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Hey.. this is my first time posting, so far I havent had any troubles in my calc class but the chapter we are on now is confusing me ( probably because i have been sick therefore missing a lot of school..)

here goes:

A 10-foot ladder is leaning against a house when its base starts to slide away. By the time it is 6 feet from the house the base is moving at a rate of 4 feet/sec. At what rate is the angle between the side of the house and the ladder changing at that time?

She had one my friends send these problems to me... the teacher said to do the following, cause it will help:
A) identify variables/constants
B) Whats given?
C) What are u trying to find at what moment?
D)Answer question with a sentence

Thank you guys so much!
 
Why are you not following the steps.

Let's see what you get for Part A.
 
well... the 10 foot ladder is a constant value. I'm confused for the dv/dt things.. so is 4 feet/sec dr/dt? and is like dh/dt 6??
 
Here's, I hope, a shove in the right direction.

You want \(\displaystyle \frac{d{\theta}}{dt}\), given \(\displaystyle \frac{dx}{dt}=4 ft/sec.\)

\(\displaystyle sin({\theta})=\frac{x}{10}\)

\(\displaystyle cos({\theta})\frac{d{\theta}}{dt}=\frac{1}{10}\frac{dx}{dt}\)

\(\displaystyle \frac{d{\theta}}{dt}=\frac{1}{10cos({\theta})}\frac{dx}{dt}\)

Also, the top of the ladder is 8 feet above the ground when x=6.
 
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