A 16 ft. ladder leans against a wall, with the top being 12 ft. from the wall when time t=0. The top is sliding down at a constant rate of 4 ft./second. Given that the top of the ladder touches the ground at 3 seconds, what is the velocity of the bottom of the ladder when t=3?

I suspect a typo somewhere. When the ladder is lying flat at t=3, the base is no longer moving.

To find the velocity of the bottom of the ladder when t=3, we need to find the rate at which the bottom is moving when the top touches the ground.

Let's first find the height of the ladder at time t.

The height of the ladder can be calculated using the Pythagorean theorem.

At time t=0, we have:

a^2 + b^2 = c^2

where a = 12 ft (distance from the wall), b is the height of the ladder at time t=0, and c = 16 ft (length of the ladder).

Solving for b, we get:

b^2 = c^2 - a^2

b^2 = 16^2 - 12^2

b^2 = 256 - 144

b^2 = 112

b = sqrt(112)

b ≈ 10.59 ft

So, at time t=0, the height of the ladder is approximately 10.59 ft.

Now, let's find the height of the ladder at time t=3.

Since the top of the ladder is sliding down at a constant rate of 4 ft/second, we can find the height at time t using the equation:

b = 10.59 - 4t

At t=3:

b = 10.59 - 4(3)

b = 10.59 - 12

b ≈ -1.41 ft

So, at t=3, the height of the ladder is approximately -1.41 ft. The negative sign indicates that the bottom of the ladder is below the ground level.

Finally, to find the velocity of the bottom of the ladder when t=3, we need to find the rate at which the bottom is moving. The velocity of the bottom can be found by taking the derivative of the height function with respect to time:

v = d/dt (10.59 - 4t)

v = -4 ft/second

Therefore, the velocity of the bottom of the ladder when t=3 is -4 ft/second.

To find the velocity of the bottom of the ladder at t=3, we can use the concept of related rates.

Let's define some variables:
y = height of the ladder on the wall
x = distance between the bottom of the ladder and the wall
t = time

Given that the ladder is 16 ft long and the top is sliding down at a constant rate of 4 ft/second, we have the following relationship:
y = 16 - 4t

We are asked to find the velocity of the bottom of the ladder (dx/dt) at t=3.

To do that, we can relate the variables x and y using the Pythagorean theorem:
x^2 + y^2 = 16^2

Differentiating both sides of this equation with respect to t (using the chain rule), we get:
2x(dx/dt) + 2y(dy/dt) = 0

Since we're interested in finding dx/dt when t=3, we need to substitute the values of x, y, and dy/dt at that time.
At t=3, the top of the ladder touches the ground, meaning y=0.
Substituting these values into our equation, we have:
2x(dx/dt) + 2(0)(dy/dt) = 0
2x(dx/dt) = 0
dx/dt = 0

Therefore, the velocity of the bottom of the ladder at t=3 is 0 ft/second.

Make a sketch:

x^2 + y^2 = 256
when t=0 , y=12, x = √112
when t = 1, y = 8 , x = √192
when t = 2 , y = 4 , x = √240
when t = 3, y = 0, x = √256 = 16

given dy/dt = -4 ft/s

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

dx/dt = -y (dy/dt) / x
= 0/16 = 0

check:
let t=1
y = 8, y = √192
dx/dt = -8(-4)/√192 = appr 2.3

let t = 2
y = 4, x = √240
dx/dt = -4(-4)/√240 = appr 1.03

slowing up ....

let t = 2.99 sec
y = 12 - 2.99(4) = .04 , x = √255.9984
dx/dt = -.04(-4)/√255.9984 = appr .01 (close to zero)