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
The ladder cannot be further away from the wall than it is tall
Originally Posted by lemontea
you forgot to take the square root when solving for x
This should help... :D