Question

You are situated 300 feet from the base of Tower Glitz Plaza watching an external elevator descend down the side of the building. At a certain instant the elevator is 500 feet away from you, and its distance from you is decreasing at a rate of 16 ft/sec. How fast is the elevator descending at that instant?

Answer 1

Answer:

18.66 ft/s

Step-by-step explanation:

The distance between you and the elevator is given by:

$h=\sqrt{x^2+y^2}$

The rate of change for the distance between you and the elevator is given by:

$\frac{dh}{dt}=\frac{dh}{dy}*\frac{dy}{dt}$

$-16=\frac{dh}{dy}*\frac{dy}{dt}$

$\frac{dh}{dy}=\frac{d}{dy} (\sqrt{x^2+y^2})$

Applying the chain rule:

$u=x^2+y^2\\\frac{dh}{dy}=\frac{d\sqrt u}{du} *\frac{du}{dy}\\\frac{dh}{dy}=\frac{1}{2\sqrt u} *2y\\\frac{dh}{dy}=\frac{y}{\sqrt {(x^2+y^2)}}$

Therefore, at x=300 and y = 500, dy/dt is:

$-16=\frac{y}{\sqrt {(x^2+y^2)}}*\frac{dy}{dt}\\-16=\frac{500}{\sqrt {(300^2+500^2)}}*\frac{dy}{dt}\\\frac{dy}{dt}=-18.66\ ft/s$

The elevator is descending at 18.66 ft/s.