Question

A 17 ft ladder leans against a wall. The bottom of the ladder is 4 ft from the wall at time t=0 and slides away from the wall at a rate of 3ft/sec.

Find the velocity of the top of the ladder at time t=3.

Answer 1

at t = 0
x^2 + y^2 = 17^2
4^2 + y^2 = 17^2
y = sqrt(289 - 16) = sqrt (273)

2x dx/dt + 2 y dy/dt = 0
dy/dt = -(x/y) dx/dt
dy/dt = -[ 4/sqrt(273) ] *(3)
That is the speed at t = 0
to get it at t = 3, you need to do that at every infintesimal step. It may hit the ground before you hit 3 seconds :)
try
y = (289-x^2)^.5
dy/dx = {.5 [(289-x^2)^-.5 ](- 2x)}
dy/dt = {.5 [(289-x^2)^-.5 ](- 2x)} dx/dt
dy/dt = - [(289-x^2)^-.5 ] x dx/dt
dy/dt = -3(289-x^2)^-.5 ] x
now at t = 3, x = 4+9 = 13
so
dy/dt = -3(289-169)^-.5 ] 13
whew :) check my arithmetic

Answer 2

-3 *13 / 10.95 = - 3.56 ft/s