Calc 1 - related rates problem

Question

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second. Find the rate (rad/sec) at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 15 feet from the wall.

Answer 1

(625 - x²)^1/2 = xcotθ; - x/(625 - x²)^1/2dx/dt = cotθdx/dt - xcsc²θdθ/dt.

Now dx/dt = 2ft/s, x = 15 ft; - 15/20·2 = 20/15·2 - 15(25/20)²dθ/dt; dθ/dt = (3/2 + 8/3)/(15·25/16) =

25/6·16/375 = 8/45 rad/s

Answer 2

Answer 3

l^2 = h^2 + b^2

take the derivative

2ll'=2hh' +2bb'

divide by 2

ll' = hh' + bb', l'=0, b' =2, b= 15, h=sqr(25^2- 15^2) = sqr(625-225)=sqr400 = 20

0 = 20h' +15(2)

h' = -30/20 = -1.5 feet per second= rate of change of the top of the ladder from the ground

angle = T= arctan(h/b) = arctan(20/15) = arctan(4/3)= about .3pi radians

T' = derivative of arctan(h/b)

derivative of (1/a)arctan(u/a) = 1/(a^2+u^2), u=h, a=b

T'= derivative of the angle with respect to time measured in seconds = (1/10) radians per second

= about pi/30 radians per second

make it negative, as the angle gets smaller as the ladder moves further away and the top tip lower