anonymous
  • anonymous
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!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
hey, how's it going?
anonymous
  • anonymous
Dont drown!! haha, hold on.
anonymous
  • anonymous
ok, you helping michael?

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anonymous
  • anonymous
i got 1/6 radians/sec as an answer. i don't even know if i was suppose to get radians
anonymous
  • 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
anonymous
  • 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
amistre64
  • 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
anonymous
  • 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
amistre64
  • amistre64
thanx for the check :)
amistre64
  • 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 :)

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