Calculus
posted by JJ .
related rates:
a ladder, 12 feet long, is leaning against a wall. if the foot of the ladder slides away from the wall along level ground,what is the rate of change of the top of the ladder with respect to the distance of the foot of the ladder from the wall when the foot is 6 ft from the wall?
I don't understand what this question is asking for.

Draw the 12 foot ladder up against the wall with the foot at 6 feet from the foot of the wall. Then the top is sqrt (144  36) feet up on the wall.
Now it starts slipping out at rate dx/dt at the bottom of the wall.
The question is, what is dy/dt, the speed of the top of the ladder down the wall.
I drew the ladder foot to the right of the wall so that dx/dt is positive :)
Of course dy/dt will change as the ladder goes down the wall even if dx/dt is constant, but the problem only asks for the vertical speed at the moment that the ladder is sqrt (108) high on the wall. 
The problem is poorly worded, and I can see why you are confused. I think they want the rate at which the elevation of the top of the ladder changes (dy/dt) in terms of how fast the bottom of the ladder slides away from the wall (dx/dt), when x = 6 ft. Let x be the distance of the bottom of the ladder from the wall, and y be the elevation of the top of the ladder above the floor.
Start with
x^2 + y^2 = 12^2 = 144
Both x and y are functions of t. Differentiate both sides of the abovew equation with respect to t.
2x dx/dt = 2y dy/dt
dy/dt =  (x/y) dx/dt
When x = 6, y = sqrt (14436)= 10.39, so
dy/dt = 0.577 dx/dt 
dave

From the foot of the building I look up at an angle 22 from horizontal in order to see the top tree.from the top building 150 m abovethe ground level i have look down At 50 from the hirizontal in order to see the top.
A.what is the heightof the tree?
B.how far from the building is the tree 
I can't understand
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