How Fast is the Angle Changing When the Ladder is 8 Feet From the Wall?

[jgens]

Homework Statement

Mr. Wilson is standing near the top of a ladder 24 feet long which is leaning against a vertical wall of his house. Dennis the little boy next door, ties a rope from his tricycle to the bottom of the ladder and starts to pull the foot of the ladder away from the house wall. The bottom end of the ladder begins to slide away from the wall at the rate of 1 foot per second. How fast is the angle between the top of the ladder and the wall changing when the foot of the ladder is 8 feet from the wall?

Homework Equations

Again, simple derivatives.

The Attempt at a Solution

Well, here's my best guess:

Listing the knowns: z = 24, x' = 1, x = 8.

One of my relating equations is x^2 + y^2 = z^2; therefore, xx' + yy' = 0. While the second relating equation is tan(theta) = x/y, and consequently sec^2(theta)(theta)' = (x'y - y'x)/y^2. Or similarly: (theta)' = (x'y - y'x)/(sec^2(theta)y^2). A few intermediate calculations give y' = -8/sqrt(512), y = sqrt(512), and theta = arctan(8/sqrt(512)). Plugging everything in produces (theta)' = 0.0405.

 

[Dick]

Now, I'm not sure I agree with that one. The angle between the wall and the ladder is theta=arcsin((8+x)/24), right? Find theta' and use x'=1. I seem to be getting a different answer from yours. But it's late, and I could be wrong.

 

[jgens]

I suppose your approach if considerably simpler. We let sin(theta) = x/z = 8/24. Therefore, cos(theta)(theta)' = x'/z; a little algebra yields (theta)' = x'/(zcos(theta)). Plugging everything in should yield, theta' = 1/sqrt(512) which is approximately 0.0442. Is this what you got Dick?

 

[jgens]

I'll try the method you presented directly. theta = arcsin(x/z); therefore, theta' = x'/(zsqrt(1 - (x/z)^2)) and plugging everything in I get the same result.

 

[Dick]

Yes, I got 0.0442. But it's late enough that I'm starting to get a different answer everytime I do it. So have faith in yourself.

 

[jgens]

That's pretty funny. Thanks.

 

[Dick]

That's pretty funny. Thanks.

Then vote for me for funniest HW helper of the year for next year. I just got 0.0442 again. So the odds are 2/3 that it's right. 3/4 if we count your post.