A 13 ft leader is leaning against a house when its base starts to slide away. When the base is 12 ft from the house, then base is moving at the rate of 5 ft/sec.
(a) How fast is the top of the ladder sliding down the wall then?
(b) At what rate is the area of the triangle formed by the ladder, wall and ground changing then ?
(c) At what rate is the angle `theta` between the ladder and the ground changing then ?

Explanation

`x^(2)+y^(2)=169`
`rArr " " 2x(dx)/(dt)+2y(dy)/(dt)=0`
`rArr "" ((dy)/(dt))=((dx)/(dt))(x-x)/(y)=5 xx(-12)/(5)=-12 ft//s`
`A=(1)/(2) xy`
`rArr " " (dA)/(dt)=(y)/(2)(dx)/(dt)+(x)/(2)(dy)/(dt)=(5)/(2) xx 5+ (12)/(2) xx-12`
`=(25-144)/(2)=-59.5 ft^(2)//s`
` y=x tan theta`
`rArr " " (dy)/(dt)=(tan theta)(dx)/(dt)+x sec^(2) theta (d theta)/(dt)`
`rArr " " -12 = 5 xx (5)/(12) + 12 xx (169)/(12^(2)) (d theta)/(dt)`
`rArr " " (169)/(12) (d theta)/(dt) =-12 -(25)/(12)`
` rArr " " (d theta)/(dt)=(144)/(169)-(25)/(169)=-1 "rad"//s`