1. ## word problem help

A weather balloon is released 60 feet from an observer with a camera. if the balloon is rising at 10 ft. per second how fast is the angle of the camera changing when the balloon is 100 ft. high?

I did the problem and got 14.29 radians per second, but I'm not too sure that i did it right. Thanks in advance

2. Originally Posted by wisezeta
A weather balloon is released 60 feet from an observer with a camera. if the balloon is rising at 10 ft. per second how fast is the angle of the camera changing when the balloon is 100 ft. high?

I did the problem and got 14.29 radians per second, but I'm not too sure that i did it right. Thanks in advance

your solution is incorrect ... what did you do to get it?

3. First I made a right triangle with th ebase as 60, the right leg as 100, the hypotnuse as s, and the angle as theta
then i made the equation tanx=(116.62/60)10
116.62 was my hyponuse
I took the derivative of the equation and got sec^2 x (dx/dt)=(116.62/60)10
I then solved for dx/dt and got my answer

4. Originally Posted by wisezeta
First I made a right triangle with th ebase as 60, the right leg as 100, the hypotnuse as s, and the angle as theta
then i made the equation tanx=(116.62/60)10
116.62 was my hyponuse
I took the derivative of the equation and got sec^2 x (dx/dt)=(116.62/60)10
I then solved for dx/dt and got my answer
the side that you made 100 (height of the balloon) is variable ...

$\frac{d}{dt}\left(\tan{\theta} = \frac{h}{60}\right)$

$\sec^2{\theta} \cdot \frac{d\theta}{dt} = \frac{1}{60} \cdot \frac{dh}{dt}$

now, try again to find $\frac{d\theta}{dt}$ when $h = 100$

5. I'm still confused. Where do I place 100 and where do i place 10?

6. Originally Posted by wisezeta
I'm still confused. Where do I place 100 and where do i place 10?
$h = 100$ ... you'll need that to figure out $\sec{\theta}$

$\frac{dh}{dt} = 10$ ft/s

8. [quote=skeeter;428168]the side that you made 100 (height of the balloon) is variable ...

$\sec^2{\theta} \cdot \frac{d\theta}{dt} = \frac{1}{60} \cdot \frac{dh}{dt}$

$\frac{d\theta}{dt} = \cos^2{\theta} \cdot \frac{1}{60} \cdot \frac{dh}{dt}$

using right triangle trig ... $\cos{\theta} = \frac{60}{\sqrt{60^2+100^2}}$

$\frac{d\theta}{dt} = \left(\frac{60}{\sqrt{60^2+100^2}}\right)^2 \cdot \frac{1}{60} \cdot 10$

try again

9. Oh i see what i did. I took the actual cosine of the ratio. Ok so is the answer .0044?

10. Originally Posted by wisezeta
Oh i see what i did. I took the actual cosine of the ratio. Ok so is the answer .0044?
you're off by a factor of 10 ...

$\frac{d\theta}{dt} \approx 0.044$

11. Ok i just wrote it down wrong on the computer. Thank you very much