A 11 foot ladder is placed against a vertical wall. The bottom of the ladder slides away from the wall at a constant rate of 2.9 ft/s. How fast is the top of the ladder sliding down the wall when the top of the ladder is 7.6 feet from the ground?
Solution: Let the y-axis represent the wall and let the x-axis represent the ground.
Let x=x(t) be the distance of the bottom of the ladder from the wall at time t and let y=y(t) be the distance of the top of the ladder from the wall at time t.
The question is to find the rate of change of y with respect to t, what is the same as to find the derivative d (x or y) y/dt, if y= and d (x y) x /dt=2.9.
By the Pythagorean theorem x^2+y^2=( 11 )^2. This implies that if y=y(t)=7.6 then x=x(t)=______63.2_____ . Therefore if we differentiate both sides of the expression
with respect to t then, for y=7.6, dx/dt=2.9, and x= ____63.2_______ , we get that dy/dt= _____-24.1_______ .
Hence, if the bottom of the ladder slides away from the wall at a constant rate of 2.9 ft/s then the top of the ladder slides down at the rate __24.1_____ ft/sec at the moment when the top of the ladder is 7.6 feet from the ground.
The answers in red are the answers i got..but something seems to be wrong and i dont know what..can anyone help me..GREATLY appreciate it