# calculus AB help with related rates

• May 22nd 2008, 05:28 AM
ballet_dansur13
calculus AB help with related rates
A hot-air balloon is rising straight up from a level field and is tracked by a range finder 500 ft from lift-off point. the balloon rises at a rate of 15 feet per second.

A) How fast is the direct distance between the range finder and the ballon changing at 5 seconds

B) how fast is the angle of elevation changing when the ballon is 500 feet high

he gave us a diagram of a right trianle with the hypotenus labeled d and the side labled h and the bottom equalling 500

i think i solved part a and i got 2.223 ft but i wasn't sure how to do part b
any help is appreciated
• May 22nd 2008, 05:36 AM
kalagota
get an equation relating the angle with the given data. and since it is a right triangle, you can use SOHCAHTOA.. in fact, you'll just need TOA.
• May 22nd 2008, 07:43 AM
galactus
Part a looks good. Good work.

For part b, use tangent. I will get you started:

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

Differentiate:

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

You know dy/dt. Find theta when y=75 and you're set.