A ladder of length 10 m rests with its lower end on a horizontal floor and with a rate of 2m/sec when the lower end is at 8m away from the wall, find: 1) the rate of sliding of the upper end. 2) the rate of change of the angle between the ladder and the floor

A ladder of length 10 m rests with its lower end on a horizontal floor and with a rate of 2m/sec when the lower end is at 8m away from the wall, find:

1) the rate of sliding of the upper end.
2) the rate of change of the angle between the ladder and the floor

x^2+y^2=10^2

y=6 when x=8

tanθ = y/x, so
y = x tanθ
tanθ=3/4 when x=8

x dx/dt + y dy/dt = 0
dy/dt = tanθ dx/dt + x sec^2θ dθ/dt

Now just plug in your numbers.

To find the rate of sliding of the upper end of the ladder, we can use related rates.

Let's denote the distance of the ladder's lower end from the wall as x. We know that dx/dt = 2 m/sec, which is the rate of change of x. We need to find dy/dt, which is the rate of sliding of the upper end of the ladder.

We can use the Pythagorean theorem to relate x, y, and the length of the ladder, L:

x^2 + y^2 = L^2

Differentiating both sides of this equation with respect to time (t), we get:

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

Plugging in the given values, we have:

2(8)(2) + 2y(dy/dt) = 0

16 + 2y(dy/dt) = 0

Dividing both sides by 2y and rearranging the equation, we get:

dy/dt = -8/y

Now we can plug in the length of the ladder, L = 10 m:

dy/dt = -8/10 = -0.8 m/sec

Therefore, the rate of sliding of the upper end of the ladder is -0.8 m/sec (since it is negative, it means that the upper end of the ladder is sliding downward).

To find the rate of change of the angle between the ladder and the floor, we can use trigonometry. Let's call this angle θ.

We have tan(θ) = y/x.

Differentiating both sides of this equation with respect to time (t), we get:

sec^2(θ)(dθ/dt) = (dy/dt)(x/y)

Plugging in the given values, we have:

(1 + (y/x)^2)(dθ/dt) = (dy/dt)(x/y)

Since we already found dy/dt = -0.8 m/sec and dx/dt = 2 m/sec, we can substitute those values in:

(1 + (y/x)^2)(dθ/dt) = (-0.8)(2/y)

Simplifying further, we have:

(1 + (y/x)^2)(dθ/dt) = -1.6/y

We know the values of x = 8 m and y = 6 m when the lower end of the ladder is 8 m away from the wall. Plugging those values in, we get:

(1 + (6/8)^2)(dθ/dt) = -1.6/6

(1 + (3/4)^2)(dθ/dt) = -4/15

(1 + 9/16)(dθ/dt) = -4/15

(25/16)(dθ/dt) = -4/15

Solving for dθ/dt, we get:

dθ/dt = (-4/15) * (16/25) = -64/375 rad/sec

Therefore, the rate of change of the angle between the ladder and the floor is approximately -0.17 rad/sec.