Question

Hi, I'd like to ask for help with the Related Rates 2 - "Sliding Ladder" question:
Why can't I use x^2 + y^2 = 20^2 ?
(Surely this relationship stays true even as the ladder moves, since it's a right-angle triangle?)
I used this equation, differentiated with respect to x, then used the chain rule to find dy/dt:
dy/dt = dy/dx * dx/dt
dy/dt = (-x/y) * 5
So at the key moment, dy/dt = (-12/16) * 5 = -15/4
The answer according to the video is -16/5.
So where am I going wrong?
Thank you! :)

The question is here: http://www.youtube.com/watch?feature=player_embedded&v=d48

Answer 1

Do you have the original problem for us? The question did not load from the link provided.

Answer 2

Hi PROSS, thanks for your reply. Here is another attempt at the link: http://www.youtube.com/watch?v=d484GRz9zjY

In case that still does not work, here is the question:
A 20 ft ladder leans over a 12 ft wall so that 5 ft project over the wall. The bottom of the ladder is pulled away from the wall at 5 ft/sec. How quickly is the top of the ladder approaching the ground?

Thanks! Kate

Answer 3

Hi Kate,
The related rates you found work once the top of the ladder reaches the top of the wall and begins falling. But the problem is looking for the rate that the top of the ladder approaches the ground when it's extended 5 ft. past the wall.
This requires a new related rate, in this case using properties of right triangles, i.e. two right triagles with the same angle measures have proportional sides.

Consider that the lenght of ladder from ground to wall is S and the dotted line is height Y. S forms a right triangle with the 12ft wall and its ratio is S/12. This is the same proportion as the lenght of the ladder (20ft) to Y, meaning S/12 = 20/Y You can then figure out how they change in relation to each other (dY/dS) by differentiating.
Then differentiate the pythagorean theorem S^2 = X^2 + 12^2 to find dS/dt.
Using chain rule, dY/dt = dY/dS * dS/dt)

Use the fact that ladder is 5ft past wall to find S, X (dist from ladder bottom to wall) and dS/dt

Answer 4

Hi Kate,

The following is the difference of the two cases you described.