a 20m ladder rests vertically against the side of a barn. a pig that has been hitched to the ladder starts to pull the base of the ladder away from the wall at a constant rate of 40 cm per second. find the rate of change of the height of the top of the ladder after 30 secs

Using Pythagorean theorm,

length^2=b^2+ h^2
where b is the base length, and h is the height.
Taking the derivative..
0=2b db/dt + 2h dh/dt
you are given db/dt, solve for dh/dt
For the second part, you will have to find b and h at 30 sec. b will be .40*30, solve for h, and again, you have it.

To find the rate of change of the height of the top of the ladder, we can use the concept of similar triangles. Let's assume that at a certain time t, the height of the top of the ladder is h and the base of the ladder is x.

We know that the ladder always forms a right triangle with the wall. So, we have two similar triangles:

1) The larger triangle formed by the ladder, the wall, and the ground with sides h, x, and 20m (the length of the ladder).

2) The smaller triangle formed by the top part of the ladder, a vertical line, and the ground with sides h - Δh, x + Δx, and 20m (the length of the ladder).

Since the triangles are similar, we can set up the following proportion:

(h - Δh) / (x + Δx) = h / x

Cross multiplying, we get:

h - Δh = (h / x) * (x + Δx)

Expanding and rearranging the equation, we get:

h - Δh = h + (Δh * h) / x

Simplifying the equation, we have:

Δh = (Δh * h) / x

Now, let's substitute the given values into the equation. The rate at which the base of the ladder is being pulled away from the wall is 40 cm/s, which is equal to 0.4 m/s. The time is given as 30 seconds.

Δx = (rate of change of the base) * (time) = 0.4 m/s * 30 s = 12 m

Plugging in the values into the equation, we have:

Δh = (12 * h) / x

Now, we need to find the value of h/x. Since we have a right triangle with sides h and x, we can use the Pythagorean theorem to relate them:

h^2 + x^2 = 20^2

Simplifying the equation, we get:

h/x = sqrt(20^2 - x^2) / x

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

(d/dt)(h/x) = (d/dt)( sqrt(20^2 - x^2) / x)

Using the chain rule, we have:

(d/dt)(h/x) = (d/dx)( sqrt(20^2 - x^2) / x) * (dx/dt)

Since dx/dt is given as 0.4 m/s, we need to find (d/dx)(sqrt(20^2 - x^2) / x).

Using the quotient rule, we find:

(d/dx)(sqrt(20^2 - x^2) / x) = ( (d/dx)(sqrt(20^2 - x^2)) * x - sqrt(20^2 - x^2) * 1) / x^2

Simplifying the equation, we get:

(d/dx)(sqrt(20^2 - x^2) / x) = -sqrt(20^2 - x^2) / x^2

Plugging in the values into the equation, we have:

(d/dt)(h/x) = (-sqrt(20^2 - x^2) / x^2) * 0.4

Finally, we substitute x = 0 and h = 20 into the equation:

(d/dt)(h/x) = (-sqrt(20^2 - 0^2) / 0^2) * 0.4

(d/dt)(h/x) = (-sqrt(20^2) / 0) * 0.4

Simplifying further, we have:

(d/dt)(h/x) = (-20 / 0) * 0.4

As the denominator is 0, the rate of change of the height of the top of the ladder is undefined.