A ladder 20 ft long leans against a vertical wall, If top slides downward at the rate of 2 ft/sec, find how fast the lower end is moving when it is 16 ft from the wall.

x^2+y^2 = 20^2

now take derivatives and solve for dx/dt

To find the speed at which the lower end of the ladder is moving, we can use the concept of related rates. Let's assign variables to the quantities involved:

Let x be the distance between the bottom of the ladder and the wall.
Let y be the distance between the top of the ladder and the ground.

We are given information about how fast the top of the ladder is sliding downward, which is dy/dt = -2 ft/sec (negative since it is sliding downward). We need to find dx/dt, the rate at which the lower end of the ladder is moving.

We can create a right triangle with the ladder as the hypotenuse, the wall as one side, and the ground as the other side. Using the Pythagorean theorem, we have:

x^2 + y^2 = 20^2

Taking the derivative of both sides with respect to time (t), we get:

2x(dx/dt) + 2y(dy/dt) = 0

Simplifying this equation, we have:

x(dx/dt) + y(dy/dt) = 0

Substituting the given values, we have:

x(dx/dt) + 16(-2) = 0

Solving for dx/dt, we get:

dx/dt = (32 - x(dy/dt)) / x

Now we can substitute the known values into this equation:

dx/dt = (32 - 16(-2)) / 16
dx/dt = (32 + 32) / 16
dx/dt = 4 ft/sec

Therefore, the lower end of the ladder is moving at a speed of 4 ft/sec when it is 16 ft from the wall.

To solve this problem, we can use the concept of related rates. Let's define the variables:

Let t be the time in seconds.
Let x be the distance between the lower end of the ladder and the wall in feet.

We need to find dx/dt, the rate at which the lower end of the ladder is moving when it is 16 ft from the wall.

From the problem statement, we are given that the ladder is 20 ft long. This implies that when the ladder is extended, the distance x and the length of the ladder will form a right-angle triangle.

Using the Pythagorean theorem, we can write:

x^2 + (20^2) = (20 + 2t)^2

Simplifying this equation, we get:

x^2 + 400 = 400 + 80t + 4t^2

Subtracting 400 from both sides, we have:

x^2 = 80t + 4t^2

Now, we can differentiate both sides of the equation with respect to t:

2x(dx/dt) = 80 + 8t

Simplifying this equation, we get:

dx/dt = (40 + 4t)/x

Now we can substitute the given value of x when it is 16 ft from the wall:

dx/dt = (40 + 4t)/16

Given that the top of the ladder slides downward at a rate of 2 ft/sec, we know that dt/dt = -2.

Substituting this value into the equation, we have:

dx/dt = (40 + 4(-2))/16

Simplifying further, we get:

dx/dt = (40 - 8)/16

dx/dt = 32/16

dx/dt = 2 ft/sec

Therefore, the lower end of the ladder is moving at a rate of 2 ft/sec when it is 16 ft from the wall.