A 15 ft ladder leans against a wall. The bottom of the ladder is 4 ft from the wall at time t=0 and slides away from the wall at a rate of 3ft/sec.

Find the velocity of the top of the ladder at time t=2.

The velocity of ladder at time t=2 is ........ft/sec.

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Well, the ladder seems to be getting quite ambitious, sliding away from the wall at a steady 3ft/sec! Now, let's figure out the velocity of the top of the ladder at time t=2.

To do that, we'll have to use a little bit of trigonometry. You see, as the ladder slides away, it forms a right triangle with the wall. The bottom of the ladder, the distance between the wall and the bottom, is given as 4 ft and the ladder itself has a height of 15 ft.

Now, let's denote the distance between the bottom of the ladder and the top of the ladder as "x". Since we know that the ladder is sliding away from the wall at a rate of 3ft/sec, the speed of x is also 3ft/sec.

Using the Pythagorean theorem, we can say that 4^2 + x^2 = 15^2. Solving for x, we get x = sqrt(15^2 - 4^2) = sqrt(225 - 16) = sqrt(209) ft.

To find the velocity of the top of the ladder at time t=2, we need to figure out how fast x is changing. Since x is changing at a rate of 3ft/sec, the velocity of the top of the ladder is also 3 ft/sec.

So, the velocity of the top of the ladder at time t=2 is... you guessed it... 3 ft/sec! The ladder might be sliding away, but at least it's doing it with some consistency. Keep it up, ladder!

To find the velocity of the top of the ladder at time t=2, we can start by visualizing the situation.

At time t=0, the bottom of the ladder is 4 ft from the wall, and it is moving away from the wall at a rate of 3 ft/sec. This means that as time passes, the bottom of the ladder will be moving further away from the wall.

Let's denote the distance between the bottom of the ladder and the wall as d, and the distance between the top of the ladder and the wall as x. Since the ladder is leaning against the wall, these distances are related by the Pythagorean theorem: x^2 + 4^2 = d^2.

Differentiate both sides of the equation with respect to time to find the rates of change:

2x * dx/dt = 2d * dd/dt

We are interested in finding dx/dt, the rate at which x is changing with respect to time.

At time t=2, we want to find dx/dt. To do this, we need to find x using the given information.

From the Pythagorean theorem, we have:

x^2 + 4^2 = d^2

Substituting the values, we have:

x^2 + 16 = d^2

Since the ladder is leaning against the wall, the height of the ladder remains constant. We know the height of the ladder is 15 ft, so d = 15.

Plugging in this value, we get:

x^2 + 16 = 15^2
x^2 + 16 = 225

Simplifying, we have:

x^2 = 209

Taking the square root of both sides, we find:

x = √209

Now that we have x, we can find dx/dt.

Differentiating both sides of the equation x^2 + 16 = 225 with respect to t, we get:

2x * dx/dt = 0

Simplifying, we have:

2x * dx/dt = 0

Now we can plug in the value of x to find dx/dt. Using x = √209, we get:

2 * √209 * dx/dt = 0

Simplifying, we find:

2 * √209 * dx/dt = 0

Dividing both sides by 2 * √209, we get:

dx/dt = 0 / (2 * √209)

Finally, evaluating this expression, we find:

dx/dt = 0 ft/sec

Therefore, the velocity of the top of the ladder at time t=2 is 0 ft/sec.

let foot of ladder be x ft from wall

let top of ladder by y ft above grouns
x^2 + y^2 = 15^2
2x dx/dt + 2y dy/dt = 0

so when t = 2, x = 4+6 = 10
y^2 = 225-100 = 125
y = √125 , and dx/dt = 3

2(10)(3) + 2(√125 dy/dt = 0
dy/dt = -60/(2√125)
= -30/√125
= appr. -.2.68 ft/sec

(the negative sign indicates the ladder is moving downwards)