Question

A ladder 13ft long rests against a vertical wall and is sliding down the wall at a rate of 3ft/s at the instant the foot of the ladder is 5 feet from the base of the wall. At this instant, how fast is the foot of the ladder moving away from the wall?

Answer 1

thanks for your help

Answer 2

Sorry I kept crashing

Answer 3

yeah i got that, and then y=12feet at the instant it is 5 feet from the base

Answer 4

\[z^2=x^2+y^2\]
\[2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}\]
\[z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}\]

Answer 5

\[\frac{dz}{dt}=0\]
as the length of the ladder doesn't change, so:
\[0=x*\frac{dx}{dt}+y*\frac{dy}{dt}\]

Answer 6

yeah i follow, and since z^2 is 144, and its asking for dy/dt then we evaluate that equation to get \[\frac{ dy }{ dt } = -\frac{ x }{ y }(\frac{ dx }{ dt })\]

Answer 7

Yea, just use the Pythagorean theorem to find y, and you know that \[\frac{dy}{dt}=-3\frac{ft}{s}\]
since it's sliding down

Answer 8

x is given

Answer 9

\[-3(\frac{ -12 }{ 5 }) = \frac{ dx }{ dt }\]

Answer 10

correct?

Answer 11

Can you post the intermediate steps? hahah I'm too lazy to evaluate.

Answer 12

\[\frac{ dx }{ dt } = \frac{ -5 }{ 4 }\]

Answer 13

y=12, because its a 13,12,5 triangle

Answer 14

looks correct to me

Answer 15

alright, just to double check the origional equation is asking for dx/dt?

Answer 16

How fast is the foot of the wall, moving away, so how fast dx/dt is increasing.. Wait it should have been positive, why did you get negative?