# related rates

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• June 22nd 2008, 09:57 AM
santatrue
related rates
1. A weather balloon rises straight up in the air at a constant speed of 5 km/h. An observer 2 km from the balloon’s starting point tracks the balloon with a telescope. The angle between the telescope’s line of sight and the horizontal ground is a.Determine the rate at which a is changing when a=

$
\frac{\pi}{3}
$
• June 22nd 2008, 10:05 AM
Jhevon
Quote:

Originally Posted by santatrue
1. A weather balloon rises straight up in the air at a constant speed of 5 km/h. An observer 2 km from the balloon’s starting point tracks the balloon with a telescope. The angle between the telescope’s line of sight and the horizontal ground is a.Determine the rate at which a is changing when a=

$
\frac{\pi}{3}
$

did you draw a diagram? you should always do this with related rates problems, where possible. see the diagram of the situation below. where would you go from here? what equation do you know that can relate a and x? it is a right-triangle, so you should immediately be thinking of things like Pythagoras' theorem and trig ratios
• June 22nd 2008, 10:07 AM
galactus
We can use tangent given that $\frac{dy}{dt}=5, \;\ {\theta}=\frac{\pi}{3}, \;\ x=2$

$tan{\theta}=\frac{y}{x}$

$y=2tan{\theta}$

$\frac{dy}{dt}=2sec^{2}{\theta}\frac{d{\theta}}{dt}$

$5=2sec^{2}(\frac{\pi}{3})\frac{d{\theta}}{dt}$
• June 22nd 2008, 10:13 AM
janvdl
Quote:

Originally Posted by santatrue
1. A weather balloon rises straight up in the air at a constant speed of 5 km/h. An observer 2 km from the balloon’s starting point tracks the balloon with a telescope. The angle between the telescope’s line of sight and the horizontal ground is a.Determine the rate at which a is changing when a=

$
\frac{\pi}{3}
$

$tan a = \frac{y}{x}$

$y = x \ tan \ a = 2 \ tan \ a$

Differentiate.

$y' = (2 sec^2 a)(a')$

$5 = 2 sec^2 (\frac{\pi}{3})(a')$

$a' = \frac{5}{2 sec^2 (\frac{\pi}{3})}$