# word problem help

• Dec 21st 2009, 12:24 PM
wisezeta
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
• Dec 21st 2009, 12:36 PM
skeeter
Quote:

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?
• Dec 21st 2009, 12:46 PM
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
• Dec 21st 2009, 01:05 PM
skeeter
Quote:

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 ...

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

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

now, try again to find $\displaystyle \frac{d\theta}{dt}$ when $\displaystyle h = 100$
• Dec 21st 2009, 01:11 PM
wisezeta
I'm still confused. Where do I place 100 and where do i place 10?
• Dec 21st 2009, 01:25 PM
skeeter
Quote:

Originally Posted by wisezeta
I'm still confused. Where do I place 100 and where do i place 10?

$\displaystyle h = 100$ ... you'll need that to figure out $\displaystyle \sec{\theta}$

$\displaystyle \frac{dh}{dt} = 10$ ft/s
• Dec 21st 2009, 01:34 PM
wisezeta
• Dec 21st 2009, 01:44 PM
skeeter
[quote=skeeter;428168]the side that you made 100 (height of the balloon) is variable ...

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

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

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

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

try again
• Dec 21st 2009, 01:47 PM
wisezeta
Oh i see what i did. I took the actual cosine of the ratio. Ok so is the answer .0044?
• Dec 21st 2009, 01:54 PM
skeeter
Quote:

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 ...

$\displaystyle \frac{d\theta}{dt} \approx 0.044$
• Dec 21st 2009, 01:55 PM
wisezeta
Ok i just wrote it down wrong on the computer. Thank you very much