Question

The top of a ladder slides down a vertical wall at a rate of 0.375 m/s. At the moment
when the bottom of the ladder is 5 m from the wall, it slides away from the wall at
a rate of 0.9 m/s. How long is the ladder?

Answer 1

does this help @Euler271

Answer 2

hm. got the coordinators wrong. i'll try again

Answer 3

i dont think the OP is here anymore. but in the mean time..

confirm in your mind that \[x^2 + y^2 = L^2\]
and that the derivative of this equation with respect to time, t is\[2x \frac{ dx }{ dt } + 2y \frac{ dy }{ dt } \]

that we are also given dy/dt as -0.375 m/s, and also dx/dt at the moment when x = 5 m. hope this helps you solve for y at this instant. knowing x and y at this instant will allow you to solve for L

Answer 4

\[2x \frac{ dx }{ dt } + 2y \frac{ dy }{ dt } = 0\]
this is what the derivative of the equation is with respect to time (don't forget the = 0 part)