ladder sliding down a wall

[suzanne1]

I am in calculus I and have a problem that is giving me trouble.
I need help setting it up and with some of the equations.

A 20 ft ladder is leaning against a wall and the bottom is being pulled along the floor at a contant rate of 2.5 ft/sec. How do I complete a table that shows t as time - and x for the horizontal change and y as the ladder moving down the wall.
I also have to solve for dy/dt.

 

[galactus]

Use Pythagoras to set up your function.

\(\displaystyle \L\\x^{2}+y^{2}=400\)

Differentiate with rspect to time:

\(\displaystyle \L\\2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\)....[1]

You can use this to find your various distances.


Example:
A 20 ft ladder is leaning against a wall and slips away from the wall at 2.5 ft/sec. at the instant the base is 12 ft from the base of the wall.
How fast is the top of the ladder moving down the wall at that instant?.

Solving [1] for dy/dt:

\(\displaystyle \L\\\frac{dy}{dt}=\frac{-x}{y}\frac{dx}{dt}\)

When x=12, it follows from Pythagoras that y=16.

We know that dx/dt=5/2=2.5

Therefore, \(\displaystyle \frac{dy}{dt}=\frac{-12}{16}(5/2)=\frac{-15}{8}ft/sec\)

See?. Hope this helps a little.

Use various times and distances in your chart. You could let the bottom of your ladder be 4.5 feet from the wall at time t=0. Then it starts to slip. Every second it moves 2.5 feet from the bottom of the wall. Use Pythagoras to find the various side lengths of the triangle. At t=0 the top of the ladder would be \(\displaystyle \sqrt{400-(\frac{9}{2})^{2}}=y=\frac{7\sqrt{31}}{2}\) feet from the bottom of the wall.

At t=1 second, the bottom would be 4.5+2.5=7 feet from the wall. Calculate accordingly.

 

[suzanne1]

I see where I can use Pythagoras to set up the problem. These type problems give me fits. I really am still pretty lost. Sorry

 

[pka]

These type problems give me fits. I really am still pretty lost.

These sorts of problems are ideal for implicit differentiation.
You should review that topic in your textbook.

 

[suzanne1]

I don't see how this solution addresses the idea that the ladder is sliding at a constant rate of 2.5 sec/ft along the floor.

pka: If I understood it, I wouldn't be on here. Simply going the chapter is not working.

 

[stapel]

If I understood it, I wouldn't be on here. Simply going the chapter is not working.

I'm afraid that it is not feasible to teach you the chapter on implicit differentiation within this context.

If reviewing the chapter material didn't refresh your memory -- that is, if you really are utterly "at sea" on this, unable even to get started, or follow hints or complete worked examples -- then I would suggest you consider hiring a tutor, local to your area, with whom you can meet face-to-face for a few hours a week (daily would be good). In this way, you can get caught up on the missing material.

My best wishes to you.

Eliz.