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**lemontea** 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

*x^*2+*y^*2=121

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