Question

Upper a 18a 18?-footfoot ladder is leaning against a building. if the bottom of the ladder is sliding along the pavement directly away from the building at 22 ?feet/second, how fast is the top of the ladder moving down when the foot of the ladder is 44 feet from the? wall?

Answer 1

The foot of the ladder cannot be 44 feet from the wall, that would be larger than the length of the ladder, which means the ladder has crawled a few feet :)

So we are assuming a distance of 4 feet, similarly a rate of change in x equal to 2ft/s.

check the picture.

let $h(x)= \sqrt{ 18^{2}- x^{2} } = (18^{2}- x^{2})^{ \frac{1}{2}}$

be the function of the height of the ladder with respect to x, the distance of the bottom of the ladder to the wall.

We want $\frac{dh}{dt}$, the rate of change of h with respect to t.

h is a function of x and x is a function of t, so we keep this in mind as we derivate h with respect to t:

$\frac{dh}{dt}= \frac{dh}{dx} \frac{dx}{dt}= \frac{1}{2} (18^{2}- x^{2})^{ -\frac{1}{2}}(-2x) \frac{dx}{dt}$

we substitute $\frac{dx}{dt}=2$ and x=4:

$\frac{dh}{dt}=\frac{1}{2} (18^{2}- 4^{2})^{ -\frac{1}{2}}(-2)*(4)*2= \frac{-8}{ \sqrt{18^{2}- 4^{2}} } = \frac{-8}{17.5}= -0.46$ ft/s