Question

A ladder 16 feet long is leaning against the wall of a tall building. The base of the ladder is moving away from the wall at a rate of 1 foot per second.

A) How fast is the top of the ladder moving down the wall when the base of the ladder is 8 feet from the wall?
B)Find the rate at which the angle theta between the ladder and the wall of the building is changing when the base of the ladder is 8 feet from the wall.
C) A right triangle is formed by the ladder, the wall, and the ground. How fast is the area of this triangle changing when the base of the ladder is 8 feet from the wall?

Answer 1

a. 0.588

b. 0.0722

c. 4.576 sqft/sec

Explanation:

Let b and h denote the base and height as indicated in the diagram. By pythagoras theorem,
`h^2 + b^2 = 16^2 = 256 \dotsc\;(1)` because it is a right angle triangle.

It is given that
`(db)/(dt) = 1`

Now differentiate (1) with respect to t (time) :

`\displaystyle{2h(dh)/(dt) + 2b(db)/(dt) = 0 \implies (dh)/(dt) = -(b)/(h) (db)/(dt)}`

`\displaystyle{=-(b)/(√(256 - b^2)) (db)/(dt) = -(8)/(13.856) * 1 = -0.588}`

The minus sign indicates that the value of h is actually decreasing. The required answer is 0.588.

b. From the diagram, infer that
`16 sin(\theta) = b`. When b = 8, then
`\theta = \arcsin{0.5} = \ang{30}`.

Differentiate the above equation w.r.t t

`\displaystyle{16 cos(\theta) (d\theta)/(dt) = (db)/(dt) \implies (d\theta)/(dt) = (1)/(16 cos(\theta)) = (1)/(13.856) = \mathbf{0.0722}}`

c. The area of the triangle is given by
`A = 0.5* h * b`. Differentiating w.r.t t,

`\displatstyle{(dA)/(dt) = 0.5 b (dh)/(dt) + 0.5 h (db)/(dt)}`

Plugging in b = 8, h = 13.856,
`(dh)/(dt) = -0.588`,

`(dA)/(dt) = -2.352 + 6.928 = \mathbf{4.576 ft^2/sec}`