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

Answer:

a. 0.588

b. 0.0722

c. 4.576 sqft/sec

Step-by-step 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 $\frac{db}{dt} = 1$

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

$\displaystyle{2h\frac{dh}{dt} + 2b\frac{db}{dt} = 0 \implies \frac{dh}{dt} = -\frac{b}{h} \frac{db}{dt}}$

$\displaystyle{=-\frac{b}{\sqrt{256 - b^2}} \frac{db}{dt} = -\frac{8}{13.856} \times 1 = -0.588}$

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

$\displaystyle{16 \cos{\theta} \frac{d\theta}{dt} = \frac{db}{dt} \implies \frac{d\theta}{dt} = \frac{1}{16 \cos{\theta}} = \frac{1}{13.856} = \mathbf{0.0722}}$

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

$\displatstyle{\frac{dA}{dt} = 0.5 b \frac{dh}{dt} + 0.5 h \frac{db}{dt}}$

Plugging in b = 8, h = 13.856, $\frac{dh}{dt} = -0.588$,

$\frac{dA}{dt} = -2.352 + 6.928 = \mathbf{4.576 ft^2/sec}$