The top of a 13 foot ladder, leaning against a vertical wall, is slipping down the wall at the rate of 4 feet per second.

Question

How fast is the bottom of the ladder sliding along the ground when the bottom of the ladder is 9 feet away from the base of the wall? (Round your answer to four decimal places. Be sure to choose the correct units.)

Answer 1

Hi Natalie!

In this situation, suppose b is the length between the base of the wall and the base of the ladder, and h represents the height from the base of the wall to the top of the ladder. Let's have L represent the length of the ladder that is placed against the wall, which would be the hypotenuse.

b = 9 ft.

h = ?

L = 13 ft.

Let's use the Pythagorean Theorem to figure out the height first when the base is 9 ft.

b² + h² = L²

9^2 + h^2 = 13^2

81 + h^2 = 169

h^2 = 88

h = sqrt(88) or 2*sqrt(22) simplified

Now, let's differentiate the Pythagorean Theorem equation with respect to time, so that we have it set up to find out how fast the base of the ladder is moving away from the base of the wall along the ground:

b² (d/dt) + h² (d/dt) = L² (d/dt)

Using the chain rule to complete the differentiation, the derivative equation is:

2b (db/dt) + 2h (dh/dt) = 2L (dL/dt)

Because the ladder is not going to change in length, dL/dt = 0, and because any value multiplied by 0 equals 0, the simplified derivative equation will be:

2b (db/dt) + 2h (dh/dt) = 0

They tell us that the ladder is slipping down the wall, which is the same as the height (h), at a constant rate of 4 ft./sec (this will be negative because it's going down). Therefore:

dh/dt = -4 ft./sec.

We now want to solve for db/dt (how fast the base of the ladder is moving away from the base of the wall will be represented by how fast the length of the base is changing). If we substitute the values we know into our simplified derivative equation, we will get:

2b (db/dt) + 2h (dh/dt) = 0 -----> As a reminder b = 9 ft., h = 2*sqrt(22), dh/dt = -4 ft./sec.

2(9) (db/dt) + 2(2*sqrt(22))( -4) = 0

18 (db/dt) + -16(sqrt(22)) = 0

18 (db/dt) = 16(sqrt(22))

db/dt = (16*sqrt(22))/18

db/dt = 4.1693 ft./sec. (Rounded to 4 decimal places)

In conclusion, the rate in which the base of the ladder is moving away from the base of the wall (db/dt) is 4.1693 ft./sec.

Let me know if you have any questions and if that helps.

Thanks!