Question

I'm drowing in calculus over here. An observer stands 200 meters from the launch site of a hot-air balloon. Balloon rises 4m/sec at a constant rate. How fast is the angle of elevation of the balloon increasing 30sec after the launch? Help mehh!

Answer 1

hey, how's it going?

Answer 2

Dont drown!! haha, hold on.

Answer 3

ok, you helping michael?

Answer 4

i got 1/6 radians/sec as an answer. i don't even know if i was suppose to get radians

Answer 5

Okay, so first we need to draw a picture. You can turn radians into degrees by multiplying 180/π and degrees into radians by multiplying by π/180.

So, here's the picture

O
- []
| |
| |
4m/s | \O/
| | |
| |____________________Θ/\
|----200m--------|

Just imagine there is a diagonal between the balloon and the person and you have the triangle.

Notice that the balloon is traveling up at a constant rate of 4m/sec, so we can use this to calculate the height of the balloon at 30sec using this equation

speed = distance/time, speed * time = distance, 4m/sec * 30sec = 120m

We will need this in order to solve for the rate of change in the angle.

Now, we can use the tan(Θ) = Opposite/Adjacent to solve for what the angle is at 30sec as well.

\[\tan (\Theta)=O/A, \tan(\Theta)=120m/200m, \tan^{-1}(120m/200m) = \Theta \]
So, Θ=30.9638 degrees @ t=30sec
Now, will make a little change to the diagram by making the opposite side of the angle a variable, like so:

O
- []
| |
| |
y | \O/
| | |
| |____________________Θ/\
|----200m--------|

because we know as the balloon rises its height relative to the ground is changing as well as the angle. So, now we have this equation:
\[\tan(\Theta)=y/200m\]Take the derivative
\[\sec ^{2}(Θ)Θ\prime=y\prime/200m\]We know y'=4m/sec and theta = 30.9638 degrees at t=30sec so all we do is plug those in and solve for the rate of change in theta, i.e. Θ'
\[Θ\prime=(4m/s) / (200m*\sec^{2}(30.9638))=(4m/s * \cos^{2}(30.9638))/200m\]
because sec^2(Θ) = 1 / cos^2(Θ)
So,\[Θ'=.0147059\]degrees/sec or \[Θ'=.0147059*π/180=0.257\]radians/sec

Answer 6

whoops i meant 1/68 radians per sec not 1/6 so i guess i got the same answer just thought it was radians not degrees which confuses me a little

Answer 7

tan(a) = y/200

(d/dt)(tan(a)) = (d/dt)(y/200)

(da/dt) sec^2(a) = (dy/dt)(1/200)

da/dt = a' and dy/dt = y'

a' sec^2(a) = y'/200

solve for a'

a' = y'/200(sec^2(a))

y' = 4

a' = 4/200(sec^2(a))

a' = 1/50(sec^2(a))

sec(a) = "c"/200 where c=sqrt(a^2 + b^2)

c = sqrt(200^2 + 120^2) ...30sec(4m/sec) = 120

c = sqrt(40000 + 14400) = sqrt(54400)

c = 233.24

sec(a) = 233.24/200

sec(a)^2 = 1.08

a' = 1/[50(1.08)]

a' = 0.0185 radians per sec if I did it right :) might need to double check meself tho

Answer 8

You can do it that way, but that's the long way.
You made an error at sec^2(a) = 1.08
(233.24/200)^2 = 1.36002
a' = 1/(50 * 1.36002) = .0147057 radians/sec

I made an error to saying that .0147057 was degrees/sec when it should have been radians/sec

Answer 9

thanx for the check :)

Answer 10

After getting some sleep, yeah....i went for a long walkabout on that one :)

tan(a) = 120/200 = 12/20 = 6/10 = 3/5

which makes the triangle 3^2 + 5^2 = c^2 and c=sqrt(34).

sec(a) = sqrt(34)/5
sec^2(a) = 34/25

a' = 4/(200(34)/25) = 4/(8(34)) = 1/(2(34)) = 1/68

a' = (1/68) radians per sec; or if we want the french version:

a' = 0.0147058823529412 radians per sec :)