# need help.. related rates

• November 5th 2006, 10:51 AM
bobby77
need help.. related rates
A small boy standing in a flat field watches a balloon ise at a distance.The ballon leaves the ground 500m away from the boy and rises vertically at a rate of 14m/ minute.At what rate is the angle of inclination of the boys line of sight (theta in radians) increases at the instant when the balloon is exactly 500m above the ground.
• November 5th 2006, 11:13 AM
CaptainBlack
Quote:

Originally Posted by bobby77
A small boy standing in a flat field watches a balloon ise at a distance.The ballon leaves the ground 500m away from the boy and rises vertically at a rate of 14m/ minute.At what rate is the angle of inclination of the boys line of sight (theta in radians) increases at the instant when the balloon is exactly 500m above the ground.

Assume the boy is of negligable height, then:

$
\tan(\theta)=\frac{h}{500}
$

where $h$ is the height of the balloon. Then:

$
\frac{d}{dt} \tan(\theta)=\frac{1}{500} \frac{dh}{dt}
$

So:

$
\sec^2(\theta) \frac{d\theta}{dt}=\frac{1}{500} \frac{dh}{dt}
$
.

When the height is $500\ m$ $\theta=\pi/4$ so:

$
\frac{d\theta}{dt}=\cos^2(\pi/2) \frac{1}{500} 14 \approx 0.0198\ \mbox{radians/s}
$

RonL