A person walks away from a pulley pulling a rope slung over it. The rope is being held at a height 10 feet below the pulley. Suppose that the weight at the opposite end of the rope is rising at 4 feet per second. At what rate is the person walking when s/he is 20 feet from being directly under the pulley

The image of this question is missing and so i have attached it.

dd/dt = 4.47 ft/s

Step-by-step explanation:

From the image attached, let's denote the following;

d = horizontal distance beneath pulley

h = height of pulley

l = diagonal from the pulley to the head of the person

v = velocity of rope rising

Using pythagoras theorem;

l² = d² + h²

Differentiating with respect to time and considering h = c^(te) gives;

2l(dl/dt) = 2d(dd/dt)

We are given;

d = 20 ft

h = 10 ft

v = 4 ft/s

We know that velocity in this case is change in diagonal distance with time. Thus;

v = dl/dt = 4 ft/s

From earlier, we saw that;

2l(dl/dt) = 2d(dd/dt)

Thus, reducing it gives

(dl/dt)(l/d) = dd/dt

Now, l² = d² + h²

l = √(d² + h²)

Also, v = dl/dt = 4

Thus;

4(√(d² + h²))/d = dd/dt

4(√(20² + 10²))/20 = dd/dt

dd/dt = 4.47 ft/s

0 is its x and y coordinate