A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second.

Question

(a) What is the velocity of the top of the ladder when the base is given below?

15 feet away from the wall

? ft/sec

20 feet away from the wall

?ft/sec

24 feet away from the wall

?ft/sec

(b) Consider the triangle formed by the side of the house, ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 20 feet from the wall.

?ft^2/sec

(c) Find the rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 20 feet from the wall.

?rad/sec

Answer 1

I will outline the method for solving part a, please ask if you need additional help

We need a relationship between the variables. Since the ladder forms a right triangle with the building we can use Pythagoras' Theorem

b² + h² = 25²

Where b is the base of the triangle, the distance of the ladder from the wall

h is the height of the triangle, the distance of the top of the ladder to the ground

By implicitly differentiating this equation, we have the relationships of how fast the base changes given a changing height

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

b (db/dt) = - h (dh/dt)

When b = 15, solving the pythagoras theorem gives h = 20. We are given that b is increasing at a rate of 2 ft/sec, (db/dt) = 2 ft/s

These can then be plugged into the above equation to find (dh/dt). Repeat this process for different values of the base of the triangle

For part b and c, use the formulas of the area and angle and apply implicit differentiation. You have all of the variables needed and can plug them in.

(Hint for part c, think of the tangent of the angle)