## anonymous 5 years ago 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!

1. anonymous

hey, how's it going?

2. anonymous

Dont drown!! haha, hold on.

3. anonymous

ok, you helping michael?

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. amistre64

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

8. anonymous

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

9. amistre64

thanx for the check :)

10. amistre64

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 :)