The top of a 5 m wheeled ladder rests against a vertical wall. If the bottom of the ladder rolls away from the base of the wall at a rate of 1/3 m/s, how fast is the top of the ladder sliding down the wall when it is 3 m above the base of the wall?

Start with the pythagorean thm:

x^2 + y^2 = 5^2

now diff wrt t

what do you mean by this?

diff part

What formula should I use?

I used
b^2+h^2=5^2
diff both sides I get
(25-h^2)^0.5+h^2=5^2
-h(25-h^2)^-0.5 dh/dt +2h dh/dt = 0
dh/dt=0..

doesnt work

differentiate "with respect to" time:

2xdx/dt + 2ydy/dt = 0

you are given dx/dt, and y. you can calculate x from x^2 + y^2 = 25

then calculate dy/dt-- the speed of the ladder down the wall

Just in case a picture helps...



... where b = horizontal distance of base of ladder from the wall, and h the vertical height of the top of the ladder above the ground, and



... is the chain rule. You need to apply this rule again (with or without a picture) when differentiating the inner (dashed balloon) function with respect to t (time).

Plug in the given values when you've got the bottom row

____________________________________
Don't integrate - balloontegrate!
http://www.ballooncalculus.org/forum/top.php

ok i still cant get the answer I have:

1/[-6h(25-h^2)^0.5] = dh/dt

and even when i plug in h=3, i dont get the answer in the back of the book which is 4/9 m/s...........
Can you please show me what i did wrong.

differentiate "with respect to" time:

2xdx/dt + 2ydy/dt = 0

you are given dx/dt, and y. you can calculate x from x^2 + y^2 = 25

then calculate dy/dt-- the speed of the ladder down the wall


4(1/3) + 3dy/dt = 0

dy/dt = -4/9

A diagram might help visualize what is going here. I can't do diagrams so hopefully I can explain this:

You are being asked to differentiate an equation with respect to another variable, in this case "t". This is no different than differentiating with respect to "x" or "y" or any other variable. We differentiate with respect to "t" because we are interested in knowing how the side of the ladder changes with respect to time.

So what do we know:

We know we have a 5M ladder resting against a wall. This forms a classic right triangle.

We know that when the "top" of the ladder (the part that is resting against the wall) is 3M above the base of the wall (or from the floor), the bottom of the ladder (that forms what we would normally refer to as the base of our triangle) is changing at a rate of .

The first thing we want to do is set up a general expressing for our triangle. As calculus said you know that:



Where X is equal to the base of our triangle (formed by the bottom of the ladder to where the wall meets the ground)
Where Y is equal to the height of our triangle (formed by the top of the ladder where it meets the wall)
Where H is the hypotenuse of our triangle (formed by the length of the ladder).

We are looking for how fast the top of the ladder is sliding down the wall, and this is why we differentiate our equation with respect to t=time: t is our variable here that x and y depend on.

I'm not sure how you got what you got but we are differentiating:



You should be able to take it from there substituting the correct rate of change given in the problem, and solving for the remainder.