct line of site distance from the observer to the heliocopter is sqrt(25^2 + 60^2) which I hope is 65 feet, so the sec(x) at that moment is 16/60 or 13/12...
the square of that is 169/144 and plugging that back into
sec^2 (x) dx/dt = 8/60 we have

169/144 dx/dt = 8/60

or dx/dt=(8/60)(144/169) which works out to be about .113 radians per second.. or about 6.47 degrees per second..

OK, bonus time... here is a way to check these kinds of problems... Calculus is about infinitesimals, so pick a really really small amount of time, and see if the answer makes sense..Let's use .01 seconds...

When the 'copter is 25 feet high the angle is 22.61986 degrees... Atan(25/60)
now in .01 seconds it will rise .08 feet, so now it is 25.08 feet high and the angle is atan(25.08/60) = 22.68492 degrees.. Ok that is a change of 22.68492 - 22.61986 = .066 degrees in .01 seconds, so in 1 second we would have 6.6 degrees per second... well that is closer than I usually get on approximations, so I'm thinking I had to be close...

Answer 2

Ok, it will help you if you first draw a right triangle with a base of 60 ft. The right side of the triangle should come up x ft., and the angle in the lower left hand corner should be theta.

Ok, so you know that helicopter is rising at 8 ft. per second (dx/dt in other words) and you know that theta depends on where the helicopter is. You can now relate the two by the following equation:

theta = tan^-1(x/60) (this should make sense if you do the trig)

Now we need to take the derivative of theta to see how theta changes with time. Since the derivative of tan^-1 is 1/(1+x^2), the derivative of theta becomes the following using the chain rule:

dtheta/dt = (dx/dt) / ( 60 (1 + (x/60)^2) )

Simplifying a little bit gives us:

dtheta/dt = 3600(8) / (216000 + 60x^2)
(I plugged in dx/dt = 8 here and multiplied the num and denom by 60^2)

Now we have a function that relates the change in angle to the height of the helicopter. So all we need to do is substitute 25 ft. in for x and we're done. I get 0.1136 rad/sec or about 6.5 degrees/sec.

I think this is correct, but it has been awhile since I have done this stuff.

Answer 3

h(t) the height the helicopter is
at a given time t.
Then h'(t)=8 ft/sec
a(t) the angle that observer sees
the helicopter.
Then tana(t)=h(t)/60
a(t)=arctan(h(t)/60)
a'(t)=[1/[h(t)/60]^2]*h'(t)/60
For h(t)=25
a'(t)=[1/[25/60]^2]*8/60=
0.1136

Answer 4

we now hights and distance, or opposite and adj of a right triangle. express h as a function of time and the angle as a function of t
h=8t
theta = arctan 8t/60 = arctan t/7.5
d theta/dt = (1/7.5) x [1/(1+{t/7.5}^2)]
t at h=25 is 3.125
plug and chug...