How fast is the top sliding down the wall when the foot of the ladder is 5 m from the wall?

Question

A ladder 13 meters long rests on horizontal ground and leans against a vertical wall. The foot of the ladder is pulled away from the wall at the rate of 0.6 m/sec.

Answer 1

Okay, so this is a pretty classic related rates Calc 1 problem. I would recommend starting by drawing the triangle formed between the ladder, ground, and wall. The diagram should look like a triangle with a 5 m base (because the foot of the ladder is 5 m from the wall) and a 13 m hypotenuse. A quick calculation with the Pythagorean Theorem will show that the height of the triangle is 12 m (the height where the other end of the ladder is touching the wall).

Once that's all sorted we can address our rates. The rate we are given is db/dt = 0.6 m/s. I call it db/dt because it is the rate at which the base of the triangle is increasing. The rate we want to find is the rate at which the ladder is falling down the wall which is dy/dt.

With any related rates problem the next step is to take the derivative of the equation that gives you a relationship between the rate you have and the rate you want. In this case we would be using the Pythagorean Thereom (b^2 + y^2 = h^2) where b is the base, y is the height, and h is the hypotenuse.

Taking the derivative implicitly we would find: 2*b*(db/dt) + 2*y*(dy/dt) = 2*h(dh/dt). This now has four values we know (b, y, h, db/dt), one we are trying to find (dy/dt), and one other we don't know (dh/dt). The variable dh/dt = 0 though, because the hypotenuse (the length of the ladder) is not changing.

From there its just substitution and algebra. You should expect dy/dt to be negative because the ladder is sliding down the wall and therefore y should be decreasing.

I hope this helps!