a ladder is 20 feet long and its top is slidding at a constant rate of 2 feet per second. How fast is the bottom of the ladder moving along the ground when the bottom is 16 feet from the wall?

Let the ladder be x ft from the wall, and y ft up along the wall

x^2 + y^2 = 20^2
2x dx/dt + 2y dy/dt = 0
x dx/dt + y dy/dt = 0
when x=16 , y^2 = 400-256 ---> y = 12
and dy/dt = -2 ft/sec

16(dx/dt) + 12(-2) = 0
dx/dt = -24/16 = -3/2

the ladder is moving away from the wall at 1.5 ft/s

last two lines should have been

16(dx/dt) + 12(-2) = 0
dx/dt = 24/16 = 3/2

To find the rate at which the bottom of the ladder is moving along the ground, we need to use related rates. Let's call the distance from the bottom of the ladder to the wall x, and the distance from the top of the ladder to the ground y.

Since the ladder is leaning against the wall, we can form a right triangle with the ladder, the wall, and the ground. The ladder acts as the hypotenuse, with x representing the bottom segment parallel to the ground and y representing the vertical segment.

Using the Pythagorean theorem, we have the equation: x^2 + y^2 = 20^2.

Differentiating both sides of this equation with respect to time (t) using implicit differentiation, we get:
2x(dx/dt) + 2y(dy/dt) = 0.

Since we want to find the rate at which x is changing (dx/dt) when x = 16, we need to find the value of dy/dt at that point.

Now, we know that the top of the ladder is sliding downward at a constant rate of 2 feet per second, so dy/dt = -2.

Substituting this value into our equation:
2(16)(dx/dt) + 2(y)(-2) = 0.

Simplifying the equation:
32(dx/dt) - 4y = 0.

But we still need the value of y. We can find it by using the Pythagorean theorem again.
When x = 16, we have: (16)^2 + y^2 = 20^2.
Simplifying: 256 + y^2 = 400.
y^2 = 400 - 256.
y^2 = 144.
y = 12.

Now, substituting the values of x = 16 and y = 12 into the equation:
32(dx/dt) - 4(12) = 0.
32(dx/dt) = 48.
(dx/dt) = 48/32.
(dx/dt) = 1.5 feet per second.

Therefore, when the bottom of the ladder is 16 feet from the wall, it is moving along the ground at a rate of 1.5 feet per second.