A ladder 20 ft long is leaning against a wall. If the foot of the ladder is being pulled away from the bottom of the wall at a rate of 4 ft per second. At what rate is the top of the ladder moving down the wall the bottom of the ladder is 12 ft from the wall?
Im not sure hot to even start this

2. This is a differential application exposed by appeal to the Pythagorean Theorem. Let's see what you get.

3. Well im sure there was a typo on the page, but im not certain if the ladder Starts 12ft from the wall, or as it was pulled away, it finished 12 ft from the wall. But lets say it started 12ft away. For some reason, I dont think theres enough info to solve this problem. I can easily draw a diagram but I dont know how to create an equation to show how its falling.

4. I think i understand. Use the theorem to find the height of the wall. But then im stuck again, I dont have how long it's been pulled for or the distance it was pulled for. Should i assume the 12ft was how far it moved? would that be better

5. Originally Posted by tonytouch311
I think i understand. Use the theorem to find the height of the wall. But then im stuck again, I dont have how long it's been pulled for or the distance it was pulled for. Should i assume the 12ft was how far it moved? would that be better
Let x be the distance from the base of the wall to where the ladder touches the ground.
Let y be the distance from the base of the wall to where the ladder touches the wall.
x^2 + y^2 = 20^2.
Take the derivative, implicitly, with respect to t.
dx/dt is given. They are asking for dy/dt when x = 12 (and when x = 12, you can solve for y).

6. G = Distance from wall.
H = Distance up the wall.

$\displaystyle G^{2} + H^{2} = 20^{2}$

G and H are functions of time. But first, a little thought should suggest that G increasing implies H decreasing. If we get the same sign, we did something wrong.

$\displaystyle G(t)^{2} + H(t)^{2} = 20^{2}$

Implicit derivatives of G and H with respect to t.

2*G*dt + 2*H*dh = 0

Solve for dh. What does that tell us?

Substitution and arithmetic after that.