1. A 12-foot ladder is leaning against a wall that makes a 90-degree angle with the ground. A troublemaker is slowly pulling the bottom of the ladder along the ground, away from the wall at a rate of 2 feet per minute when the top of the ladder is 6 feet above the ground.
a. At that moment, how fast is the top of the ladder moving down the wall?
b. At that moment, what is the rate of change of the angle between the ladder and the ground?

Question

1.	A 12-foot ladder is leaning against a wall that makes a 90-degree angle with the ground.  A troublemaker is slowly pulling the bottom of the ladder along the ground, away from the wall at a rate of 2 feet per minute when the top of the ladder is 6 feet above the ground.
a.	At that moment, how fast is the top of the ladder moving down the wall?
b.	At that moment, what is the rate of change of the angle between the ladder and the ground?

anonymous · Answer

another related rate ladder problem
start with 
$$x^2+y^2=9^2$$ then 
$$2xx'+2yy'=0$$ plug in the numbers you know to find the number you want

anonymous · Answer

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anonymous · Answer

in my picture $x'=2$ and you are asked for $y'$ when $y=6$

anonymous · Answer

a) triangle is formed by the ladder and the wall. So, let $$h=(12^2-b^2)^{1/2}$$. Then the rate at which h changes is $$\frac{dh}{dt} = \frac{1}{2} (12^-b^2)^{-1/2} (-2b)\frac{db}{dt}$$. The above comes from the derivative of the intial equation stated at the start with respect to time. Then solveing for when h=6 for b and knowing $$\frac{db}{dt} = 2$$, we can solve for $$\frac{dh}{dt}$$ when h =6. This is a good start, and i think the answer should be 1/6.

anonymous · Answer

b) Is similar to part a but instead of using pythagoreans equation use $$\cos(\theta)=b/12$$. Aswell i beleave the answer the answer is -1/3.

anonymous · Answer

so is the answer 8.5 ft/sec