Answer to Question #248922 in Calculus for Sam

Question

A 10-foot ladder leans against the side of a building as in example 7.2. If the bottom of the ladder is pulled away from the wall at the rate of 3 ft/s and the ladder remains in contact with the wall, (a) find the rate at which the top of the ladder is dropping when the bottom is 6 feet from the wall.
(b) Find the rate at which the angle between the ladder and the horizontal is changing when the bottom of the ladder is 6 feet from the wall.

Answer 1

$\displaystyle \text{By pythagoras theorem }\\ 10^2 = x^2 + y^2\\ \text{Differentiating, we have }\\ 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\\ \implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}\\ \text{where $y = \sqrt{10^2-6^2}=8$}\\ \therefore \frac{dy}{dt}= -\frac{6}{8}\cdot 3 = -\frac{-9}{4}\\ \tan \theta(t) = \frac{x}{y}\\ \text{Differentiating, we have }\\ \sec^2 \theta(t) \theta'(t) = \frac{y(t)x'(t)- x(t)y'(t)}{y^2}\\ \sec^2 \theta(t) \theta'(t) = \frac{8\cdot 3-6\cdot -\frac{9}{4}}{8^2}=0.586\\ = \theta'(t) = 0.586 \cos^2(36.87) = 0.375$