Differential calculus

Question

A ladder 20 feet long leans against a vertical building.If the bottom of the ladder slides away from the building horizontally at a rate of 3ft/sec,how fast is the ladder sliding down the building when the top of the ladder is 8ft from the ground?

Answer 1

let $x$ be the distance of the ladder from the wall and $y$ its height from the floor. then you have $$x^2 + y^2 = 20^2, \frac{dx}{dt} = 3\,ft/sec$$ you want to find $\frac{dy}{dt}$ when $y = 8\, ft.$

diffrencing you get $$0 = 2x\frac{dx}{dt} + 2y\frac{dy}{dt} \to 0 = \sqrt{20^2 - 8^2} \times 3 + 8\frac{dy}{dt} $$

you can solve the equation for $\frac{dy}{dt}.$ you should get a negative answer. reason why.