A 10-ft ladder rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft/sec, how fast, in ft/sec, is the top of the ladder sliding down the wall, at the instant when the bottom of the ladder is 6 ft from the wall? Answer with 2 decimal places. Type your answer in the space below. If your answer is a number less than 1, place a leading "0" before the decimal pointif you can please add work if not correct answer is fine too

Interesting question!

First, we need to come up with an equation that describes the relationship between x and y, or the bottom and top of the ladder, respectively. Think of the ladder, the wall, and the floor forming a right triangle. The ladder is the hypotenuse, and the floor and the wall are the sides of the triangle. 
We know that:
a^2 + b^2 = c^2.
c is the ladder, which we know is 10. a is x, and y is b
Therefore:
x^2 + y^2 = 10^2          Let's get y by itself....
y^2 = 100 - x^2             Take the square root of both sides
y = (100 - x^2) ^(1/2) 
We have an equation that relates y, the height of the top of the ladder, to x, the distance from the wall! Now we just take the derivative to figure out the rate of change in y for any given x:
y= (100-x^2)^(1/2)   
Take the derivative (For help on this step, check out the chain rule for derivatives)
y' = (1/2) (-2x) (100 - x^2) ^(-1/2) 
y' = -x(100 - x^2) ^(-1/2)

We want to find y' when x=6, so plug in 6 for x. 
y'= -6(100 - 36)^(-1/2)
y'= -6 ÷ (64^(1/2))
y' = -6 ÷ 8
y' = -6/8 = -3/4

Remember that this is a negative number because as x, or the distance from the wall, increases, y, the height of top of the ladder, decreases. 

When the ladder is 6 feet from the wall, and moving at a constant 1ft/second, the ladder is falling at a speed of 3/4, or 0.75 ft/second.