# Relate Rates Problem

• Dec 3rd 2012, 07:50 PM
mjo
Relate Rates Problem
Here is the question I am having trouble with:
The sun is passing over a 100 m tall building. The angle θ made by the sun with the ground is increasing at a rate of pi/20 rads/min. At what rate is the length of the shadow of the building changing when the shadow is 60 m long? Give your answer in exact values.
So far I have got:
100cosfata=x

dx/dt=-100(cosfata)(dfata/dx)
Dont no where to get cosfata from.
• Dec 4th 2012, 10:05 AM
SujiCorp12345
Re: Relate Rates Problem
"Increasing at a rate of rads/min?"

• Dec 4th 2012, 01:28 PM
mjo
Re: Relate Rates Problem
Oh my gosh I have read this problem so many times I can just picture the numbers when they are not even there. Ugh.
• Dec 4th 2012, 03:25 PM
skeeter
Re: Relate Rates Problem
Quote:

The sun is passing over a 100 m tall building. The angle θ made by the sun with the ground is increasing at a rate of pi/20 rads/min. At what rate is the length of the shadow of the building changing when the shadow is 60 m long?
$\displaystyle \cot{\theta} = \frac{x}{100}$

$\displaystyle -\csc^2{\theta} \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot \frac{dx}{dt}$
• Dec 5th 2012, 01:23 PM
mjo
Re: Relate Rates Problem
But how do I find a value for -csc^2fata
• Dec 5th 2012, 02:19 PM
skeeter
Re: Relate Rates Problem
Quote:

Originally Posted by mjo
But how do I find a value for -csc^2fata

review your basic right triangle trig ...

$\displaystyle \csc{\theta} = \frac{hypotenuse}{opposite} = \frac{\sqrt{60^2+100^2}}{100}$

square the result and change its sign to get the value of
$\displaystyle -\csc^2{\theta}$

btw ... $\displaystyle \theta$ is prounounced "theta" , not "fata"
• Dec 6th 2012, 11:00 AM
mjo
Re: Relate Rates Problem
I found my original problem. Thanks guys